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I need to prove that $\log_{10}{2}$ is irrational. I understand the way this proof was done using contradiction to show that the even LHS does not equal the odd RHS, but I did it a different way and wanted to check its validity!

Prove by contradiction: Suppose that $\log{2}$ is rational - that is, it can be written as $$\log{2} = \frac{a}{b}$$ where $a$ and $b$ are integers. Then

$$2 = 10^{\frac{a}{b}}$$

$$2 = 10^a10^{\frac{1}{b}}$$

$$\frac{2}{10^{a}} = 10^{\frac{1}{b}}$$

Log both sides:

$$\log(\frac{2}{10^{a}}) = \frac{1}{b}$$

$$\log{2} - \log(10^a) = \frac{1}{b}$$

$$\log{2} = \frac{1}{b} + a$$

$$\log{2} = \frac{ab+1}{b}$$

However we assumed that $\log(2)=\frac{a}{b}$ and thus we have a contradiction.

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    $\begingroup$ Your step $10^{\frac{a}{b}} = 10^a 10^{\frac{1}{b}}$ is incorrect. $\endgroup$ – Cameron Williams Dec 14 '14 at 5:19
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    $\begingroup$ I guess you really want to prove that it is irrational. Since it is. $\endgroup$ – MPW Dec 14 '14 at 5:19
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    $\begingroup$ To a computer scientist, $\log 2$ is rational! $\endgroup$ – David Richerby Dec 14 '14 at 11:58
  • $\begingroup$ @DavidRicherby why does it need to be rational? $\endgroup$ – Ooker Dec 14 '14 at 13:57
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    $\begingroup$ @Ooker Because computer scientists most often take base-2 logs, for example to calculate the height of a binary tree or for recurrence relations involving divide-and-conquer algorithms that chop the input in half and operate recursively on one or both of the halves (which is, essentially, looking at the height of a binary tree again). $\endgroup$ – David Richerby Dec 14 '14 at 14:12
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As has been pointed out in comments and in another answer, $10^{a/b}\neq 10^{a}10^{\frac{1}b}$. This is a rather subtle error, however there's a notable warning flag that could alert you to it: Your proof does not use the hypothesis that $a$ and $b$ are integers. This is a serious issue, because it means you've proved the (false) statement that $\log(2)$ cannot be written as a fraction $\frac{a}b$ - even if we let $a$ and $b$ be real, but: $$\frac{\log(2)}1=\log(2)$$

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A proof can be carried out after modifying your calculation a bit.

$$2=10^{\frac{a}{b}}\implies \color{blue}{2^b=10^a}\implies2^{b-a}=5^a$$

Which is a contradiction when both $a$ and $b$ are non-zero integers. Check the colored step carefully and you will understand in which step you have made a mistake.

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    $\begingroup$ From where did you get that $5^b$? I think I'll make it $5^a$. $\endgroup$ – aaaaaaaaaaaa Dec 14 '14 at 20:01
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Be careful, $10^{\frac{a}{b}}$ equals $(10^{a})^{\frac{1}{b}}$ not $10^{a}10^{\frac{1}{b}}$

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Others have already pointed out that your proof was wrong. A different way to see that your proof is wrong is as follows: if I would replace 2 in your proof everywhere by 10, I would get the result that log(10) is also irrational.

Any proof you have for the irrationality of log(2) should not work for log(10). (If you take 10 as the base of your logarithm.)

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The proof for irrationality of $\log 2$ can be done in few lines using elementary divisibility properties or natural numbers. Assume $a, b \in \mathbb{N}$, $\text{gcd}(a,b) = 1$, and $a < b$. Thus: $\log 2 = \dfrac{a}{b} \to 2 = 10^{\frac{a}{b}} \to 2^b = 10^a = 2^a\cdot 5^a \to 2^{b-a} = 5^a$. We see a contradiction here because the $LHS$ is even while the $RHS$ is odd.

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  • $\begingroup$ Who said that $b>a$? If $a>b$ then your LHS is not even integer. $\endgroup$ – Marc van Leeuwen Dec 14 '14 at 10:06
  • $\begingroup$ log(2) < 1 <=> b > a $\endgroup$ – Guillaume Poussel Dec 14 '14 at 12:11
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The equality $2=10^a10^{\frac{1}{b}}$ is not true, because $10^a10^{\frac{1}{b}}=10^{a+\frac{1}{b}}.$

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