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8.6: Solutions for Chapter6


Exercise 6.3

Let A = {a, b}, and consider the preorders shown here: (egin{array}{l} a ullet end{array}) (egin{array}{l} b ullet end{array}), (egin{array}{l} a ullet end{array}) → (egin{array}{l} b ullet end{array}), (egin{array}{l} a ullet end{array}) (leftrightarrows) (egin{array}{l} b ullet end{array}).

  1. The left-most (the discrete preorder on A) has no initial object, because a (nleq) b and b (nleq) a.
  2. The middle one has one initial object, namely a.
  3. The right-most (the co-discrete preorder on A) has two initial objects.

Exercise 6.6

Recall that the objects of a free category on a graph are the vertices of the graph, and the morphisms are paths. Thus the free category on a graph G has an initial object if there exists a vertex v that has a unique path to every object. In 1. and 2., the vertex a has this property, so the free categories on graphs 1. and 2. have initial objects. In graph 3., neither a nor b have a path to each other, and so there is no initial object. In graph 4., the vertex a has many paths to itself, and hence its free category does not have an initial object either.

Exercise 6.7

1. The remaining conditions are that f (1(_{R})) = 1(_{S}), and that f (r1 ∗(_{R}) r2) = f (r1) ∗(_{S}) f (r2).

2. The initial object in the category Rig is the natural numbers rig ((mathbb{N}), 0, +, 1, ∗).

The fact that is initial means that for any other rig R = (R, 0(_{R}), +(_{R}), 1(_{R}), ∗(_{R})), there is a unique rig homomorphism f : (mathbb{N}) → R.

What is this homomorphism? Well, to be a rig homomorphism, f must send 0 to 0(_{R}), 1 to 1(_{R}). Furthermore, we must also have f (m + n) = f (m) +(_{R}) f (n), and hence

So if there is a rig homomorphism f : (mathbb{N}) → R, it must be given by the above formula. But does this formula work correctly for multiplication?
It remains to check f (m n) = f (m) ∗(_{R}) f (n), and this will follow from distributivity. Noting that f (m n) is equal to the sum of m n copies of 1(_{R}), we have

Thus ((mathbb{N}), 0, +, 1, ∗) is the initial object in Rig.

Exercise 6.8

In Definition 6.1, it is the initial object Ø (in) C that is universal. In this case, all objects c (in) C are ‘comparable objects’. So the universal property of the initial object is that to any object c (in) C, there is a unique map Ø → c coming from the initial object.

Exercise 6.10

If c(_{1}) is initial then by the universal property, for any c there is a unique morphism c(_{1}) → c; in particular, there is a unique morphism c(_{1}) → c(_{2}), call it f . Similarly, if c(_{2}) is initial then there is a unique morphism c(_{2}) → c(_{1}), call it g. But how do we know that f and g are mutually inverse? Well since c(_{1}) is initial there is a unique morphism c(_{1}) → c(_{1}). But we can think of two: idc1 and f ; g. Thus they must be equal. Similarly for c(_{2}), so we have f ; g = id(_{c_{1}}) and g ; f = id(_{c_{2}}), which is the definition of f and g being mutually inverse.

Exercise 6.13

Let (P, ≤) be a preorder, and p, q (in) P. Recall that a preorder is a category with at most one morphism, denoted ≤, between any two objects. Also recall that all diagrams in a preorder commute, since this means any two morphisms with the same domain and codomain are equal.
Translating Definition 6.11 to this case, a coproduct p + q is P is an element of P such that p p + q and q p + q, and such that for all elements x (in) P with maps p x and q x, we have p + q x. But this says exactly that p + q is a join: it is a least element above both p and q. Thus coproducts in preorders are exactly the same as joins.

Exercise 6.16

The function [f , g] is defined by

[f , g] : A B (longrightarrow) T

apple1 (mapsto) a

banana1 (mapsto) b

pear1 (mapsto) p

cherry1 (mapsto) c

orange1 (mapsto) o

apple2 (mapsto) e

tomato2 (mapsto) o

mango2 (mapsto) o.

Exercise 6.17

1. The equation ι(_{A}) ; [f , g] = f is the commutativity of the left hand triangle in the commutative diagram (6.12) defining [f , g].

2. The equation ι(_{B}) ; [f , g] = g is the commutativity of the right hand triangle in the commutative diagram (6.12) defining [f , g].

3. The equation [f , g] ; h = [f ; h, g ; h] follows from the universal property of the coproduct. Indeed, the diagram

commutes, and the universal property says there is a unique map [f ; h, g ; h]: A + B D for which this occurs.

Hence we must have [f , g] ; h = [f ; h, g ; h]. 4. Similarly, to show [ι(_{A}), ι(_{B})] = id(_{A + B}), observe that the diagram

trivially commutes. Hence by the uniqueness in (6.12), [ι(_{A}) , ι(_{B})] = id(_{A + B}).

Exercise 6.18

This exercise is about showing that coproducts and an initial object give a symmetric monoidal category. Since all we have are coproducts and an initial object, and since these are defined by their universal properties, the solution is to use these universal properties over and over, to prove that all the data of Definition 4.45 can be constructed.

1. To define a functor + : C × C → C we must define its action on objects and morphisms. In both cases, we just take the coproduct. If (A, B) is an object of C × C, its image A + B is, as usual, the coproduct of the two objects of C.

If (f, g): (A, B) → (C, D) is a morphism, then we can form a morphism f + g = [f ; ι(_{C}) , g ; ι(_{D})] : A + B C + D, where ι(_{C}) : C C + D and ι(_{D}) : D C + D are the canonical morphisms given by the definition of the coproduct A + B.

Note that this construction sends identity morphisms to identity morphisms, since by Exercise 6.17 4 we have

id(_{A}) + id(_{B}) = [id(_{A}) ; ι(_{A}), id(_{B}) ; ι(_{B})] = [ι(_{A}), ι(_{B})] = id(_{A + B}).

To show that + is a functor, we need to also show it preserves composition. Suppose we also have amorphism(h, k): (C, D) → (E, F) in C × C. We need to show that (f + g) ; (h + k) = (f ; h) + (g ; k). This is a slightly more complicated version of the argument in Exercise 6.17 3. It follows from the fact the diagram below commutes:

Indeed, we again use the uniqueness of the copairing in (6.12), this time to show that (f ; h) + (g ; k) = [f ; h ; ι(_{E}), k ; ι(_{F})] = (f + g) ; (h + k), as required.

2. Recall the universal property of the initial object gives a unique map !(_{A}) : Ø → A. Then the copairing [id(_{A}), !(_{A})] is a map A + Ø → A. Moreover, it is an isomorphism with inverse ι(_{A}) : A A + Ø.

Indeed, using the properties in Exercise 6.17 and the universal property of the initial object, we have ι(_{A}) ; [id(_{A}), !(_{A})] = id(_{A}), and

[id(_{A}), !(_{A})] ; ι(_{A}) = [id(_{A}) ; ι(_{A}), !(_{A}) ; ι(_{A})] = [ι(_{A}), !(_{A + Ø})] = [ι(_{A}), ι(_{Ø})] = id(_{A + Ø}).

An analogous argument shows [!(_{A}) , id(_{A})] : Ø + A A is an isomorphism.

3. We’ll just write down the maps and their inverses; we leave it to you, if you like, to check that they indeed are inverses

a) The map [id(_{A + ι_{B},ι_{C}})] = [[ι(_{A}), ι(_{B}) ; ι(_{B + C})], ι(_{C}) ; ι(_{B + C})]: (A + B) + C A + (B + C) is an isomorphism, with inverse [ι(_{A}), ι(_{B}) + id(_{C})]: A + (B + C) → (A + B) + C.

b) The map [ι(_{A}), ι(_{B})]: A + B B + A is an isomorphism.

Note our notation here is slightly confusing: there are two maps named ι(_{A}),(i) ι(_{A}): A A + B, and (ii) ι(_{A}): A B + A, and similarly for ι(_{B}). In the above we mean the map (ii). It has inverse [ι(_{A}), ι(_{B})]: B + A A + B, where in this case we mean the map (i).

Exercise 6.24

1. Suppose given an arbitrary diagram of the form B A C in Disc(_{S}); we need to show that it has a pushout. The only morphisms in Disc(_{S}) are identities, so in particular A = B = C, and the square consisting of all identities is its pushout.

2. Suppose Disc(_{S}) has an initial object s. Then S cannot be empty! But it also cannot have more than one object, because if s′ is another object then there is a morphism s s′, but the only morphisms in S are identities so s = s′. Hence the set S must consist of exactly one element.

Exercise 6.26

The pushout is the set (underline{4}), as depicted in the top right in the diagram below, equipped also with the depicted functions:

We want to see that this checks out with the description from Example 6.25, i.e. that it is the set of equivalence classes in (underline{5}) ⊔ 3 generated by the relation {f (a) ∼ g(a) | a (in) (underline{4})}. If we denote elements of 5 as {1, ..., 5} and those of (underline{3}) as {1′, 2′, 3′}, we can redraw the functions f, g:

which says we take the equivalence relation on (underline{5}) ⊔ (underline{3}) generated by: 1 ∼ 1′, , 3 ∼ 1′, 5 ∼ 2′, and 5 ∼ 3′. The equivalence classes are {1, 1′, 3}, {2}, {4}, and {5, 2′, 3′}. These four are exactly the four elements in the set labeled ‘pushout’ in Eq. (A.1).

Exercise 6.28

1. The diagram to the left commutes because Ø is initial, and so has a unique map Ø → X + Y. This implies we must have f ; ι(_{X}) = g ; ι(_{Y}).

2. There is a unique map X + Y T making the diagram in (6.21) commutes imply by the universal property of the coproduct (6.12) applied to the maps x : X T and y : Y T.

3. Suppose X + (_{Ø})Y exists. By the universal property of Ø, given any pair of arrows x : X T and y : Y T, the diagram

commutes. This means, by the universal property of the pushout X + (_{Ø})Y, there exists a unique map t : X + (_{Ø})Y T such that ι(_{X}) ; t = x and ι(_{Y}) ; t = y.

Thus X + (_{Ø})Y is the coproduct X + Y.

Exercise 6.35

We have to check that the colimit of the diagram shown left really is given by taking three pushouts as shown right:

That is, we need to show that S, together with the maps from A, B, X, Y, and Z, has the required universal property. So suppose given an object T with two commuting diagrams as shown:

We need to show there is a unique map S T making everything commute. Since Q is a pushout of X A Y, there is a unique map Q T making a commutative triangle with Y, and since R is the pushout of Y B Z, there is a unique map R T making a commutative triangle with Y. This implies that there is a commuting (Y, Q, R, T) square, and hence a unique map S T from its pushout making everything commute. This is what we wanted to show.

Exercise 6.41

The formula in Theorem 6.37 says that the pushout X +(_{N}) Y is given by the set of equivalence classes of X N Y under the equivalence relation generated by x n if x = f (n), and y n if y(n), where x (in) X, y (in) Y, n (in) N. Since for every n (in) N there exists an x (in) X such that x = f (n), this set is equal to the set of equivalence classes of X Y under the equivalence relation generated by x y if there exists n such that x = f (n) and y = g(n). This is exactly the description of Example 6.25.

Exercise 6.48

The monoidal product is

Exercise 6.49

Let x and y be composable cospans in Cospan(_{FinSet}). In terms of wires and connected components, the composition rule in Cospan(_{FinSet}) says that (i) the composite cospan has a unique element in the apex for every connected component of the concatenation of the wire diagrams x and y, and (ii) in the wire diagram for x ; y, each element of the feet is connected by a wire to the element representing the connected component to which it belongs.

Exercise 6.57

Morphisms 1, 4, and 6 are equal, and morphisms 3 and 5 are equal. Morphism 3 is not equal to any other depicted morphism. This is an immediate consequence of Theorem 6.55.

Exercise 6.59

1. The input to h should be labelled B.

2. The output of g should be labelled D, since we know from the labels in the top right that h is a morphism B D D.

3. The fourth output wire of the composite should be labelled D too!

Exercise 6.62

We draw the function depictions above, and the wiring depictions below. Note that we depict the empty set with blank space.

Exercise 6.63

The special law says that the composite of cospans

is the identity. This comes down to checking that the square

is a pushout square. It is trivial to see that the square commutes. Suppose now that we have maps f : X Y and g : X Y such that

Write ι(_{1}) : X X + X for the map into the first copy of X in X + X, given by the definition of coproduct. Then, using the fact that ι(_{1}) ; [id, id] = id from Exercise 6.17 1, and the commutativity of the above square, we have f = ι(_{1}) ; [id, id] ; f = ι(_{1}) ; [id, id] ; g = g. This means that f : X T is the unique map such that

commutes, and so (A.2) is a pushout square.

Exercise 6.67

The missing diagram is

Exercise 6.70

Let A (subseteq) S and B (subseteq) T. Then

(egin{aligned}
varphi_{S^{prime}, T^{prime}}left(left(mathrm{im}_{f} imes mathrm{im}_{g} ight)(A imes B) ight) &=varphi_{S^{prime}, T^{prime}}({f(a) mid a in A} imes{g(b) mid b in B})
&={(f(a), g(b)) mid a in A, b in B}
&=operatorname{im}_{f imes g}(A imes B)
&=operatorname{im}_{f imes g}left(varphi_{S, T}(A, B) ight)
end{aligned})

Thus the required square commutes.

Exercise 6.78

They mean that every category Cospan(_{C}) is equal to a category Cospan(_{F}), for some well-chosen F. They also tell you how to choose this F: take the functor F : C → Set that sends every object of C to the set {∗}, and every morphism of C to the identity function on {∗}. Of course, you will have to check this functor is a lax symmetric monoidal functor, but in fact this is not hard to do.
To check that Cospan(_{C}) is equal to Cospan(_{F}) , first observe that they have the same objects: the objects of C. Next, observe that a morphism in Cospan(_{F}) is a cospan X N Y in C together with an element of FN = {∗}. But FN also has a unique element, ∗! So there’s no choice here, and we can consider morphisms of Cospan(_{F}) just to be cospans in C.

Moreover, composition of morphisms in Cospan(_{F}) is simply the usual composition of cospans via pushout, so Cospan(_{F}) = Cospan(_{C}).
(More technically, we might say that Cospan(_{C}) and Cospan(_{F}) are isomorphic, where the isomorphism is the identity-on-objects functor Cospan(_{C}) → Cospan(_{F}) that simply decorates each cospan with ∗, and its inverse is the one that forgets this ∗. But this is close enough to equal that many category theorists, us included, don’t mind saying equal in this case.)

Exercise 6.79

We can represent the circuit in Eq. (6.71) by the tuple (V, A, s, t, l) where V = {ul, ur, dl, dr}, A = {r1, r2, r3, c1, i1}, and s, t, and l are defined by the table

Exercise 6.80

The circuit Circ(f )(c) is

Exercise 6.82

The circuit ψ(_{underline{2, 2}})(b, s) is the disjoint union of the two labelled graphs b and s :

Exercise 6.83

The cospan is the cospan (underline{1} stackrel{f}{ ightarrow} underline{2} stackrel{g}{leftarrow} underline{1}), where f(1) and g(1) = 2. The decoration is the C-ciruit ((underline{2}), {a}, s, t, l), where s(a) = 1, t(a) = 2 and l(a) = battery.

Exercise 6.86

Recall the circuit C = (V, A, s, t, l) from the solution to Exercise 6.79. Then the first decorated cospan is given by the cospan (underline{1} stackrel{f}{ ightarrow} V stackrel{g}{leftarrow} underline{2}), f (1) = ul, (1) = ur, and g(2) = ur, decorated by circuit C. The second decorated cospan is given by the cospan (underline{1} stackrel{f'}{ ightarrow} V' stackrel{g'}{leftarrow} underline{2}) and the circuit C′:= (V′, A′, s′, t′, l′), where V′ = {l, r, d}, A′ = {r1′, r2′}, and the functions are given by the tables

To compose these, we first take the pushout of (V stackrel{g}{leftarrow} underline{2} stackrel{f'}{leftarrow} V').

This gives the a new apex V ′′ = {ul, dl, dr, m, r} with five elements, and composite cospan (underline{1} stackrel{h}{ ightarrow} V'' stackrel{k}{leftarrow} underline{2}) given by h(1) = ul, k(1) = r and k(2) = m. The new circuit is given by (V,′′ A + A′, s,′′ t,′′ l′′) where the functions are given by

This is exactly what is depicted in Eq. (6.74).

Exercise 6.88

Composing η and x we have

and composing the result of (mathcal{E}) gives

Exercise 6.96

1. The cospan shown left corresponds to the wiring diagram shown right:

It has two inner circles, each with two ports. One port of the first is wired to a port of the second. One port of the first is wired to the outside circle, and one port of the second is wired to the outside circle. This is exactly what the cospan says to do.

2. The cospan shown left corresponds to the wiring diagram shown right:

3. The composite g ◦(_{1}) f has arity (2, 2, 2, 2; 0); there is a depiction on the left:

4. The associated wiring diagram is shown on the right above. One can see that one diagram has been substituted in to a circle of the other.


NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.1

Ex 6.1 Class 8 Maths Question 1.
What will be the unit digit of the squares of the following numbers?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 20387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Solution:
(i) Unit digit of 81 2 = 1
(ii) Unit digit of 272 2 = 4
(iii) Unit digit of 799 2 = 1
(iv) Unit digit of 3853 2 = 9
(v) Unit digit of 1234 2 = 6
(vi) Unit digit of 26387 2 = 9
(vii) Unit digit of 52698 2 = 4
(viii) Unit digit of 99880 2 = 0
(ix) Unit digit of 12796 2 = 6
(x) Unit digit of 55555 2 = 5

Ex 6.1 Class 8 Maths Question 2.
The following numbers are not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050
Solution:
(i) 1057 ends with 7 at unit place. So it is not a perfect square number.
(ii) 23453 ends with 3 at unit place. So it is not a perfect square number.
(iii) 7928 ends with 8 at unit place. So it is not a perfect square number.
(iv) 222222 ends with 2 at unit place. So it is not a perfect square number.
(v) 64000 ends with 3 zeros. So it cannot a perfect square number.
(vi) 89722 ends with 2 at unit place. So it is not a perfect square number.
(vii) 22000 ends with 3 zeros. So it can not be a perfect square number.
(viii) 505050 ends with 1 zero. So it is not a perfect square number.

Ex 6.1 Class 8 Maths Question 3.
The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution:
(i) 431 2 is an odd number.
(ii) 2826 2 is an even number.
(iii) 7779 2 is an odd number.
(iv) 82004 2 is an even number.

Ex 6.1 Class 8 Maths Question 4.
Observe the following pattern and find the missing digits.
11 2 = 121
101 2 = 10201
1001 2 = 1002001
100001 2 = 1𔅾𔅽
10000001 2 = ………
Solution:
According to the above pattern, we have
100001 2 = 10000200001
10000001 2 = 100000020000001

Ex 6.1 Class 8 Maths Question 5.
Observe the following pattern and supply the missing numbers.
11 2 = 121
101 2 = 10201
10101 2 = 102030201
1010101 2 = ……….
………. 2 = 10203040504030201
Solution:
According to the above pattern, we have
1010101 2 = 1020304030201
101010101 2 = 10203040504030201

Ex 6.1 Class 8 Maths Question 6.
Using the given pattern, find the missing numbers.
1 2 + 2 2 + 2 2 = 3 2
2 2 + 3 2 + 6 2 = 7 2
3 2 + 4 2 + 12 2 = 13 2
4 2 + 5 2 + …. 2 = 21 2
5 2 + …. 2 + 30 2 = 31 2
6 2 + 7 2 + ….. 2 = …… 2
Solution:
According to the given pattern, we have
4 2 + 5 2 + 20 2 = 21 2
5 2 + 6 2 + 30 2 = 31 2
6 2 + 7 2 + 42 2 = 43 2

Ex 6.1 Class 8 Maths Question 7.
Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
We know that the sum of n odd numbers = n 2
(i) 1 + 3 + 5 + 7 + 9 = (5) 2 = 25 [∵ n = 5]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10) 2 = 100 [∵ n = 10]
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12) 2 = 144 [∵ n = 12]

Ex 6.1 Class 8 Maths Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Solution:
(i) 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)
(ii) 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)

Ex 6.1 Class 8 Maths Question 9.
How many numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100.
Solution:
(i) We know that numbers between n 2 and (n + 1) 2 = 2n
Numbers between 12 2 and 13 2 = (2n) = 2 × 12 = 24
(ii) Numbers between 25 2 and 26 2 = 2 × 25 = 50 (∵ n = 25)
(iii) Numbers between 99 2 and 100 2 = 2 × 99 = 198 (∵ n = 99)


8.6: Solutions for Chapter6

If $V$ is the space of all polynomials of degree less than or equal to $n$ over a field $F$, prove that the differentiation operator on $V$ is nilpotent.

Solution: This is clear. Let $D$ be the differentiation operator, then $deg (Df)=deg f-1$ for polynomial $f$ such that $deg fgeqslant 1$. Hence we have[D^f=D(D^f)=Dc=0,]where $c$ is a constant number as $deg(D^f)=0$.

If $deg f=0$, then we also have $D^f=Dc=0$ since $f$ is a constant.

Therefore, $D^$ is a zero operator on the space of all polynomials of degree less than or equal to $n$ over a field $F$.

Exercise 6.8.4

(a) Let $S_i$ be the set of all vectors $alpha$ in such that $(T-c_iI)^malpha=0$ for some $m>0$. It is clear that $W_isubset S_i$. We would like to show $S_isubset W_i$. Take any $alphain S_i$. Then we have $(T-c_iI)^malpha=0$ for some $m>0$. We write[alpha=alpha_1+cdots+alpha_k]where $alpha_jin W_j$. Then[0=(T-c_iI)^malpha_1+cdots+(T-c_iI)^malpha_k.]Since $W_j$ is invariant under $T$, we have $(T-c_iI)^malpha_jin W_j$ for all $j$. But $V=W_1oplus cdotsoplus W_k$, we have $(T-c_iI)^malpha_j=0$ for all $j$.

If $i e j$, then $(T-c_iI)^m$ and $(T-c_jI)^$ are relatively prime. Hence there exist polynomials $f_$ and $g_$ such that $f_(T)(T-c_iI)^m+g_(T-c_jI)^=I$. Hence we have[alpha_j=Ialpha_j=f_(T)(T-c_iI)^malpha_j+g_(T-c_jI)^alpha_j=0.]Here $(T-c_iI)^malpha_j=0$ as we proved above and $(T-c_jI)^alpha_j=0$ by the definition of $W_j$. Hence all $alpha_j=0$ if $j e i$. It follows that $alpha=alpha_iin W_i$. Part (a) is done.

(b) See the hint and Exercise 6.8.15. We remark that[det(xI-(A-cI))=det((x+c)I-A).]Hence the characteristic polynomial of $A-cI$ is equal to $f_A(x+c)$, where $f_A$ is the characteristic polynomial of $A$.

Exercise 6.8.5

Solution: We write $T=D+N$ as in Theorem 13. Since $D$ is diagonalizable, so is $g(D)$. Note that $D$ is a polynomial in $T$, so is $g(D)$. Thus $g(T)$ commutes with $g(D)$. It follows that $g(D)$ commutes with $g(T)-g(D)$. Therefore, by Theorem 13, it suffices to show that $g(T)-g(D)$ is nilpotent.

Since $T$ and $D$ commute, we can write $g(T)-g(D)=(T-D)h(T,D)=Nh(T,D)$for some polynomial $h$ in $T$ and $D$. Clearly, $N$ commutes with $h(T,D)$. Because $N$ is nilpotent, there exists $m>0$ such that $(g(T)-g(D))^m=(Nh(T,D))^m=N^mh^m(T,D)=0.$We are done.

Exercise 6.8.6

Solution: We prove it by contradiction. Suppose $T$ is diagonalizable and nilpotent. By Theorem 13, [T=0+T=T+0.]Clearly, $ is also diagonalizable and nilpotent. Moreover, $ and $T$ commute. Therefore, both +T$ and $T+0$ are the decomposition in the sense of Theorem 13. Since the decomposition is unique, we must have $T=0$ which contradicts with the condtion $mathrm(T)=1$.

Exercise 6.8.7

Solution: Let $vin V$ be nonzero. Consider a projection $E$ onto the subspace spanned by $v$. Then $E$ is diagonalizable and hence commutes with $T$. We have[TEv=ETvLongrightarrow Tv=lambda v]for some $lambdain F$. Hence $v$ is an characteristic vector of $T$. Therefore, all nonzero vectors are characteristic vectors of $T$.

Take a basis $v_1,dots,v_n$, then we have $Tv_i=lambda_i v_i$ for some $lambda_iin F$. We show that $lambda_i=lambda_j$ for $i e j$. Since $v_i+v_j$ is nonzero, there exists $lambdain F$ such that $T(v_i+v_j)=lambda(v_i+v_j)$. Therefore, we have[lambda(v_i+v_j)=Tv_i+Tv_j=lambda_iv_i+lambda_j v_jiff (lambda-lambda_i)v_i+(lambda-lambda_j)v_j=0.]But $v_i,v_j$ is linearly independent, we have $lambda=lambda_i$ and $lambda=lambda_j$. It follows that $lambda_i$ are all the same. Hence $T$ is a scalar multiple of identity matrix.

Exercise 6.8.8

Solution: The following formula is well-known:[T^n(B)=sum_^n (-1)^k inom A^BA^k.]It is can be proved by induction on $n$.

Suppose $A^m=0$, then if $n=2m$, we must have[max<2m-k,k>geqslant m.]Hence $A^<2m-k>=0$ or $A^k=0$. Thus $A^<2m-k>BA^=0$ for all $k$. Hence $T^<2m>=0$ and $T$ is also nilpotent.

Exercise 6.8.9

Solution: Consider $A=egin0&1 & 0 & 0 0&0& 0 & 0&0 & 0 & 1&0 & 0 & 0end$and $B=egin0&0 & 0 & 0 0&0& 0 & 0&0 & 0 & 1&0 & 0 & 0end.$Then is clear that the characteristic polynomials of $A$ and $B$ are $x^4$ while the minimal polynomials are the same which is $x^2$. But $A$ and $B$ are not similar. Since $2=mathrm(A) e mathrm(B)=1.$

Exercise 6.8.10

Solution: See Lemma of page 263.

Exercise 6.8.11

Solution: The matrices $D$ and $N$ may not commute. Let $A=egin1&1 0&2end$, then $D=egin1&0 0&2end$ and $Negin0&1 0&0end$. It is easy to check that $DN e ND$.

Exercise 6.8.12

Exercise 6.8.13

Let $T$ be a linear operator on $V$ with minimal polynomial of the form $p^n$, where $p$ is irreducible over the scalar field. Show that there is a vector $alpha$ in $V$ such that the $T$-annihilator of $alpha$ is $p^n$.

Solution: (Let us assume $V$ is finite-dimensional!) Let $v_1,dots,v_m$ be a basis of $V$. Then for each $v_i$, the $T$-annihilator of $v_i$ is a divisor of $p^n$ and hence has the form $p^$ for some positive integer $k_i$. Clearly, $k_ileqslant n$.

We consider the set $$. If $max<n,$then $k_ileqslant n-1$ for all $i$. Hence $p(T)^v_i=0$ for all $i$. Thus $P(T)^v=0$ for all $vin V$ as $v_1,dots,v_m$ is a basis of $V$. Then the minimal polynomial of $T$ cannot be $p^n$, we get a contradiction. Therefore, we must have $max=n$. That it, $k_i=n$ for some $i$. We are done.

Exercise 6.8.14

Solution: Here, we use the notation from Theorem 12. By Theorem 12 (Primary Decomposition Theorem) and Exercise 6.8.13, we can take $alpha_iin W_i$ such that the $T$-annihilator of $alpha_i$ is $p_i^$.

Let $alpha=alpha_1+cdots+alpha_k$. Suppose $f(T)alpha=0$ for some polynomial $f$. Then we have[0=f(T)alpha=f(T)alpha_1+cdots+f(T)alpha_k.]Since $W_i$ is invariant under $T$, we have $f(T)alpha_iin W_i$. Note that[V=W_1oplus cdotsoplus W_k.]We have $f(T)alpha_i=0$ for all $i$. Since the $T$-annihilator of $alpha_i$ is $p_i^$, we conclude that $p_i^|f$ for all $i$. Hence $p|f$. Therefore, the $T$-annihilator of $alpha$ is $p$.

Exercise 6.8.15

If $N$ is a nilpotent linear operator on an $n$-dimensional vector space $V$, then the characteristic polynomial for $N$ is $x^n$.

Solution: Let us work on algebraically closed field. In general, we can enlarge the filed $F$ to its algebraic closure. So this argument works even $F$ is not algebraically closed. But I don't like this argument. This statement is clear if we know the Jordan form.

Since $N$ is nilpotent, we have $N^m=0$ for some positive integer $m$. Hence the minimal polynomial of $N$ is a monic divisor of $x^m$ and so has the form $x^k$. Recall Theorem 3 of page 193, the characteristic polynomial and the minimal polynomial has the same roots. Hence characteristic polynomial must be of the form $x^ell$. But we also know that the degree of characteristic polynomial is exactly $dim V=n$. Hence the characteristic polynomial for $N$ is $x^n$.


NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.4

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.4

Ex 6.4 Class 8 Maths Question 1.
Find the square root of each of the following numbers by Long Division method.
(i) 2304
(ii) 4489
(iii) 3481
(iv) 529
(v) 3249
(vi) 1369
(vii) 5776
(viii) 7921
(ix) 576
(x) 1024
(xi) 3136
(xii) 900
Solution:

Ex 6.4 Class 8 Maths Question 2.
Find the number of digits in the square root of each of the following numbers (without any calculation)
(i) 64
(ii) 144
(iii) 4489
(iv) 27225
(v) 390625
Solution:
We know that if n is number of digits in a square number then
Number of digits in the square root = (frac < n >< 2 >) if n is even and (frac < n+1 >< 2 >) if n is odd.
(i) 64
Here n = 2 (even)
Number of digits in √64 = (frac < 2 >< 2 >) = 1
(ii) 144
Here n = 3 (odd)
Number of digits in square root = (frac < 3+1 >< 2 >) = 2
(iii) 4489
Here n = 4 (even)
Number of digits in square root = (frac < 4 >< 2 >) = 2
(iv) 27225
Here n = 5 (odd)
Number of digits in square root = (frac < 5+1 >< 2 >) = 3
(iv) 390625
Here n = 6 (even)
Number of digits in square root = (frac < 6 >< 2 >) = 3

Ex 6.4 Class 8 Maths Question 3.
Find the square root of the following decimal numbers.
(i) 2.56
(ii) 7.29
(iii) 51.84
(iv) 42.25
(v) 31.36
Solution:

Ex 6.4 Class 8 Maths Question 4.
Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained.
(i) 402
(ii) 1989
(iii) 3250
(iv) 825
(v) 4000
Solution:
(i)

Here remainder is 2
2 is the least required number to be subtracted from 402 to get a perfect square
New number = 402 – 2 = 400
Thus, √400 = 20

(ii)

Here remainder is 53
53 is the least required number to be subtracted from 1989.
New number = 1989 – 53 = 1936
Thus, √1936 = 44

(iii)

Here remainder is 1
1 is the least required number to be subtracted from 3250 to get a perfect square.
New number = 3250 – 1 = 3249
Thus, √3249 = 57

(iv)

Here, the remainder is 41
41 is the least required number which can be subtracted from 825 to get a perfect square.
New number = 825 – 41 = 784
Thus, √784 = 28

(v)

Here, the remainder is 31
31 is the least required number which should be subtracted from 4000 to get a perfect square.
New number = 4000 – 31 = 3969
Thus, √3969 = 63

Ex 6.4 Class 8 Maths Question 5.
Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.
(i) 525
(ii) 1750
(iii) 252
(iv) 1825
(v) 6412
Solution:
(i)

Here remainder is 41
It represents that square of 22 is less than 525.
Next number is 23 an 23 2 = 529
Hence, the number to be added = 529 – 525 = 4
New number = 529
Thus, √529 = 23

(ii)

Here the remainder is 69
It represents that square of 41 is less than in 1750.
The next number is 42 and 42 2 = 1764
Hence, number to be added to 1750 = 1764 – 1750 = 14
Require perfect square = 1764
√1764 = 42

(iii)

Here the remainder is 27.
It represents that a square of 15 is less than 252.
The next number is 16 and 16 2 = 256
Hence, number to be added to 252 = 256 – 252 = 4
New number = 252 + 4 = 256
Required perfect square = 256
and √256 = 16

(iv)

The remainder is 61.
It represents that square of 42 is less than in 1825.
Next number is 43 and 43 2 = 1849
Hence, number to be added to 1825 = 1849 – 1825 = 24
The required perfect square is 1848 and √1849 =43

(v)

Here, the remainder is 12.
It represents that a square of 80 is less than in 6412.
The next number is 81 and 81 2 = 6561
Hence the number to be added = 6561 – 6412 = 149
The require perfect square is 6561 and √6561 = 81

Ex 6.4 Class 8 Maths Question 6.
Find the length of the side of a square whose area = 441 m 2
Solution:
Let the length of the side of the square be x m.
Area of the square = (side) 2 = x 2 m 2
x 2 = 441 ⇒ x = √441 = 21

Thus, x = 21 m.
Hence the length of the side of square = 21 m.

Ex 6.4 Class 8 Maths Question 7.
In a right triangle ABC, ∠B = 90°.
(a) If AB = 6 cm, BC = 8 cm, find AC
(b) If AC = 13 cm, BC = 5 cm, find AB
Solution:
(a) In right triangle ABC

AC 2 = AB 2 + BC 2 [By Pythagoras Theorem]
⇒ AC 2 = (6) 2 + (8) 2 = 36 + 64 = 100
⇒ AC = √100 = 10
Thus, AC = 10 cm.
(b) In right triangle ABC

AC 2 = AB 2 + BC 2 [By Pythagoras Theorem]
⇒ (13) 2 = AB 2 + (5) 2
⇒ 169 = AB 2 + 25
⇒ 169 – 25 = AB 2
⇒ 144 = AB 2
AB = √144 = 12 cm
Thus, AB = 12 cm.

Ex 6.4 Class 8 Maths Question 8.
A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs more for this.
Solution:
Let the number of rows be x.
And the number of columns also be x.
Total number of plants = x × x = x 2
x 2 = 1000 ⇒ x = √1000

Here the remainder is 39
So the square of 31 is less than 1000.
Next number is 32 and 32 2 = 1024
Hence the number to be added = 1024 – 1000 = 24
Thus the minimum number of plants required by him = 24.
Alternative method:

The minimum number of plants required by him = 24.

Ex 6.4 Class 8 Maths Question 9.
There are 500 children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?
Solution:
Let the number of children in a row be x. And also that of in a column be x.
Total number of students = x × x = x 2
x 2 = 500 ⇒ x = √500

Here the remainder is 16
New Number 500 – 16 = 484
and, √484 = 22
Thus, 16 students will be left out in this arrangement.


NCERT Solutions for Class 8 English Honeydew Chapter 6 This is Jody’s Fawn

NCERT Solutions for Class 8 English Honeydew Chapter 6 This is Jody’s Fawn

NCERT Solutions for Class 8 English Honeydew Chapter 6 This is Jody’s Fawn

Comprehension Check (Page 90)

  1. What had happened to Jody’s father?
  2. How did the doe save Penny’s life?
  3. Why does Jody want to bring the fawn home?
  4. How does Jody know that the fawn is a male?
  1. Jody’s father had been bitten by a rattlesnake.
  2. Jody’s father killed the doe or she dear. He used her heart and liver to draw out the snake’s poison. In this way the doe saved Penny’s life.
  3. Jody’s father had killed the doe. Without the mother-deer, the fawn was likely to starve to death in the forest. So Jody wanted to bring the young fawn home.
  4. The spots on the fawn’s body made Jody know that it was a male.

Comprehension Check (Page 91)

  1. Jody didn’t want Mill-wheel with him for two reasons. What were they?
  2. Why was Mill-wheel afraid to leave Jody alone?
  1. Jody didn’t want Mill-wheel to join him in the search for the fawn. The reason was that he was not sure about the fawn’s safety. He didn’t want Mill-wheel to see his disappointment.
  2. Mill-wheel was afraid that Jody might be lost in the jungle.

Comprehension Check (Page 94)

  1. How did Jody bring the fawn back home?
  2. Jody was filled with emotion after he found the fawn. Can you find at least three words or phrases which show how he felt?
  3. How did the deer drink milk from the gourd?
  4. Why didn’t the fawn follow Jody up the steps as he had thought it would?
  1. Jody picked up the fawn into his arms and proceeded to home. After some dis­tance, he kept the fawn down and took rest. Later on, the fawn followed him. Thus he brought the fawn back home.
  2. (i) (The fawn) shook him through with the stare of its liquid eye.
    (ii) The touch of the fawn made him delirious.
    (iii) As though the fawn were a china deer.
  3. Jody dipped his fingers in the milk. Then he left the fawn suck his fingers. He did so several times. Finally, the fawn drank off all the milk from the gourd.
  4. The fawn didn’t know how to raise its feet to climb the steps.

Working With the Text (Page 94)

Questions 1:
Why did Penny Baxter allow Jody to go find the fawn and raise it?
Answer:
Penny was convinced by Jody’s argument that it would be ungrateful if they left the fawn in the forest to starve. He realised that Jody was right.

Question 2:
What did Doc Wilson mean when he said, “Nothing in the world ever comes quite free”?
Answer:
Doc Wilson meant that Penny must pay back to the doe whom he had killed for his own gain by bringing up her fawn.

Question 3:
How did Jody look after the fawn, after he accepted the responsibility for doing this?
Answer:
Jody looked after the faWh like a mother. He made it drink milk with his fingers dipped in milk. This is how a mother feeds her baby. Jody was glad that he had found the fawn.

Question 4:
How does Jody’s mother react when she hears that he is going to bring the fawn home? Why does she react in this way?
Answer:
Jody’s mother turned her nose when she heard that he was going to bring back the fawn. She gasped with surprise because she didn’t want to see an animal in her home.

Working With Language (Page 94)

Question 1:
Look at these pairs of sentences.

Penny said to Jody, “ Will you be back before dinner?”
Penny asked Jody if he would be back before dinner.
“How are you feeling, Pa? ” asked Jody.
Jody asked his father how he was feeling.

Here are some questions in direct speech. Put them into reported speech.

  1. Penny said, “Do you really want it son?”
  2. Mill-wheel said, “Will he ride back with me?”
  3. He said to Mill-wheel, “Do you think the fawn is still there?”
  4. He asked Mill-wheel, “Will you help me find him?”
  5. He said, “Was it up here that Pa got bitten by the snake?”
  1. Penny asked his son if he really wanted the fawn.
  2. Mill-wheel enquired if Jody would ride back with him.
  3. Jody asked Mill-wheel if he thought the fawn was still there.
  4. He asked Mill-wheel if he would help him find the fawn.
  5. Mill-wheel wanted to know if that was the place where Pa had got bitten by the snake.

Question 2:
Look at these two sentences.

The first sentence has an intransitive verb, a verb without an object. The second sentence has a transitive verb. It has a direct object. We can ask: “What did it turn?” You can answer. “Its head. It turned its head.”
Say whether the verb in each sentence below is transitive or intransitive. Ask yourself a “what’ question about the verb, as in the example above. (For some verbs, the object is a person, so ask the question ‘who’ instead of ‘what’).

(i) Jody then went to the kitchen.
(ii) The fawn wobbled after him.
(iii) You found him.
(iv) He picked it up.
(v) He dipped his fingers in the milk.
(vi) It bleated frantically and butted
(vii) The fawn sucked his fingers.
(viii) He lowered his fingers slowly into the milk.
(ix) It stamped its small hoofs impatiently.
(x) He held his fingers below the level of the milk,
(xi) The fawn followed
(xii) He walked all day.
(xiii) He stroked its sides.
(xiv) The fawn lifted its nose.
(xv) Its legs hung limply.

Question 3:
Here are some words from the lesson. Working in groups, arrange them in the order in which they would appear in the dictionary. Write down some idioms and phrasal verbs connected to these words. Use the dictionary for more idioms and phrasal verbs.

Answer:

Idioms or phrasal verbs connected to the above words.
Clearing: clearing, campaign
Close: close shave, close up, close quarters
Draw: draw the curtain on/over, draw a blank
Light: in the light of, bring to light
Make: make the most of, make up
Part: part with, parted comparing
Pick: pick up, pick and choose
Scrawny: the scrawny neck
Sweet: have a sweet tooth, sweet seventeen, sweet tongued, sweet nothings
Wonder: wonder world, wonder load, nine day’s wonder, wonder about, do wonders.

Speaking (Page 96)

Question 1:
Do you think it is right to kill an animal to save a human life? Give reasons for your answer.
Answer:
Most of the animals are our friends. Dogs, horses, elephants, cows are a few such animals that serve us. But man has been killing codfish or the whales for oil. Tigers are killed for their skin and bones. This is not fair. But there is no harm if any of them are killed strictly to save human life, properly and agriculture.
However, killing animals is a crime. It is wrong to kill wild life for their hide or for pleasure.

Question 2:
Imagine you wake up one morning and find a tiny animal on your doorstep. You want to keep it as a pet but your parents are not too happy about it. How would you persuade them to let you keep it? Discuss it in groups and present your argu­ments to the class.
Answer:
The young ones of cats, dogs and some birds attract us as does a human child. When I was a child, I wanted to adopt a kitten or a puppy as pet. I found a good breed puppy at my doorstep one day. But it created a commotion in the house. My mother got irritated at the veiy presence of pets in the house. They bite and bark, enter the kitchen or sit on our beds and make things dirty. But I assured her that I would look after my puppy and train it. The loyal dog would act as security guard and a playmate. My parents finally relented and let me have the poor puppy as a pet.

Writing (Page 96)

Question 1:
Imagine you have a new pet that keeps you busy. Write a paragraph describing your pet, the things it does, and the way it makes you feel. Here are some words and phrases that you could use.

frisky, smart, disobedient, loyal, happy, enthusiastic, companion, sharing, friend, rolls in mud, dirties the bed, naughty, lively, playful, eats up food, hides the newspaper, drinks up milk, runs away when called, floats on the water as if dead.

Answer:
I have taken a kitten as my pet. It is female with silky fur and skin. She keeps me busy. My mother does not take interest in my pet. She curses the little one for doing mischief, for moving about in the house, for making the bed and floor dirty. The kitten enters the kitchen and drinks up milk. She is naughty and disobedient also. She is most unlike a dog which is loyal, obedient and strong. Still I like my pet because it is lively, playful and frisky.

Question 2:
Human life is dependent on nature (that’s why we call her Mother Nature). We take everything from nature to live our lives. Do we give back anything to nature?
(i) Write down some examples of the natural resources that we use.
(ii) Write a paragraph expressing your point of view regarding our relationship with nature.
Answer:
(i)
Man and nature are complementary to each other. Man for ages has been using forests, minerals and chemicals for his survival. Earth and nature are our lifelines. They help us directly or indirectly. Take for example the paper we print, our books and newspapers. They are products of trees. We get fruits, flowers and fodder from nature. We get water and air free from nature. It is unfortunate that we are over using the limited resources and are also polluting them.

Nature is our Mother. We must not use up anything to the extent that it is not restored naturally. By cutting down trees or killing whales we are, in a way, depriving our children of their share. Let us give back to nature for the benefits we get from it.

(ii) Some of the natural resources that we use are water, coal, mineral oil, etc.

Question 3:
In This is Jody’s Fawn, Jody’s father uses a “home remedy’ for a snake bite. What
should a person now do if he or she is bitten by a snake? Are all snakes poisonous?
With the help of your teacher and others, find out answers to such questions. Then write a short paragraph on—What to do if a snake chooses to bite you.
Answer:
Snakes are the most dreaded of wild creatures. This is why we use sticks to kill them. There are many poisonous snakes. Green snakes or water snakes are not poisonous. Still we cannot be sure of it. So we don’t take a chance. We call in a snake charmer to draw the cobra out of the house. A snake-bite can kill the victim in a few minutes. But the victim can be saved if he gets the first aid in the form of blood-letting and anti-venom serum. The cure for snake bite is prepared front the snake’s poison.

In case I am bitten by a poisonous snake, the first thing I would do is to put a band tightly over the bitten part. Then I shall use a blade or knife to make a small cut on the bitten part, and press the poisonous blood out. Then I shall go to hospital for medical help. I shall not go to sleep until I feel better and safe.

MORE QUESTIONS SOLVED

I. SHORT ANSWER TYPE QUESTIONS

Question 1:
When and why does Jody’s father need a remedy?
Answer:
Penny, Jody’s father, is bitten by a poisonous rattlesnake. Instead of going to a doctor, he kills a she deer and uses her liver to draw out the poison.

Question 2:
How does Jody react to the cruelty of his father?
Answer:
Jody, the small boy, tells his father that he had left the fawn alone and defenceless to die. So it is their moral duty to save the innocent and hungry young one of the doe.

Question 3:
How does Penny take his son’s argument?
Answer:
Penny agreed with Jody’s argument that it would be ungrateful to leave the fawn to starve.

Question 4:
What did Doc Wilson say about Jody’s suggestion?
Answer:
Doc Wilson said that they had to pay the price for everything. He justified the plan of Jody and Penny about the fawn.

Question 5:
Why did Jody see only vultures and kites feeding on the dead body of the doe?
Answer:
The sand showed large footprints of tigers or leopards but they did not eat up the dead doe. The reason was that the big cats killed an animal themselves to eat its flesh. Vultres and kites are birds of prey. They also feed on the dead bodies.

Question 6:
How did Jody approach and win the trust of the fawn?
Answer:
The fawn shook with fear as Jody drew near. It lifted its nose and scented the
visitor. Jody moved forward on all fours and put his arms around its body.

Question 7:
How did Jody feel as he touched the fawn’s skin?
Answer:
Jody found the fawn’s skin very soft and clean. He stroked its sides gently as though it were made of clay and would break soon.

Question 8:
How did Jody feed the fawn?
Answer:
Jody decided to give away his share of milk to the fawn. He poured the milk into a small pot. Then he dipped his fingers in the milk and put them into its mouth. The fawn sucked slowly until the milk vanished.

Question 9:
What message does the story of the fawn convey to the readers?
Answer:
The story highlights two things. It is not fair to kill an animal for its use as a cure. Secondly, one should have pity and love for the animals.

II. LONG ANSWER TYPE QUESTIONS

Question 1:
How did Jody persuade his father to go to the forest to bring back the fawn?
Answer:
Jody was a small, brave and sensitive boy. He was with his father when he (his father) was bitten by a rattlesnake. His father quickly killed a doe and used its heart and liver to draw out the snake’s poison. Jody was happy to see that his father got a new life but at the same time he was worried for the little fawn who was left alone without its mother. He wanted to bring back the fawn. He requested his father to allow him to go to the forest to find the fawn. He told him that he didn’t need to drink milk because he was now a big boy. He would give the milk to the fawn. He also said that it was ungrateful to leave the fawn to starve. His father was in a fix. He couldn’t say “no’ to his son. And finally allowed him (Jody) to go to the forest to find the fawn.

Question 2:
How did Jody feed the little fawn?
Answer:
Jody poured milk into a small gourd. He dipped his fingers in the milk and thrust them into the fawn’s soft wet mouth. It sucked greedily. When he withdrew them, it bleated frantically and butted him. He dipped his fingers again and as the fawn sucked, he lowered them slowly into the milk. The fawn blew and sucked and snorted. It stamped its small hoofs impatiently. As long as he held his fingers below the level of the milk, the fawn was content.


2009 Alabama Code Title 8 &mdash COMMERCIAL LAW AND CONSUMER PROTECTION. Chapter 6 &mdash SECURITIES. Section 8-6-19 Civil liabilities of sellers, agents, etc. remedies of purchasers.

(2) Sells or offers to sell a security by means of any untrue statement of a material fact or any omission to state a material fact necessary in order to make the statements made, in the light of the circumstances under which they are made, not misleading, the buyer not knowing of the untruth or omission, and who does not sustain the burden of proof that he did not know and in the exercise of reasonable care could not have known of the untruth or omission,

is liable to the person buying the security from him who may bring an action to recover the consideration paid for the security, together with interest at six percent per year from the date of payment, court costs and reasonable attorneys' fees, less the amount of any income received on the security, upon the tender of the security, or for damages if he no longer owns the security. Damages are the amount that would be recoverable upon a tender less the value of the security when the buyer disposed of it and interest at six percent per year from the date of disposition.

(b)(1) Any person who engages in the business of advising others, for compensation, either directly or through publications or writings, as to the value of securities or as to the advisability of investing in, purchasing, or selling securities, or who, for compensation and as part of a regular business, issues or promulgates analyses or reports concerning securities in violation of subsection (b), (c), (d), (e), or (f) of Section 8-6-17, subsection (b) or (c) of Section 8-6-3, Section 8-6-14, is liable to that person, who may bring an action to recover the consideration paid for such advice and any loss due to such advice, together with interest at six percent per year from the date of payment of the consideration plus costs and reasonable attorney's fees, less the amount of any income received from such advice.

No person may maintain an action hereunder pursuant to a violation of subsection (c) of Section 8-6-3 based solely on the fact that an investment adviser representative other than the one from whom the person received advice is unregistered.

(2) Any person who receives, directly or indirectly, any consideration from another person for advice as to the value of securities or their purchase or sale, whether through the issuance of analyses, reports, or otherwise and employs any device, scheme, or artifice to defraud such other person or engages in any act, practice, or course of business which operates or would operate as a fraud or deceit on such other person, is liable to that person, who may bring an action to recover the consideration paid for such advice and any loss due to such advice, together with interest at six percent per year from the date of payment of the consideration plus costs and reasonable attorney's fees, less the amount of any income received from such advice.

An action based on a violation of subsection (c) of Section 8-6-17 and this section may not prevail where the person accused of the violation sustains the burden of proof that he did not know, and in the exercise of reasonable care, could not have known of the existence of the facts by reason of which the liability is alleged to exist.

(c) Every person who directly or indirectly controls a person liable under subsections (a) or (b) of this section, including every partner, officer, or director of such a person, every person occupying a similar status or performing similar functions, every employee of such a person who materially aids in the conduct giving rise to the liability, and every dealer or agent who materially aids in such conduct is also liable jointly and severally with and to the same extent as the person liable under subsection (a) or (b), unless he is able to sustain the burden of proof that he did not know, and in exercise of reasonable care could not have known, of the existence of the facts by reason of which the liability is alleged to exist.

(d) Any tender specified in this section may be made at any time before entry of judgment.

(e) Every cause of action under this section survives the death of any person who might have been a plaintiff or defendant.

(f) No person may obtain relief under this section in an action involving the failure to register unless suit is brought within two years from the date of sale. All other actions for relief under this section must be brought within the earlier of two years after discovery of the violation or two years after discovery should have been made by the exercise of reasonable care. No person may bring an action under subsection (a) of this section:

(1) If the buyer received a written offer, before the action and at a time when he owned the security, to refund the consideration paid together with interest at six percent per year from the date of payment, less the amount of any income received on the security, and he failed to accept the offer within 30 days of its receipt, or

(2) If the buyer received such an offer before the action and at a time when he did not own the security, unless he rejected the offer in writing within 30 days of its receipt.

(g) No person who has made or engaged in the performance of any contract in violation of any provision of this article or any rule or order hereunder or who has acquired any purported right under any such contract with knowledge of the facts by reason of which its making or performance was in violation, may base any action on the contract.

(h) Any condition, stipulation, or provision binding any person acquiring any security or receiving any investment advice to waive compliance with any provision of this article or any rule or order hereunder is void.

(i) The rights and remedies provided by this article are in addition to any other rights or remedies that may exist.

(j)(1) The commission may by order, if it finds such order to be in the public interest, impose an administrative assessment upon any person who violates any provision of this article or any rule or order issued under this article.

(2) Any administrative assessment imposed under this section shall not exceed ŭ,000 for each act or omission that constitutes the basis for an order issued under this section, except that the amount of the administrative assessment may not exceed โ,000 for any person subject to the order.

(3) For the purposes of determining the amount or extent of an administrative assessment, if any, to be imposed under this section, the commission shall consider among other factors, the frequency, persistence, and willfulness of the conduct constituting a violation of any provision of this article or any rule or order issued under this article, and the number of persons adversely affected by the conduct.

(4) The administrative assessment under this section is in addition to any other penalty, remedy, or sanction that may be imposed under this article.

(5) All assessments collected under this subsection (j) of Section 8-6-19 shall be deposited in the general fund of the state.

(k)(1) The commission may charge, in addition to any administrative assessment, fine, penalty, remedy, or sanction imposed under this article, the actual cost of any investigation resulting from any violation of any provision of this article or any violation of any rule or order issued under this article or the actual cost of any examination made by the commission pursuant to this article, to the party or parties subject to such investigation or examination. Such charge may include, but is not limited to, a per diem prorated upon the salary cost of any employee of the commission together with actual travel, housing and any and all other reasonable expenses incurred as a result of such investigation or examination.

(2) All charges assessed for costs involved pursuant to subdivision (1) of subsection (k) of Section 8-6-19 shall be deposited in the Alabama Securities Commission Fund in the State Treasury to be drawn upon by the commission for its use in the administration of this article.

(Acts 1959, No. 542, p. 1318, §18 Acts 1990, No. 90-527, p. 772, §1.)

Disclaimer: These codes may not be the most recent version. Alabama may have more current or accurate information. We make no warranties or guarantees about the accuracy, completeness, or adequacy of the information contained on this site or the information linked to on the state site. Please check official sources.


NCERT Solutions for Class 8 Social Science History Chapter 6

NCERT Solutions for Class 8 Social Science History Chapter 6 Weavers, Iron Smelters and Factory Owners (बुनकर, लोहा बनाने वाले और फैक्ट्री मालिक) to Study online or download in PDF form free for session 2021-22.

Download latest NCERT Books 2021-2022 and Offline Apps based on these books free of cost. Get the Answers of your questions with your friends and experts through Discussion Forum.

NCERT Solutions for Class 8 Social Science History Chapter 6

Class 8 History Chapter 6 Question Answers

CBSE NCERT Solutions for Class 8 Social Science History Chapter 6 Weavers, Iron Smelters and Factory Owners is given below in updated form for session 2021-22. Download these solutions for offline use and Offline apps also based on latest CBSE Syllabus 2021-22.

8th History Chapter 6 Question Answers

Describe the Story of Crafts and Industry in British rule?

The crafts and industries of India during British rule by focusing on two industries, namely, textiles and iron and steel. Both these industries were crucial for the industrial revolution in the modern world. Mechanised production of cotton textiles made Britain the foremost industrial nation in the nineteenth century. The industrialisation of Britain had a close connection with the conquest and colonisation of India. With the growth of industrial production, British industrialists began to see India as a vast market for their industrial products and over time manufactured goods from Britain began flooding India.

Write a short note on Indian Textiles?

The textile production in 1750, before the British conquered Bengal, India was by far the world’s largest producer of cotton textiles. Indian textiles had long been renowned both for their fine quality and exquisite craftsmanship. They were extensively traded in Southeast Asia (Java, Sumatra and Penang) and West and Central Asia. From the sixteenth century European trading companies began buying Indian textiles for sale in Europe.
There are many other words which point to the popularity of Indian textiles in Western markets. The
English East India Company sent to its representatives in Calcutta in 1730. The order that year was for 5,89,000 pieces of cloth. These were known by their common name in the European trade as piece goods – usually woven cloth pieces that were 20 yards long and 1 yard wide.

Important Notes on 8th History Chapter 6

The pieces ordered in bulk were printed cotton cloths called chintz, cossaes (or khassa) and bandanna. The word chintz comes from the Hindi word chhint, a cloth with small and colourful flowery designs. From the 1680s there started a craze for printed Indian cotton textiles in England and Europe mainly for their exquisite floral designs, fine texture and relative cheapness. Rich people of England including the Queen herself wore clothes of Indian fabric. Similarly, the word bandanna now refers to any brightly coloured and printed scarf for the neck or head. Originally, the term derived from the word “bandhna” (Hindi for tying), and referred to a variety of brightly coloured cloth produced through a method of tying and dying.


NCERT Solutions for Class 8 Maths Chapter 6

NCERT Solutions for Class 8 Maths Chapter 6 SQUARE AND SQUARE ROOTS Exercise 6.1, Exercise 6.2, Exercise 6.3 and Exercise 6.4 in English Medium as well as Hindi Medium updated for new academic session 2021-22.

Download Prashnavali 6.1, Prashnavali 6.2, Prashnavali 6.3 and Prashnavali 6.4 in Hindi Medium free to download in PDF file. NCERT Solutions 2021-22 are for all the students using NCERT Books in the new academic session 2021-2022. NCERT Solutions for class 8 other subjects are also available in PDF format.

NCERT Solutions for Class 8 Maths Chapter 6

Class 8 Maths Chapter 6 all Exercises Solution

Class 8 all Subjects Solutions App

Class 8 Maths Chapter 6 Solutions

Class 8 Maths Chapter 6 Square and Square Roots all exercises question answers in English as well as Hindi Medium or View in Video Format updated for new academic session 2021-22. Download Class 8 Maths Offline App for offline use. It works without internet, once downloaded.

8 Maths Chapter 6 Solutions in English Medium

8 Maths Chapter 6 Solutions in Hindi Medium

Class 8 Maths Exercise 6.1 Solutions in Videos

Class 8 Maths Exercise 6.2 Solutions in Videos

Class 8 Maths Exercise 6.3 Solutions in Videos

Class 8 Maths Exercise 6.4 Solutions in Videos

Important Terms about Class 8 Maths Chapter 6

What will be the unit digit of the squares of the number 81?

The number 81 contains its unit’s place digit 1. So, square of 1 is 1.
Hence, unit’s digit of square of 81 is 1.

Without adding, find the sum of 1 + 3 + 5 + 7 + 9.

Here, there are five odd numbers.
Therefore square of 5 is 25.
1 + 3 + 5 + 7 + 9 = 5^2 = 25

In chapter 6 Square and Square Roots, we will find the square roots of different numbers with various methods. Testing a number, whether it is a perfect square or not. Expressing a perfect square number as a sum of odd numbers. Finding the square of a number using a particular identity and finding the Pythagoras triplet when one of the triplet number is given.
There are two methods, one with prime factorization method and other is long division method, to find the square root of a number. In chapter 6, the square root of decimal number using long division method is also given which is very useful tool for further classes also.

Do you know?

There are several ways to find the square root of any number. All methods have their own different importance, so students should learn all methods well. It is also frequently used in the coming classes.


Precipitation Reactions and Solubility Rules

A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter).

The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility , defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble . A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble , and these are the substances that readily precipitate from solution. More information on these important concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that have been observed for many ionic compounds (Table (PageIndex<1>)).

  • group 1 metal cations (Li + , Na + , K + , Rb + , and Cs + ) and ammonium ion (left(ce ight))
  • the halide ions (Cl &minus , Br &minus , and I &minus )
  • the acetate bicarbonate nitrate and chlorate (ce<(ClO3- )>) ions
  • the sulfate (ce<(SO4- )>) ion
  • halides of Ag + , (ce) and Pb 2+
  • sulfates of Ag + , Ba 2+ , Ca 2+ , (ce) Pb 2+ , and Sr 2+
  • carbonate chromate phosphate (ce<(PO4^3- )>) and sulfide (S 2&minus ) ions
  • hydroxide ion (OH &minus )
  • compounds of these anions with group 1 metal cations and ammonium ion
  • hydroxides of group 1 metal cations and Ba 2+

A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:

This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.

The net ionic equation representing this reaction is:

Lead iodide is a bright yellow solid that was formerly used as an artist&rsquos pigment known as iodine yellow (Figure (PageIndex<1>)). The properties of pure PbI2 crystals make them useful for fabrication of X-ray and gamma ray detectors.

Figure (PageIndex<1>): A precipitate of PbI2 forms when solutions containing Pb 2+ and I &minus are mixed. (credit: Der Kreole/Wikimedia Commons)

The solubility guidelines in Table (PageIndex<1>) may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag + , (ce) Na + , and F &minus ions. Aside from the two ionic compounds originally present in the solutions, AgNO3 and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO3 and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:

Example (PageIndex<1>): P redicting Precipitation Reactions

Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.

  1. potassium sulfate and barium nitrate
  2. lithium chloride and silver acetate
  3. lead nitrate and ammonium carbonate

(a) The two possible products for this combination are KNO3 and BaSO4. The solubility guidelines indicate BaSO4 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

(b) The two possible products for this combination are LiC2H3O2 and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

(c) The two possible products for this combination are PbCO3 and NH4NO3. The solubility guidelines indicate PbCO3 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

Which solution could be used to precipitate the barium ion, Ba 2+ , in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate?


Free PDF download of RD Sharma Solutions for Class 8 Maths Chapter 6 - Algebraic Expressions and Identities solved by Expert Mathematics Teachers on Vedantu.com. All Chapter 6 - Algebraic Expressions and Identities Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced), NEET, Engineering and Medical entrance exams.

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