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Question-31596

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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
such type of questions are easily done by contradiction !
$${such}\:{type}\:{of}\:{questions}\:{are}\:{easily}\:{done}\: \\ $$$${by}\:{contradiction}\:! \\ $$
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 = (a/b), where a, b ∈ Z, b ≠ 0 log_2 5 = (a/b) → 2^(a/b) = 5 → 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
$$\mathrm{If}\:\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:\mathrm{is}\:\mathrm{rational},\:\mathrm{then}\:\mathrm{there}\:\mathrm{is}\:\mathrm{number}\:{a}\:\mathrm{and}\:{b} \\ $$$$\mathrm{satisfy}\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:=\:\frac{{a}}{{b}},\:\mathrm{where}\:{a},\:{b}\:\in\:\mathbb{Z},\:{b}\:\neq\:\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:=\:\frac{\mathrm{a}}{\mathrm{b}}\:\:\rightarrow\:\:\mathrm{2}^{\frac{{a}}{{b}}} \:=\:\mathrm{5}\:\:\rightarrow\:\:\mathrm{2}^{{a}} \:=\:\mathrm{5}^{{b}} \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{number}\:\mathrm{that}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{because}\:\mathrm{LHS}\:\mathrm{always}\:\mathrm{even}\:\mathrm{and}\:\mathrm{RHS}\:\mathrm{always}\:\mathrm{odd} \\ $$$$\mathrm{Hence},\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:\mathrm{is}\:\mathrm{irrational} \\ $$
Answered by mrW2 last updated on 11/Mar/18
assume log_2 5 is rational, i.e. it can be expressed as log_2 5=(m/n) where m, n are integers and gcd(m,n)=1. ⇒2^(m/n) =5 ⇒2^m =5^n since 2 and 5 are co−prime, ⇒m=n=0. i.e. log_2 5 can not be expressed as (m/n), it′s irrational.
$${assume}\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:{is}\:{rational},\:{i}.{e}.\:{it}\:{can} \\ $$$${be}\:{expressed}\:{as}\: \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{5}=\frac{{m}}{{n}} \\ $$$${where}\:{m},\:{n}\:{are}\:{integers}\:{and}\:{gcd}\left({m},{n}\right)=\mathrm{1}. \\ $$$$\Rightarrow\mathrm{2}^{\frac{{m}}{{n}}} =\mathrm{5} \\ $$$$\Rightarrow\mathrm{2}^{{m}} =\mathrm{5}^{{n}} \\ $$$${since}\:\mathrm{2}\:{and}\:\mathrm{5}\:{are}\:{co}−{prime}, \\ $$$$\Rightarrow{m}={n}=\mathrm{0}. \\ $$$${i}.{e}.\:\mathrm{log}_{\mathrm{2}} \:\mathrm{5}\:{can}\:{not}\:{be}\:{expressed}\:{as}\:\frac{{m}}{{n}}, \\ $$$${it}'{s}\:{irrational}. \\ $$

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