# Can you prove log_10 2  is irrational?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2

### This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Rhys Share
Dec 2, 2017

We can prove this by contradiction;

#### Explanation:

if $x = \mathbb{Q}$ then $x = \frac{p}{q}$ for $p , q \in \mathbb{Z}$

so hence let, ${\log}_{10} 2 = \frac{p}{q}$
then $2 = {10}^{\frac{p}{q}}$

so hence ${2}^{q} = {10}^{p}$

$\implies {2}^{q} = {5}^{p} {2}^{p}$

$\implies {2}^{q - p} = {5}^{p}$

we know ${\log}_{10} 1 < {\log}_{10} 2 < {\log}_{10} 10$

so hence $0 < {\log}_{10} 2 < 1$

so hence $0 < \frac{p}{q} < 1$

$p < q$

so $q - p > 0$ and let $m = q - p$

But we can see that ${2}^{m}$ is even and ${5}^{p}$ is odd, hence there are no values for $p \mathmr{and} q$ that satisify this

$\implies$ Hence proven by contradiction

• 9 minutes ago
• 10 minutes ago
• 12 minutes ago
• 15 minutes ago
• 34 seconds ago
• 2 minutes ago
• 3 minutes ago
• 4 minutes ago
• 4 minutes ago
• 8 minutes ago
• 9 minutes ago
• 10 minutes ago
• 12 minutes ago
• 15 minutes ago