# Math Help - Prrove common log of 3 is irrational

1. ## Prrove common log of 3 is irrational

This is what I've done so far:

Suppose it is rational, then log3 = a/b where a and b are in lowest terms. Hence 10^(a/b) = 3 which implies 10^a = 3^b which implies 2^a * 5^a = 3^b. This says that 2^a|3^b. Since 2|2^a it follows that 2|3^b which implies 2|3 because 2 is prime.

So what do I do from here? Can I just say I have a contradiction and thus log3 is irrational?

EDIT: Here's another one, thanks!

Prove that if r is a rational number such that r>1 and r /=/ 10^n for any positive integer n, then common log(r) is irrational.

Suppose logr is rational, then logr = a/b (a,b in lowest terms). This implies 10^(a/b) = r which implies that 10^a = r^b which implies 5^a * 2^a = r^b. This says that 2|r and 5|r, thus the only prime factors of r are 2 and 5. Hence 2^x * 5^y = r which implies 2^xb * 2^yb = r^b. Thus xb=a and yb=a which says x=y. Thus 2^x * 5^x = r, which says 10^x = r, a contradiction. Thus logr is irrational.

Is this right?

2. Hello, superevilcube!

I have a proof . . . I hope it's acceptable.

Prove that $\log_{10}(3)$ is irrational.

Assume $\log_{10}(3)$ is rational.

Then: . $\log_{10}(3) = \frac{a}{b}\;\;\text{ where }a\text{ and }b\text{ are positive integers}$

And we have: . $10^{\frac{a}{b}} \:=\:3$

Raise both sides to the power $b\!:\;\;10^a \:=\:3^b$

Since . $\begin{Bmatrix}10^a\text{ always ends in 0} \\3^b\text{ ends in 1, 3, 7, or 9}\end{Bmatrix}\quad\text{then: }\:10^a \neq 3^b \quad\hdots\;\text{a contradiction}$

Therefore, $\log_{10}(3)$ is irrational.

3. Originally Posted by superevilcube
10^a = 3^b which implies 2^a * 5^a = 3^b.
This is not true in general. 10^a is not equal to 2^a * 5^a for a>1.

4. Originally Posted by frankdent1
This is not true in general. 10^a is not equal to 2^a * 5^a for a>1.
How disappointing. To discover that the formula $(a \, b)^n = a^n \, b^n$ is wrong. After all these years of me believing in it. *sigh* Consigned to the same bin of broken dreams as Father Christmas and the Easter Bunny ...... I just can't believe it ...... Either $10^2 \neq 2^2 \, 5^2$ or 2 < 1 ..... I feel so ..... foolish

5. Originally Posted by mr fantastic
Consigned to the same bin of broken dreams as ... the Easter Bunny
Hey ... what's that about the Easter Bunny?