Question

prove that root 5 is irrational​

Answer 1

Step-by-step explanation:

Let √5 is rational number

then,

√5=a/b(b not equal to 0 and a,b are co prime

5=a^2/b^2

5b^2=a^2

5 is the factor of a ---equation 1

Now let a=5c

5b^2=5c^2

b^2=5c^2

5 is the factor of b----- equation 2

but 5 is the factor of a and b both , so our statement is wrong.

So √5 is irrational number.

I hope it helps you.

Please mark me as brainliest answer.

Answer 2

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$\sf{On\:Our\:contratry,\:let\:us\:assume\:that\:\sqrt{5}\:is\:a}$

$\sf{rational\:number.}$

$\therefore\:\:$$\sf{\sqrt{5}\:=\:\frac{p}{q}\:(q\:not\:equals\:to\:0)}$

$\sf{If\:p,\:q\:have\:a\:common\:factor,\:on\:cancelling}$

$\sf{the\:common\:factor\:let\:it\:be\:reduces}$

$\sf{to\:\frac{a}{b},\:where\:a,\:b\:are\:co-primes.}$

$\sf{Now\:\sqrt{5}\:=\:\frac{a}{b},\:where\:HCF\:(a,\:b)\:=\:1}$

$\sf{Squaring \:on\:both\:sides\:we\:get}$

$\sf{(\sqrt{5})^2 \: = \: (\:\frac{a}{b}\:)^2}$

$\sf{5\:=\:\frac{a^2}{b^2}$

$\longrightarrow$$\sf{5b^2 \: = \: a^2}$

$\longrightarrow$$\sf{5\:divides\:a^2\:and\:thereby\:5\:divides\:5}$

$\sf{\:\:\:Now,\:take\:a\:=\:5c}$

$\sf{\:\:\:then,\:a^2\:=\:25c^2}$

$\sf{I.e.,\:\:5b^2\:=\:25c^2}$

$\longrightarrow$$\sf{b^2\:=\:5c^2}$

$\longrightarrow$$\sf{5\:divides\:b^2\:and\:thereby\:b.}$

$\longrightarrow$$\sf{5\:divides\:both\:b\:and\:a.}$

$\sf{This\:contradicts\:that\:a\:and\:b\:are}$

$\sf{co-primes}$

$\sf{This\:contradiction \:arised\:due\:to\:our}$

$\sf{assumption\:that\:\sqrt{5}\:is\:a\:rational\:number.}$

$\sf{Hence\:our\:assumption\:is\:wrong}$

$\sf{i.e.,\:\sqrt{5}\:is\:an\:irrational \:number}$

✍️ Ⓣhank Ⓨou ‼️

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