9

Prove that $\log_{2}(7)$ is an irrational number.

My Attempt:

Assume that $\log_{2}(7)$ is a rational number. Then we can express $\log_{2}(7)$ as $\frac{p}{q}$, where $p,q\in \mathbb{Z}$ and $q\neq 0$. This implies that $7^q = 2^p$, where either $p,q>0$ or $p,q<0$.

My question is this: why can't we count for $p,q<0$? My textbook's author counts only for $p,q>0$. Could someone explain the reasoning being used by the author?

Fabrosi
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juantheron
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3 Answers3

6

Since $\log_2(7) > 0$, we know that $p$ and $q$ both have the same sign. Therefore both can be taken as positive.

Neal
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  • But Neal why we can not count for $p,q<0$ bcz here $p$ and $q$ both have same sign.Thanks – juantheron Sep 12 '13 at 18:51
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    @juantheron: If $p$ and $q$ are both negative, then consider $\frac{-p}{-q}$ (which is the same rational) instead, and continue in the case where numerator and denominator are both positive. – hmakholm left over Monica Sep 12 '13 at 18:55
4

Consider the case $\log_27=p/q$ where $p, q<0$. You'll still have $$ 7^q=2^p $$ and thus you'll have $$ \frac{1}{7^{-q}}=\frac{1}{2^{-p}} $$ which is the same as saying $$ 7^{-q}=2^{-p} $$ and since $-q$ and $-p$ are both positive, you're back in the case where both exponents are positive.

Rick Decker
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1

Saying what Neal has already said a bit different: First you realize that $\log_2(7) > 0$. Then if $\log_2(7)$ is rational you can find positive integers $p$ and $q$ such that $$\log_2(7) = \frac{p}{q}.$$ You could, of course, also have chosen $p$ and $q$ both negative, but you can definitely pick then both positive. So then, as you note, $7^q = 2^p$. Now since both $7$ and $2$ are prime, you know that this will never happen when $p$ and $q$ are positive integers.

Thomas
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