Question

Prove that log 10^2 is an irrational number

Answer 1

Question:

prove that

$\log_{10}(2)$

is irrational

Answer:

Assuming it as rational

log 2 = p/q

Where p and q are integers

Remove the logarithm

${10}^{ \frac{p}{q} } = 2$

${10}^{p} . {10}^{ \frac{1}{q} } = 2$

${10}^{ \frac{1}{q} } = \frac{2}{ {10}^{ p } }$

Apply logarithm on both sides

$\frac{1}{q} = \log( \frac{2}{ {10}^{p} } )$

Using

log (a/b) = log( a ) - log( b )

$\frac{1}{q} = \log(2) - \log( {10}^{p} ) \\ \\ \frac{1}{q} = \log(2) - p \log(10)$

Using

log_a (a) = 1

$\dfrac{1}{q} = \log(2) - p$

$\dfrac{1}{q} + p = \log(2)$

$\frac{qp + 1}{p} = \log(2)$

By contradiction log 2 is irrational

Answer 2

Answer:

irrational number

Step-by-step explanation:

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