Question

prove that √5 is irrational​

Answer 1

5 is common factor of a and b. so √5 is rational

but the contradiction the fact is √5 is irrational

Our assumption is wrong

Hence ,√5 is irrational

Answer refer to attachment h

★Hope it helps you

Answer 2

$\huge{\underline{\bf{\red{Solution:-}}}}$

Let √5 be a rational number

So,

√5 = a/b

where a and b are intigers and b ≠ 0.

Then,

√5= a/b

squaring on both sides

$: \implies \sf5=\frac{a^2}{b^2}$

$: \implies \sf5b^2=a^2..........(i)$

$: \implies \sf5\:\:divides \:\:a^2[\therefore\:5\:divides\:5b^2]$

$: \implies \sf5\:divides \:a$

So,

5 is a prime and divides b²

Then, 5 also divides b.

Let a = 5c for some intiger c.

putting a = 5c in (i)

$: \implies \sf5b^2=25c^2$

$: \implies \sf\:b^2=5c^2$

$: \implies \sf5\: divides\:b^2$so,5 also divides 5c².

$: \implies \sf5\: divides\:b$

So,

5 is a prime and 5 divides b² and b also.

Then, 5 is a common factor of a and b.

but, this contradicts the fact that a and b have no common factor other then 1.

So, this contradiction is arissen because we assume that √5 is rational .

Hence,

√5 is irrational.

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