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Question Number 31596 by Nayon.Sm last updated on 11/Mar/18

Commented by Nayon.Sm last updated on 11/Mar/18
Proof that it is irrational
Commented by rahul 19 last updated on 11/Mar/18

$s u c h t y p e o f q u e s t i o n s a r e e a s i l y d o n e$ $b y c o n t r a d i c t i o n!$
	Answered by Joel578 last updated on 11/Mar/18

	$If \log_{2} 5 is rational, then there is number a and b$ $satisfy \log_{2} 5 = \frac{a}{b}, where a, b \in Z, b \neq 0$ $\log_{2} 5 = \frac{a}{b} \to 2^{\frac{a}{b}} = 5 \to 2^{a} = 5^{b}$ $There is no number that satisfy the equation$ $because LHS always even and RHS always odd$ $Hence, \log_{2} 5 is irrational$
	Answered by mrW2 last updated on 11/Mar/18

	$a s s u m e \log_{2} 5 i s r a t i o n a l, i . e . i t c a n$ $b e e x p r e s s e d a s$ $\log_{2} 5 = \frac{m}{n}$ $w h e r e m, n a r e i n t e g e r s a n d g c d (m, n) = 1 .$ $\Rightarrow 2^{\frac{m}{n}} = 5$ $\Rightarrow 2^{m} = 5^{n}$ $s i n c e 2 a n d 5 a r e c o - p r i m e,$ $\Rightarrow m = n = 0 .$ $i . e . \log_{2} 5 c a n n o t b e e x p r e s s e d a s \frac{m}{n},$ ${i t}^{'} s i r r a t i o n a l .$
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