Math, asked by aliahmad46, 6 months ago

prove √5 is irrational

Answers

Answered by shreyaa14
3

Answer:

Let

5

be a rational number.

then it must be in form of

q

p

where, q

=0 ( p and q are co-prime)

5

=

q

p

5

×q=p

Suaring on both sides,

5q

2

=p

2

--------------(1)

p

2

is divisible by 5.

So, p is divisible by 5.

p=5c

Suaring on both sides,

p

2

=25c

2

--------------(2)

Put p

2

in eqn.(1)

5q

2

=25(c)

2

q

2

=5c

2

So, q is divisible by 5.

.

Thus p and q have a common factor of 5.

So, there is a contradiction as per our assumption.

We have assumed p and q are co-prime but here they a common factor of 5.

The above statement contradicts our assumption.

Therefore,

5

is an irrational number.

Answered by Anonymous
3

Given: √5

We need to prove that √5 is irrational

Proof:

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number

Hence proved..

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