I'm an English major, now doubling in computer science. The first course I'm taking is Discrete Mathematics for Computer Science, using the MIT 6.042 textbook.

Within the first chapter of the book's practice problems, they ask us multiple times to prove that some log function is either rational or irrational.

Specific cases make more sense than others, but I would really appreciate any advice on how to approach these problems. Not how to carry them out algebraically, but what thought constructs are necessary to consider a log being (ir)rational.

For example, in the case of $\sqrt{2}^{2\log_2 3}$, proving that $2\log_23$ is irrational (and therefore $a^b$, when $a=\sqrt{2}$ and $b=2\log_23$, is rational) is not an easily solvable problem. I understand the methods of proofs, but the rules of logs are not intuitive to me.

A section from my TF's solution is not something I would know myself to construct:

Since $2 < 3$, we know that $\log_23$ is positive (specifically it is greater than $1$), and hence so is $2\log_23$. Therefore, we can assume that $a$ and $b$ are two positive integers. Now $2\log_2 3=a/b$ implies $2^{2\log_2 3}=2^{a/b}$. Thus $$2^{a/b}=2^{2\log_2 3} = 2^{\log_2 3^2} =3^2 =9\text{,}$$ and hence $2^a = 9^b$.

Any advice on approaching thought construct to logs would be greatly appreciated!

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    $\begingroup$ It looks as if they may be talking about proving certain values of logarithm functions are irrational, but I'm not sure what it would mean to say that the log function itself is irrational. It does not make sense to say "$b$ is irrational; therefore $a^b$ is rational. No matter what positive number $a$ is, as long as it's not $1$, there will be irrational numbers $b$ such that $a^b$ is rational, and other irrational numbers $b$ such that $a^b$ is irrational. I wonder if you're actually conveying what it is they want you to do. $\endgroup$ – Michael Hardy Oct 26 '13 at 20:20
  • $\begingroup$ The task itself is stated as: Let a=sqrt(2), b=2log(base2)3. We know that sqrt(2) is irrational, and a^b=3. Finish the proof by showing that 2log(base2)3 is irrational. $\endgroup$ – greenso Oct 26 '13 at 20:25
  • $\begingroup$ You seem to have changed the values of $a$ and $b$ in midstream without mentioning it. Where you say "Therefore, we can assume that $a$ and $b$ are two positive integers", that doesn't make sense UNLESS at some point you've told us that $a/b$ is supposed to be $\log_2 9$ rather than what $a$ and $b$ were earlier in your question. $\endgroup$ – Michael Hardy Oct 26 '13 at 20:27
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    $\begingroup$ "Finish the proof" of WHAT? $\endgroup$ – Michael Hardy Oct 26 '13 at 20:28
  • $\begingroup$ Could you pin down a more specific question? "Advice on approaching thought construct to logs" is very broad and vague. $\endgroup$ – Jack M Aug 31 '14 at 23:11

To say that $\log_2 3$ is rational would mean that there are integers $m$, $n$ such that $\log_2 3=\dfrac mn$. That would imply that $2^{m/n}=3$, so that $2^m=3^n$. But that says an even number equals an odd number, which is impossible, so $\log_2 3$ cannot be rational.

But that doesn't help much in figuring out whether $2^{2\log_2 3}$ is rational. One can write $$ 2^{2\log_2 3} = \left(2^{\log_2 3}\right)^2 = 3^2 = 9,\tag1 $$ so that is rational. But in doing that you don't need to know anything at all about rational or irrational numbers until that final step where you observe that $9$ is rational. Line $(1)$ above is just application of standard rules governing exponents and logarithms.

  • $\begingroup$ You assume that $m$ and $n$ are integers. Then, when we get to the point that $2^m = 3^n$, you say that it says that even number = odd number, which is impossible. I agree. But what if we choose $a=0$ and $b=0$? ($0$ is an integer, right?). Then there's $0$ factors of $2$ on the left, and $0$ factors of $3$ on the right, that is, we get $1=1$, which is true, and your proof falls down. Am I right? $\endgroup$ – SasQ Jan 23 '15 at 14:29
  • $\begingroup$ @SasQ : Yes, for completeness perhaps we should add that $\log_2 3$ cannot be $0$, so the numerator cannot be $0$. But the non-zero nature of the denominator is already assumed when we suppose that $\log_2 3$ is rational, since rational numbers have non-zero denominators. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 24 '15 at 0:34
  • $\begingroup$ @SasQ : $\log_2 3$ certainly cannot be $0/0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 12 '15 at 15:51
  • $\begingroup$ Where did I say that $\log_2 3 = 0/0$? $\endgroup$ – SasQ Jun 13 '15 at 11:11

I came up with what I believe to be an original formulation of proving that $\log_ab $ is irrational.

If $\log_ab$ is rational, it is equal to some $a/b$, where $ a,b \in \mathbb{Z} $, where $ \mathbb{Z} $ is the set of integers.

If $\log_2 3=a/b$, then

Thus $$2^{a} =3^b = N\text{,}$$

Where $N$ is some number $ \in \mathbb{Z} $

Now, since 2 and 3 are both prime, if this N exists, it has a prime factorization of $2 \cdotp 2 \cdotp 2 \cdotp ... \cdotp 2 $ and also of $3\cdotp3\cdotp3\cdotp3\cdotp...\cdotp3 $

( $a$ number of 2's and $b$ number of 3's)

But any number $ N $ can only have one unique prime factorization, so no such $ a,b $ can exist.

This shows that any $\log_x y$ where $a, b \in \mathbb{Z}$ and $a, b $ are mutually prime is irrational, since $x^{a} =y^b $ cannot be true, and all prime numbers (including 2 and 3) are mutually prime by definition.

  • $\begingroup$ If $a$ is negative, then $2^a$ is not an integer. (The integers $\mathbb Z$ includes negative numbers, while $\mathbb N$ does not.) Maybe you should use $a,b\in\mathbb N$ instead, knowing that $\log_23>0$. $\endgroup$ – mr_e_man Oct 18 '18 at 4:01
  • $\begingroup$ Actually, the rational numbers also have unique prime factorizations, so $a$ and $b$ may be negative integers. Then $N$ is some number $\in\mathbb Q$, and your proof is still good. $\endgroup$ – mr_e_man Oct 18 '18 at 4:07

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