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3.2: Proofs - Mathematics


Investigate!

Decide which of the following are valid proofs of the following statement:

If (a b) is an even number, then (a) or (b) is even.

  1. Suppose (a) and (b) are odd. That is, (a=2k+1) and (b=2m+1) for some integers (k) and (m ext{.}) Then egin{align*} ab & =(2k+1)(2m+1) & =4km+2k+2m+1 & =2(2km+k+m)+1. end{align*}

    Therefore (ab) is odd.

  2. Assume that (a) or (b) is even - say it is (a) (the case where (b) is even will be identical). That is, (a=2k) for some integer (k ext{.}) Then egin{align*} ab & =(2k)b & =2(kb). end{align*}

    Thus (ab) is even.

  3. Suppose that (ab) is even but (a) and (b) are both odd. Namely, (ab = 2n ext{,}) (a=2k+1) and (b=2j+1) for some integers (n ext{,}) (k ext{,}) and (j ext{.}) Then egin{align*} 2n & =(2k+1)(2j+1) 2n & =4kj+2k+2j+1 n & = 2kj+k+j+frac{1}{2}. end{align*}

    But since (2kj+k+j) is an integer, this says that the integer (n) is equal to a non-integer, which is impossible.

  4. Let (ab) be an even number, say (ab=2n ext{,}) and (a) be an odd number, say (a=2k+1 ext{.}) egin{align*} ab & =(2k+1)b 2n & =2kb+b 2n-2kb& =b 2(n-kb)& =b. end{align*}

    Therefore (b) must be even.

Anyone who doesn't believe there is creativity in mathematics clearly has not tried to write proofs. Finding a way to convince the world that a particular statement is necessarily true is a mighty undertaking and can often be quite challenging. There is not a guaranteed path to success in the search for proofs. For example, in the summer of 1742, a German mathematician by the name of Christian Goldbach wondered whether every even integer greater than 2 could be written as the sum of two primes. Centuries later, we still don't have a proof of this apparent fact (computers have checked that “Goldbach's Conjecture” holds for all numbers less than (4 imes 10^{18} ext{,}) which leaves only infinitely many more numbers to check).

Writing proofs is a bit of an art. Like any art, to be truly great at it, you need some sort of inspiration, as well as some foundational technique. Just as musicians can learn proper fingering, and painters can learn the proper way to hold a brush, we can look at the proper way to construct arguments. A good place to start might be to study a classic.

Theorem (PageIndex{1})

There are infinitely many primes.

Proof

Suppose this were not the case. That is, suppose there are only finitely many primes. Then there must be a last, largest prime, call it (p ext{.}) Consider the number

egin{equation*} N = p! + 1 = (p cdot (p-1) cdot cdots 3cdot 2 cdot 1) + 1. end{equation*}

Now (N) is certainly larger than (p ext{.}) Also, (N) is not divisible by any number less than or equal to (p ext{,}) since every number less than or equal to (p) divides (p! ext{.}) Thus the prime factorization of (N) contains prime numbers (possibly just (N) itself) all greater than (p ext{.}) So (p) is not the largest prime, a contradiction. Therefore there are infinitely many primes.

(square)


Indiana Once Tried to Change Pi to 3.2

Any high school geometry student worth his or her protractor knows that pi is an irrational number, but if you’ve got to approximate the famed ratio, 3.14 will work in a pinch. That wasn’t so much the case in late-19th-century Indiana, though. That’s when the state’s legislators tried to pass a bill that legally defined the value of pi as 3.2.

The very notion of legislatively changing a mathematical constant sounds so crazy that it just has to be an urban legend, right? Nope. As unbelievable as it sounds, a bill that would have effectively redefined pi as 3.2 came up before the Indiana legislature in 1897.

The story of the “Indiana pi bill” starts with Edward J. Goodwin, a Solitude, Indiana, physician who spent his free time dabbling in mathematics. Goodwin’s pet obsession was an old problem known as squaring the circle. Since ancient times, mathematicians had theorized that there must be some way to calculate the area of a circle using only a compass and a straightedge. Mathematicians thought that with the help of these tools, they could construct a square that had the exact same area as the circle. Then all one would need to do to find the area of the circle was calculate the area of the square, a simple task.

Sounds like a neat trick. The only problem is that it’s impossible to calculate the area of a circle in this way. It just won’t work. Furthermore, when Goodwin was toying with this problem, mathematicians already knew it was impossible Ferdinand von Lindemann had proven that the task was a fool’s errand in 1882.

Goodwin wasn’t going to let something trivial like the proven mathematical impossibility of his task deter his efforts, though. He persevered, and in 1894 he even convinced the upstart journal American Mathematical Monthly to print the proof in which he “solved” the squaring-the-circle problem. Goodwin’s proof didn’t explicitly deal with approximating pi, but when you’re quite literally trying to fit a square peg in a round hole, weird things happen. One of the odd side effects of Goodwin’s machinations was that the value of pi morphed into 3.2.

Let's Make a Deal

Although Goodwin’s “proof” was anything but, he was pretty cocky about its infallibility. He didn’t just publish his faulty method in journals he copyrighted it. Goodwin figured everyone would be lining up to use his revolutionary new trick, and his plan was to collect royalties from businesses and mathematicians who sought to exploit his method.

Goodwin wasn’t totally greedy, though, and that’s where the Indiana legislature entered the picture. Goodwin couldn’t bear the thought of Hoosier schoolchildren being deprived of the fruits of his brilliance just because the state couldn’t foot the bill for his royalties. So he magnanimously offered to let the state use his masterpiece free of charge.

Indiana wasn’t going to get such an awesome deal totally for free, though. The state could avoid paying royalties if and only if the legislature would accept and adopt this “new mathematical truth” as state law. Goodwin convinced Representative Taylor I. Record to introduce House Bill 246, which outlined both this bargain and the basics of his method.

Again, Goodwin’s method and the accompanying bill never mention the word “pi,” but on the topic of circles, it clearly states, “[T]he ratio of the diameter and circumference is as five-fourths to four.” Yup, that ratio is 3.2. Goodwin isn’t afraid to lambaste the old approximation of pi, either. The bill angrily condemns 3.14 as “wholly wanting and misleading in its practical applications.”

Goodwin’s blasting of the old approximation isn’t even the funniest part of the bill’s text. The third and final section extols his other mathematical breakthroughs, including solving the similarly impossible problems of angle trisection and doubling the cube, before reminding any reader who wasn’t sufficiently awestruck at his magnificence, “And be it remembered that these noted problems had been long since given up by scientific bodies as insolvable mysteries and above man's ability to comprehend.“

Math Problem

To anyone who passed the aforementioned high school geometry class, this bill was patently absurd. Apparently Indiana legislators weren’t a pack of math whizzes, though. After the bill bounced around between committees, the Committee on Education finally sent it out for a vote, and the bill passed the House unanimously. No, not a single one of Indiana’s 67 House members raised an eyebrow at a proof that effectively redefined pi as 3.2.

Luckily the state’s senators had a bit more numerical acumen. Well, some of them did. Eventually. After sailing through the House, the bill first went to the Senate’s Committee on Temperance, which also recommended that it pass. By this point, news of Indiana attempting to legislate a new value of pi and endorse an airtight solution to an unsolvable math problem had become national news, and papers all over the country were mocking the legislature’s questionable calculations.

All this attention ended up working in Indiana’s favor. While the state’s lawmakers couldn’t follow Goodwin’s bizarre brand of mathe-magic well enough to refute his proof, there were other smart Hoosiers who could. Professor C.A. Waldo of Purdue University was in Indianapolis while the pi hoopla was unfolding, and after watching part of the debate at the statehouse he was so thoroughly horrified that he decided to intervene.

The legislators may have been nearly bamboozled by Goodwin’s pseudo-math, but Waldo certainly wasn’t. Waldo got the ear of a group of senators after watching the absurd debate and explained why Goodwin’s theory was nonsense. (It seemed that most of the legislators didn’t really understand what was going on in the bill they just knew that by approving it the state would get to use a new theory for free.)

After receiving Waldo’s coaching, the Senate realized that the new bill was a very, very bad idea. Senator Orrin Hubbel moved that a vote on the bill be postponed indefinitely, and Goodwin’s new math died a quiet legislative death. The Indiana legislature hasn’t tried to rewrite the basic principles of math in all the years since.

This post originally appeared in 2011. Image credit: Instructables member hertzgamma. You can read the text of the bill here.


3.2: Proofs - Mathematics

It was one of the most surprising discoveries of the Pythagorean School of Greek mathematicians that there are irrational numbers. According to Courant and Robbins in "What is Mathematics": This revelation was a scientific event of the highest importance. Quite possibly it marked the origin of what we consider the specifically Greek contribution to rigorous procedure in mathematics. Certainly it has profoundly affected mathematics and philosophy from the time of the Greeks to the present day.

Specifically, the Greeks discovered that the diagonal of a square whose sides are 1 unit long has a diagonal whose length cannot be rational. By the Pythagorean Theorem, the length of the diagonal equals the square root of 2. So the square root of 2 is irrational!

The following proof is a classic example of a proof by contradiction: We want to show that A is true, so we assume it's not, and come to contradiction. Thus A must be true since there are no contradictions in mathematics!


3.2: Proofs - Mathematics

In this section we are going to prove some of the basic properties and facts about limits that we saw in the Limits chapter. Before proceeding with any of the proofs we should note that many of the proofs use the precise definition of the limit and it is assumed that not only have you read that section but that you have a fairly good feel for doing that kind of proof. If you’re not very comfortable using the definition of the limit to prove limits you’ll find many of the proofs in this section difficult to follow.

The proofs that we’ll be doing here will not be quite as detailed as those in the precise definition of the limit section. The “proofs” that we did in that section first did some work to get a guess for the (delta )and then we verified the guess. The reality is that often the work to get the guess is not shown and the guess for (delta )is just written down and then verified. For the proofs in this section where a (delta ) is actually chosen we’ll do it that way. To make matters worse, in some of the proofs in this section work very differently from those that were in the limit definition section.

So, with that out of the way, let’s get to the proofs.

Limit Properties

In the Limit Properties section we gave several properties of limits. We’ll prove most of them here. First, let’s recall the properties here so we have them in front of us. We’ll also be making a small change to the notation to make the proofs go a little easier. Here are the properties for reference purposes.

Assume that (mathop limits_ fleft( x ight) = K) and (mathop limits_ gleft( x ight) = L) exist and that (c) is any constant. Then,

Note that we added values ((K), (L), etc.) to each of the limits to make the proofs much easier. In these proofs we’ll be using the fact that we know (mathop limits_ fleft( x ight) = K) and (mathop limits_ gleft( x ight) = L) we’ll use the definition of the limit to make a statement about (left| ight|) and (left| ight|) which will then be used to prove what we actually want to prove. When you see these statements do not worry too much about why we chose them as we did. The reason will become apparent once the proof is done.

Also, we’re not going to be doing the proofs in the order they are written above. Some of the proofs will be easier if we’ve got some of the others proved first.

Proof of 7

This is a very simple proof. To make the notation a little clearer let’s define the function (fleft( x ight) = c) then what we’re being asked to prove is that (mathop limits_ fleft( x ight) = c). So let’s do that.

Let (varepsilon > 0) and we need to show that we can find a (delta > 0) so that

The left inequality is trivially satisfied for any (x) however because we defined (fleft( x ight) = c). So simply choose (delta > 0) to be any number you want (you generally can’t do this with these proofs). Then,

Proof of 1

There are several ways to prove this part. If you accept 3 And 7 then all you need to do is let (gleft( x ight) = c) and then this is a direct result of 3 and 7. However, we’d like to do a more rigorous mathematical proof. So here is that proof.

First, note that if (c = 0) then (cfleft( x ight) = 0) and so,

[mathop limits_ left[ <0fleft( x ight)> ight] = mathop limits_ 0 = 0 = 0fleft( x ight)]

The limit evaluation is a special case of 7 (with (c = 0)) which we just proved Therefore we know 1 is true for (c = 0) and so we can assume that (c e 0) for the remainder of this proof.

Let (varepsilon > 0) then because (mathop limits_ fleft( x ight) = K) by the definition of the limit there is a (> > 0) such that,

Now choose (delta = >) and we need to show that

and we’ll be done. So, assume that (0 < left| ight| < delta )and then,

Proof of 2

Note that we’ll need something called the triangle inequality in this proof. The triangle inequality states that,

[left| ight| le left| a ight| + left| b ight|]

We’ll be doing this proof in two parts. First let’s prove (mathop limits_ left[ ight] = K + L).

Let (varepsilon > 0) then because (mathop limits_ fleft( x ight) = K) and (mathop limits_ gleft( x ight) = L) there is a (> > 0) and a ( > 0) such that,

Now choose (delta = min left< <>,>> ight>). Then we need to show that

Assume that (0 < left| ight| < delta ). We then have,

In the third step we used the fact that, by our choice of (delta ), we also have (0 < left| ight| < >) and (0 < left| ight| < >) and so we can use the initial statements in our proof.

Next, we need to prove (mathop limits_ left[ ight] = K - L). We could do a similar proof as we did above for the sum of two functions. However, we might as well take advantage of the fact that we’ve proven this for a sum and that we’ve also proven 1.

Proof of 3

This one is a little tricky. First, let’s note that because (mathop limits_ fleft( x ight) = K) and (mathop limits_ gleft( x ight) = L) we can use 2 and 7 to prove the following two limits.

[eginmathop limits_ left[ ight] & = mathop limits_ fleft( x ight) - mathop limits_ K = K - K = 0 mathop limits_ left[ ight] & = mathop limits_ gleft( x ight) - mathop limits_ L = L - L = 0end]

Now, let (varepsilon > 0). Then there is a (> > 0) and a ( > 0) such that,

Choose (delta = min left< <>,>> ight>). If (0 < left| ight| < delta ) we then get,

So, we’ve managed to prove that,

This apparently has nothing to do with what we actually want to prove, but as you’ll see in a bit it is needed.

Before launching into the actual proof of 3 let’s do a little Algebra. First, expand the following product.

[left[ ight]left[ ight] = fleft( x ight)gleft( x ight) - Lfleft( x ight) - Kgleft( x ight) + KL]

Rearranging this gives the following way to write the product of the two functions.

[fleft( x ight)gleft( x ight) = left[ ight]left[ ight] + Lfleft( x ight) + Kgleft( x ight) - KL]

With this we can now proceed with the proof of 3.

Fairly simple proof really, once you see all the steps that you have to take before you even start. The second step made multiple uses of property 2. In the third step we used the limit we initially proved. In the fourth step we used properties 1 and 7. Finally, we just did some simplification.

Proof of 4

This one is also a little tricky. First, we’ll start of by proving,

Let (varepsilon > 0) . We’ll not need this right away, but these proofs always start off with this statement. Now, because (mathop limits_ gleft( x ight) = L) there is a (> > 0) such that,

Now, assuming that (0 < left| ight| < >) we have,

Now, there is also a ( > 0) such that,

Choose (delta = min left< <>,>> ight>). If (0 < left| ight| < delta ) we have,

Now that we’ve proven (mathop limits_ frac<1><> = frac<1>) the more general fact is easy.

Proof of 5. for (n) an integer

As noted we’re only going to prove 5 for integer exponents. This will also involve proof by induction so if you aren’t familiar with induction proofs you can skip this proof.

For (n = 2) we have nothing more than a special case of property 3.

[mathop limits_ ight]^2> = mathop limits_ fleft( x ight)fleft( x ight) = mathop limits_ fleft( x ight)mathop limits_ fleft( x ight) = KK = ]

So, 5 is proven for (n = 2). Now assume that 5 is true for (n - 1), or (mathop limits_ ight]^> = <>>). Then, again using property 3 we have,

Proof of 6

As pointed out in the Limit Properties section this is nothing more than a special case of the full version of 5 and the proof is given there and so is the proof is not give here.

Proof of 8

This is a simple proof. If we define (fleft( x ight) = x) to make the notation a little easier, we’re being asked to prove that (mathop limits_ fleft( x ight) = a).

Let (varepsilon > 0) and let (delta = varepsilon ). Then, if (0 < left| ight| < delta = varepsilon ) we have,

So, we’ve proved that (mathop limits_ x = a).

Proof of 9

This is just a special case of property 5 with (fleft( x ight) = x) and so we won’t prove it here.

Facts, Infinite Limits

Given the functions (fleft( x ight)) and (gleft( x ight)) suppose we have,

[mathop limits_ fleft( x ight) = infty hspace<0.5in>hspace<0.25in>mathop limits_ gleft( x ight) = L]

for some real numbers (c) and (L). Then,

Partial Proof of 1

We will prove (mathop limits_ left[ ight] = infty ) here. The proof of (mathop limits_ left[ ight] = infty ) is nearly identical and is left to you.

Let (M > 0) then because we know (mathop limits_ fleft( x ight) = infty ) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Also, because we know (mathop limits_ gleft( x ight) = L) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

[0 < left| ight| < 1hspace<0.25in>,, o hspace <0.25in>- 1 < gleft( x ight) - L < 1hspace <0.25in> o ,,,,,,,L - 1 < gleft( x ight) < L + 1]

Now, let (delta = min left< <>,>> ight>) and so if (0 < left| ight| < delta ) we know from the above statements that we will have both,

[fleft( x ight) > M - L + 1hspace<0.5in>hspace<0.25in>gleft( x ight) > L - 1]

[eginfleft( x ight) + gleft( x ight) & > M - L + 1 + L - 1 & = Mhspace<0.5in>hspace <0.25in>Rightarrow hspace<0.5in>fleft( x ight) + gleft( x ight) > Mend]

Proof of 2

Let (M > 0) then because we know (mathop limits_ fleft( x ight) = infty ) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Also, because we know (mathop limits_ gleft( x ight) = L) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Note that because we know that (L > 0) choosing (frac<2>) in the first inequality above is a valid choice because it will also be positive as required by the definition of the limit.

Now, let (delta = min left< <>,>> ight>) and so if (0 < left| ight| < delta ) we know from the above statements that we will have both,

[eginfleft( x ight)gleft( x ight) & > left( >> ight)left( <2>> ight) & = Mhspace<0.5in>hspace <0.25in>Rightarrow hspace<0.5in>fleft( x ight)gleft( x ight) > Mend]

So, we’ve proved that (mathop limits_ fleft( x ight)gleft( x ight) = infty ).

Proof of 3

Let (M > 0) then because we know (mathop limits_ fleft( x ight) = infty ) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Note that because (L < 0) in the case we will have (frac<< - 2M>> > 0) here.

Next, because we know (mathop limits_ gleft( x ight) = L) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Again, because we know that (L < 0) we will have ( - frac <2>> 0). Also, for reasons that will shortly be apparent, multiply the final inequality by a minus sign to get,

Now, let (delta = min left< <>,>> ight>) and so if (0 < left| ight| < delta ) we know from the above statements that we will have both,

[fleft( x ight) > frac<< - 2M>>hspace<0.5in>hspace <0.25in>- gleft( x ight) > - frac<2>]

This may seem to not be what we needed however multiplying this by a minus sign gives,

[fleft( x ight)gleft( x ight) < - M]

and because we originally chose (M > 0) we have now proven that (mathop limits_ fleft( x ight)gleft( x ight) = - infty ).

Proof of 4

We’ll need to do this in three cases. Let’s start with the easiest case.

Case 1 : (L = 0)

Let (varepsilon > 0) then because we know (mathop limits_ fleft( x ight) = infty ) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Next, because we know (mathop limits_ gleft( x ight) = 0) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Now, let (delta = min left< <>,>> ight>) and so if (0 < left| ight| < delta ) we know from the above statements that we will have both,

In the second step we could remove the absolute value bars from (fleft( x ight)) because we know it is positive.

Case 2 : (L > 0)

Let (varepsilon > 0) then because we know (mathop limits_ fleft( x ight) = infty ) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Next, because we know (mathop limits_ gleft( x ight) = L) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Also, because we are assuming that (L > 0) it is safe to assume that for (0 < left| ight| < >) we have (gleft( x ight) > 0).

Now, let (delta = min left< <>,>> ight>) and so if (0 < left| ight| < delta ) we know from the above statements that we will have both,

In the second step we could remove the absolute value bars because we know or can safely assume (as noted above) that both functions were positive.

Case 3 : (L < 0).

Let (varepsilon > 0) then because we know (mathop limits_ fleft( x ight) = infty ) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Next, because we know (mathop limits_ gleft( x ight) = L) there exists a (> > 0) such that if (0 < left| ight| < >) we have,

Next, multiply this be a negative sign to get,

Also, because we are assuming that (L < 0) it is safe to assume that for (0 < left| ight| < >) we have (gleft( x ight) < 0).

Now, let (delta = min left< <>,>> ight>) and so if (0 < left| ight| < delta ) we know from the above statements that we will have both,

In the second step we could remove the absolute value bars by adding in the negative because we know that (fleft( x ight) > 0) and can safely assume that (gleft( x ight) < 0) (as noted above).

Fact 1, Limits At Infinity, Part 1

Proof of 1

This is actually a fairly simple proof but we’ll need to do three separate cases.

Case 1 : Assume that (c > 0). Next, let (varepsilon > 0) and define

Note that because (c) and (varepsilon ) are both positive we know that this root will exist. Now, assume that we have

Given this assumption we have,

So, provided (c > 0) we’ve proven that (mathop limits_ frac<<>> = 0).

Case 2 : Assume that (c = 0). Here all we need to do is the following,

Case 3 : Finally, assume that (c < 0). In this case we can then write (c = - k) where (k > 0). Then using Case 1 and the fact that we can factor constants out of a limit we get,

Proof of 2

This is very similar to the proof of 1 so we’ll just do the first case (as it’s the hardest) and leave the other two cases up to you to prove.

Case 1 : Assume that (c > 0). Next, let (varepsilon > 0) and define

Note that because (c) and (varepsilon ) are both positive we know that this root will exist. Now, assume that we have

Note that this assumption also tells us that (x) will be negative. Give this assumption we have,

So, provided (c > 0) we’ve proven that (mathop limits_ frac<<>> = 0). Note that the main difference here is that we need to take the absolute value first to deal with the minus sign. Because both sides are negative we know that when we take the absolute value of both sides the direction of the inequality will have to switch as well.

Case 2, Case 3 : As noted above these are identical to the proof of the corresponding cases in the first proof and so are omitted here.

Fact 2, Limits At Infinity, Part I

If (pleft( x ight) = + <>><>> + cdots + x + ) is a polynomial of degree (n) (i.e. ( e 0)) then,

Proof of (mathop limits_ pleft( x ight) = mathop limits_ )

We’re going to prove this in an identical fashion to the problems that we worked in this section involving polynomials. We’ll first factor out () from the polynomial and then make a giant use of Fact 1 (which we just proved above) and the basic properties of limits.

Now, clearly the limit of the second term is one and the limit of the first term will be either (infty ) or ( - infty ) depending upon the sign of (). Therefore by the Facts from the Infinite Limits section we can see that the limit of the whole polynomial will be the same as the limit of the first term or,

Proof of (mathop limits_ pleft( x ight) = mathop limits_ )

The proof of this part is literally identical to the proof of the first part, with the exception that all (infty )’s are changed to ( - ,infty ), and so is omitted here.

Fact 2, Continuity

If (fleft( x ight)) is continuous at (x = b) and (mathop limits_ gleft( x ight) = b) then,

[mathop limits_ fleft( ight) = fleft( limits_ gleft( x ight)> ight) = fleft( b ight)]

Proof

Let (varepsilon > 0) then we need to show that there is a (delta > 0) such that,

Let’s start with the fact that (fleft( x ight)) is continuous at (x = b). Recall that this means that (mathop limits_ fleft( x ight) = fleft( b ight)) and so there must be a (> > 0) so that,

Now, let’s recall that (mathop limits_ gleft( x ight) = b). This means that there must be a (delta > 0) so that,

But all this means that we’re done.

Let’s summarize up. First assume that (0 < left| ight| < delta ). This then tells us that,

But, we also know that if (0 < left| ight| < >) then we must also have (left| ight| < varepsilon ). What this is telling us is that if a number is within a distance of (>) of (b) then we can plug that number into (fleft( x ight)) and we’ll be within a distance of (varepsilon ) of (fleft( b ight)).

So, [left| ight| < >] is telling us that (gleft( x ight)) is within a distance of (>) of (b) and so if we plug it into (fleft( x ight)) we’ll get,


3.2: Proofs - Mathematics

Texts :
Understanding Analysis (second edition) by Stephen Abbott.

Materials to be covered: I plan to cover chapters 1 to 6 of the book.

Exam 1 , Friday, October 12, noon-1:20pm in the class (LSH-B269 (LIV)).
You need to know the proofs of the following theorems for the exam: Lemma 1.3.8, Thm 1.4.1, Thm 1.4.2, Thm 1.4.3, Thm 1.4.5, Thm 1.5.6, Thm 1.5.7, Thm 1.5.8, Thm 2.3.2, Thm 2.3.3, Thm 2.3.4, Thm 2.4.2, Thm 2.5.5, Thm 2.6.2, Thm 2.6.4, Thm 2.7.2, Thm 2.7.3, Thm 2.7.4, Thm 2.7.6, Thm 2.7.7, Thm 2.7.10.

Exam 2 , Friday, November 16, 10:20-11:40am, in the class (BE 011 (LIV)).
You need to know the proofs of the following theorems for the exam: Thm 3.2.3, Thm 3.2.5, Thm 3.2.12, Thm 3.2.13, Thm 3.3.4, Thm 3.3.5, Thm 3.3.8, Thm 3.4.6, Thm 3.4.7, Thm 4.2.3, Corollary 4.2.5, Corollary 4.3.3, Thm 4.3.4, Thm 4.3.9, Thm 4.4.1, Thm 4.4.2, Thm 4.4.5, Thm 4.4.7, Thm 4.5.1, Thm 4.5.2.
You need to be familiar and know all the definitions we went over in the class.

Final Exam: December 20, 8-11am, LSH-B269 You need to know the proofs of the following theorems for the exam: 1.4.1, 1.4.3, 1.5.6, 1.5.8, 2.3.3, 2.4.2, 2.5.5, 2.6.4, 2.7.6, 2.7.7, 2.7.10, 3.2.3, 3.2.12, 3.2.13, 3.3.4, 3.3.8, 3.4.7, 4.2.3, 4.3.4, 4.3.9, 4.4.1, 4.4.7, 4.5.2, 5.2.4, 5.2.5, 5.2.6, 5.2.7, 5.3.1, 5.3.2, 5.3.5, 5.3.6, 5.3.8, 6.2.6, 6.3.1, 6.3.2
You need to be familiar and know all the definitions we went over in the class.

Exams : There will be two closed book midterm exams tentatively scheduled for the above dates. You are not allowed to use a calculator during the midterm and final exams. The final exam will be cumulative (covers all material). It will count for 35 % of your final grade.

Homework : Homeworks will be assigned weekly. Two lowest homework grades will be dropped and therefore no late homeworks will be accepted. Required homework assignments will be found here.
Homework is due on Fridays in the class.
The first homework is due on Friday, September 21st. The two lowest homework grades will be dropped.


Abel's Proof: A Gentle Introduction to Mathematics

Mathematics is not something that we normally associate with sublime beauty. The way mathematics is taught, it is a subject that many if not most people would prefer to avoid. Just open any Dover math book that you can find your local book store and you may be hard pressed to find any passage that makes any sense. With its barrage of definitions, lemmas, theorems, and QED&aposs, it is a wonder that it claims to be written in English.

And yet, if you learned the details behind all those proofs. If you start being able to distinguish between textbook lemmas and the good stuff, there is a surprising excitement that emerges. It is like looking at a Picasso, getting annoyed by the abstraction, and then suddenly realizing that the artwork represents a maximum of effect with a minimum of detail. In other words, it is elegance of the highest level.

In today&aposs entry, I will walk you through one of the highest achievements of mathematics: Abel&aposs Impossibility Proof. I will work hard to make the journey pleasant and gentle. If you have any questions, feel free to post them to this hub. To provide the context for the proof, it is necessary to explore that some important ideas from the past.

The Ishango Bone discovered in 1960 in Zaire

A Very Short History of Math

All human societies have words for simple numbers. It takes deep mathematical insight to realize that 0 is also a number. With this realization, it becomes possible to represent (or at least to approximate) all numerical values with a finite set of digits. We call 0 to 9, the Arabic numerals but they were pretty much invented in India (in fact, they should be called Hindu-Arabic numerals). They were later popularized by the great Arab mathematicians (the word algebra after all is Arabic in origin). The Greeks and Romans had a plenitude of numbers and this made even basic multiplication almost impossible. The Arabic numerals greatly changed the possibilities for doing advanced mathematics.

Another important innovation was the invention of algebraic notation where we represent an unknown value by a letter such as x. The ability to state mathematical relationships as formulas greatly improves the ability to introduce abstractions. Consider the state of algebra before the use of 0 and the use of algebraic notation. The ancient Greek mathematician Diophantus (around 200AD) is often considered the father of algebra. He wrote a classic book (the translated book is available online here) where he asked questions about unknowns values. Since he didn&apost have the arabic numerals and the Greeks were partial to geometry, he asked these questions using geometric language. For example, here&aposs a problem that he posed: find three squares where one is equal to the sum of the other two.

Diophantus&aposs question becomes easier to solve if we use algebraic notation. Just as 0 enabled us to simplify the amount of numbers we need, algebraic notation enables us to restate mathematical questions in a simpler way.

First, we can treat unknown values as numbers. We can call them x,y,z. The first person to do this was Muhammad ibn Musa al-Khwarizmi (780 - 850 AD) in his famous book Al-Jabr (Although the full name in English is The Compendious Book on Calculations by Completion and Balancing). Al-Khwarizmi is also considered by many to be the father of Algebra. In addition to x,y,z, he used notation for exponents. An unknown square becomes x 2 . If you would like a gentle introduction to powers and exponents, review here before continuing.

So, now, armed with letters for unknowns and notation for exponents, we can solve Diophantus&aposs problem. "Find three suqare where one is equal to the sum of the other two" can be restated as find x,y,z such that x 2 + y 2 = z 2 . (By the way, the answer is 3,4,5 since 9 + 16 = 25.

Any expression like x 2 + 3x + 2 is called a polynomial. Each grouping of x and a number is called a monomial. When you match a polynomial with an equal sign and another value, you get a mathematical equation.

Al-Khwarizmi on a Soviet Union stamp issued on September 6, 1983

Using Algebraic Notation

With this notation, it now became possible for mathematics to take off. Al-Khwarizmi showed the world how to solve what are known as the quadratic equations. These are equation of one unknown value where no power is greater than 2. A typical example is: 5x 2 + 3x + 2 = 10. Find x (the answer in this case is x=1). Al-Khwarizmi showed that there is a formula that solves all possible quadratic equations.

If you raise the power to 3 or 4, the problem is significantly harder. Among Western mathematicians, it was not until the very famous Italian mathematician Girolamo Cardano (1501 - 1576 AD) that a solution is found for equations of one unknown where no power is greater than 3. This equation, known as the cubic equation, turns out to be more difficult than the quadratic. But its solution led very quickly to the solution of quartic equations by Lodovico Ferrari (1522 - 1565 AD). Quartic equations are equations of one unknown value where no power is greater than 4. For mathematicians, these were very exciting times.

But if there&aposs a general solution for equations with a power of 2, power of 3, and power 4, shouldn&apost there also be a general solution for equations with a power of 5? Despite the efforts of great geniuses of mathematics (Francois Viete, Rene Descartes, Isaac Newton, Leonhard Euler, Joseph Lagrange, and Carl Friedrich Gauss), no general solution to the 5th power (known as the quintic) was found.

The Reason It&aposs So Hard: It Can&apost Be Done

In 1799, an Italian medical doctor trained in mathematics proposed that quintic equations and any equations with powers greater than 5 had no general solution. Paolo Ruffini (1765 - 1822 AD) presented his argument in a book which he sent to all the famous mathematicians of his day. Not surprisingly, mathematicians were not ready to believe that the problem was impossible. They ignored his argument, his methods, and continued to look for a general solution.

In retrospect, poor Ruffini did have a small gap in his reasoning but other than this small gap which would later be filled by Niels Abel, his argument was sound. He was an outsider but he was right and no one listened. In fairness to the mathematicians, his book was long and used nonstandard terms.

The foolproof that there is no general solution to the quintic equation would have to await the entry of a young genius from Norway named Niels Abel (1802 - 1829 AD). Independently of Ruffini, he came up with his own proof and this one was eventually accepted by the mathematical community. Unfortunately for Abel, he died before he received the full credit that he deserved. Today, one of the highest prizes a person achieve in mathematics is called the Abel Prize.

OK: Now, We Are Ready to talk about Abel&aposs Insight

Using algebraic notation, Abel&aposs proof says that there is no general solution to an equation such as: 5x 5 + 4x 4 + 3x 3 + 2x 2 + x = 15. Yes, we can figure out that the answer is x=1 but we can&apost do it in the same way that we can for the quadratic equation. If our equation is x 2 + x + 1 = 3, then we can solve it using Al-Kwarizmi&aposs solution to the quadratic equation.

To understand the solution to the quadratic equation, we also need to know about square roots. A square root is just the inverse of a square. If 3 2 = 9, then sqrt(9) = 3. We can speak about square roots, cube roots, quadriatic roots, etc. A root is just the inverse of an exponent. Mathematicians call all these roots "radicals" or "nth roots". For purposes of my discussion today, I will stick to radicals.

Now, with the idea of square roots, we can state the solution to all quadratic equations. For any equation ax 2 + bx + c = 0, the solution is:


3.2: Proofs - Mathematics

Via StumbleUpon, I came across this short text page which lists three mathematical ‘proofs’ which seem to violate common sense, listed below. The first is:

Each of these proofs is (intentionally) wrong! They highlight classic fallacies in mathematical thinking. See if you can figure out where, in each of them, the proof goes wrong, and then look for the answers below the fold…

(Note: the third proof involves the ‘imaginary number’ . If you’re not familiar with it, you can safely skip that problem, as it is closely related to one of the others.)

The mistake in the first proof is highlighted in red below:

The mistake lies in the assumption that we can divide by . Why is this a problem? Because, according to the equation we started with, ! We are not allowed to divide by zero, and therefore the result highlighted in red is invalid.

The algebra ‘muddies the waters’ significantly, and it can be helpful to see the same proof, but with . The first two steps suggest that . The third step suggests that . The fourth step suggests that . We run into problems in the next step, in trying to divide both sides of this equation by zero!

The next ‘proof’ is a little more subtle. Again, we highlight the problem step in red:

The problem here is that, in order to get to the step in red, there is an implicit square root taken of both sides of the equation. However, there are potentially two roots to any square e.g. has possible solutions or . The ‘proof’ above assumes that the positive root is correct, which leads to the erroneous answer . If one looks at the negative root, one finds that , which leads right back to the starting equation . This proof, incidentally, might have been helpful to legislators in Indiana back in the late 1900s 1800s.

The third ‘proof’ suffers from essentially the same problem as the second, except that complex numbers are involved:

The step highlighted in red assumes that we break up the square roots as , and then take individually the positive root of each term. It is wrong, however, to assume that the positive roots are always the correct ones when dealing with the square root of an equation. If we have an equation of the form

we can at best say that , and not necessarily both are true. Nothing changes if we take the square root of an equation of the form

in which case we can at best conclude that

For our ‘proof’ above, we find that the statement in red should be replaced with “ or , but not necessarily both.”

There’s a nice classic book that covers such mathematical errors, and other more important mathematical paradoxes: Bryan Bunch, Mathematical Fallacies and Paradoxes (Dover, New York, 1997).

If you really want to be certain that you understand such mistakes, try writing your own ‘proof’ for !Feel free to post yours in the comments. I’ll add my own ‘proof’ at a later date!


3.2: Proofs - Mathematics

We need to argue two things. First, we need to show that $q$ and $r$ exist. Then, we need to show that $q$ and $r$ are unique.

To show that $q$ and $r$ exist, let us play around with a specific example first to get an idea of what might be involved, and then attempt to argue the general case.

Recall that if $b$ is positive, the remainder of the division of $b$ by $a$ is the result of subtracting as many $a

How to write a two column proof?

So what should we keep in mind when tackling two-column proofs?

Always start with the given information and whatever you are asked to prove or show will be the last line in your proof, as highlighted in the above example for steps 1 and 5, respectively.

Remember when you are presented with a word problem it’s imperative to write down what you know, as it helps to jumpstart your brain and gives you ideas as to where you need to end up?

The same thing is true for proofs.

Start with what you know (i.e., given) and this will help to organize your statements and lead you to what you are trying to verify.

Sometimes it is easier to first write down the statements first, and then go back and fill in the reasons after the fact. Other times, you will simply write statements and reasons simultaneously. There is no one-set method for proofs, just as there is no set length or order of the statements.

As long as the statements and reasons make logical sense, and you have provided a reason for every statement, as ck-12 accurately states. As seen in the above example, for every action performed on the left-hand side there is a property provided on the right-hand side. These steps and accompanying reasons make for a successful proof.

Proofs take practice! The more your attempt them, and the more you read and work through examples the better you will become at writing them yourself.

Additionally, it’s important to know your definitions, properties, postulates, and theorems.

Consequently, I highly recommend that you keep a list of known definitions, properties, postulates, and theorems and have it with you as you work through these proofs. Again, the more you practice, the easier they will become, and the less you will need to rely upon your list of known theorems and definitions.

In the video below, we will look at seven examples, and begin our journey into the exciting world of geometry proofs.


Divisibility by 7 and Its Proof

This is the 6th post in the Divisibility Rules Series. In this post, we discuss divisibility by 7.

Simple steps are needed to check if a number is divisible by 7. First, multiply the rightmost (unit) digit by 2 , and then subtract the product from the remaining digits. If the difference is divisible by 7, then the number is divisible by 7.

Example 1 : Is 62 3 divisible by 7?

3 x 2 = 6
62 – 6 = 56
56 is divisible by 7, so 623 is divisible by 7.

If after the process above, the number is still large, and it is difficult if to know if it is divisible by 7, the steps can be repeated. We take the difference as the new number, we multiply the rightmost digit by 2, and then subtract from the remaining digits.

Example 2 : Is 342 3 divisible by 7?

We repeat the process for 33 6. We multiply 6 by 2 and then subtract it from 33 .

6 x 2 = 12
33 – 12 = 21
21 is divisible by 7, so 3423 is divisible by 7.

Note that if the number is still large, this process can be repeated over and over again, until it is possible to determine if the remaining digits is divisible by 7.

Delving Deeper (for the adventurous)

The following portion are for students who have basic knowledge on proofs. In particular, we will be proving an if and only if statement. A if and only if B requires to prove that A implies B and B implies A.

Let be the number that we want divide by 7. Let be the unit’s digit and be the rest of the digit. Then N = 10 a + b .

Explanation: All whole numbers N can be expressed as the product of 10 and a number added to its units digit. For example 983 = 10( 98 ) + 3, 5896 = 10( 598 ) + 6 , and so on.

We assign the following statements to A and B.

A: a – 2 b is divisible by 7.
B: N is divisible by 7.

As we have mentioned above, we have to show that (1) A implies B and (2) B implies A. This means that we have to show that if is divisible by , then is divisible by . The statement is the step where we multiplied the unit’s digit by 2, and then subtracted from the remaining digits .

For (1) We have to show that A implies B. That is, we have to show that if is divisible by , then is divisible by .

If is divisible by , then we can find a natural number such that (Can you see why?).

Multiply both sides by , we have . Adding on both sides, we have . Now, . Notice that the left hand side of our equation is and the right hand side can be divided by . Therefore, is divisible by . That proves our first statement that If is divisible by , is divisible by .#

For (2), we have to show that B implies A. That is, we have to show that if is divisible by 7, is divisible by .

If is divisible by , then is divisible by . This means we can find a natural number such that . Subtracting from both sides, we have . This means that . Factoring, we have

Now, since is not divisible by , is divisible by . This proves the second statement if is divisible by , then is divisible by #

From above, we have shown that A implies B and B implies A. We have shown that the process that we have done above will hold for all cases.


Watch the video: Class 11th maths ncert in hindi exercise (December 2021).