# 2 is greater than 3?

Consider the following situation. Be:

1/4 > 1/8

but this same inequality can be written in another way where the sign of inequality will be the same:

(1/2)2 > (1/2)3

Applying the logarithms to both members and since the logarithm is a growing function, that is, a larger number corresponds to a larger logarithm, we have:

log ((1/2)2)> log ((1/2)3) ,

so by the properties of logarithms we have:

2.log (1/2)> 3.log (1/2)

In conclusion, if we divide both members by log (1/2) we have:

2 > 3

It is evident that at first sight all reasoning is correct. But if we look closely, we find the flaw: When we divide by log (1/2), we are dividing by a negative value (-0.3010…), which reverses the sign of inequality (ie 2 < 3).

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