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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.

Bart Michels
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Tminus2
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    Your step $10^{\frac{a}{b}} = 10^a 10^{\frac{1}{b}}$ is incorrect. – Cameron Williams Dec 14 '14 at 05:19
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    I guess you really want to prove that it is irrational. Since it is. – MPW Dec 14 '14 at 05:19
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    To a computer scientist, $\log 2$ is rational! – David Richerby Dec 14 '14 at 11:58
  • @DavidRicherby why does it need to be rational? – Ooker Dec 14 '14 at 13:57
  • @DavidRicherby is it because they often work with binary numbers? – Ooker Dec 14 '14 at 14:08
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    @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). – David Richerby Dec 14 '14 at 14:12
  • @DavidRicherby Ya, the question should specify that we're talking about base-10 here. – Elliot Gorokhovsky Dec 14 '14 at 23:31
  • @DavidRicherby By writing $\log_{10}{2}$. – Elliot Gorokhovsky Dec 15 '14 at 00:35
  • Duplicate of http://math.stackexchange.com/questions/656138/help-me-to-prove-that-log2-3-is-irrational (See Bill Dubuque's answer.) – Alan Dec 18 '14 at 01:52
  • @DavidRicherby: To a computer scientist, every number is rational, since this is how IEEE 754 standard for storing floating-point numbers works. It stores every number as a finite approximation by a binary fraction. There is no digital machine capable of storing an irrational number, and there never will. It is an inherent limitation of digital notation of numbers. – SasQ Jan 23 '15 at 13:51
  • @SasQ I'm a computer scientist and I most certainly don't consider every number to be rational. Computer science is not programming so the fact that programmers have to use rational approximations doesn't have a whole lot of bearing on computer science. Also, there are plenty of ways to represent irrational numbers algebraically: one can perfectly well compute over the field $\mathbb{Q}(\sqrt{2})$, for example. – David Richerby Jan 23 '15 at 17:28
  • @DavidRicherby: Blame the original poster, then, not me, since I just generalized his claim to show that $log 2$ isn't any more special than any other number being represented in a digital computer. – SasQ Jan 23 '15 at 18:29
  • Whoops, the original poster is you. Then are you going to contradict yourself now or something? I don't get it... – SasQ Jan 23 '15 at 18:30
  • @SasQ There is no contradiction. I already explained the joke: computer scientists think $\log 2$ is rational because they usually take base-$2$ logs. – David Richerby Jan 24 '15 at 22:25
  • Oh, so that was your case... I've never seen the binary logarithm being written this way. lg yes, but log without the base specified explicitly is usually meant to be decimal by default where I live. I have this problem with Wolfram Alpha, too, which uses log for the natural logarithm instead of ln I used do. These disagreements on notation could be very confusing sometimes... – SasQ Jan 26 '15 at 02:15

6 Answers6

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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: $$\frac{\log(2)}1=\log(2)$$

Milo Brandt
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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.

aaaaaaaaaaaa
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Be careful, $10^{\frac{a}{b}}$ equals $(10^{a})^{\frac{1}{b}}$ not $10^{a}10^{\frac{1}{b}}$

TheZeroTheorem
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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.

DeepSea
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The equality $2=10^a10^{\frac{1}{b}}$ is not true, because $10^a10^{\frac{1}{b}}=10^{a+\frac{1}{b}}.$

user201043
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