Question

Prove that √5 is irrational
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Answer 1

$\huge\boxed{\underline{\underline{\green{Solution}}}} </p><p>$

Let us assume that √5 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√5 = p/q { where p and q are co- prime}

√5q = p

Now, by squaring both the side

we get,

(√5q)² = p²

5q² = p² ........ ( i )

So,

if 5 is the factor of p²

then, 5 is also a factor of p ..... ( ii )

=> Let p = 5m { where m is any integer }

squaring both sides

p² = (5m)²

p² = 25m²

putting the value of p² in equation ( i )

5q² = p²

5q² = 25m²

q² = 5m²

So,

if 5 is factor of q²

then, 5 is also factor of q

Since

5 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

Hence √5 is an irrational number

$</p><p></p><p>\displaystyle\textcolor{red}{Please \: Mark \: it \: Brainliest}$

Answer 2

$\sqrt{5}$

METHOD OF CONTRADICTION

let √5 be a rational no. that is it can be expressed in the form of p by q where p and q are co primes and q ≠0

now ,

$\sqrt{5} = \frac{p}{q}$

$\sqrt{5} \times q = p$

sq. b/s

$5 {q}^{2} = {p}^{2}$

hence ,by this we can say that p is a multiple of 5

so , let P = 5m

$5 {q}^{2} ={ (5m) }^{2}$

$5 {q}^{2} = 25 {m}^{2}$

${q}^{2} = \frac{25 {m}^{2} }{5}$

${q}^{2} = 5 {m}^{2}$

therefore , q is also a multiple of 5

But, p and q were co primes

hence , our contradiction is wrong

$\longrightarrow$$\sf\pink{ √5\: is \: an \: irrational \: number }$