Question

prove that√5 is irrational
​

Answer 1

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.

Answer 2

Answer:

$\sqrt{5} \: is \: irrational$

$it \: is \: possible \: \sqrt{5} \: is \: rational \\ 5 = \frac{p {}^{2} }{q {}^{2} } \\ (q \: equal \: to \: not \: 0 \: where \: co - prime \: number) \\ 5 = \frac{p {}^{2} }{q {}^{2} } = 5q {}^{2} = p {}^{2} \\ p {}^{2} = 5q {}^{2} \\ 5 \: divides \: p {}^{2} = 5 \: divides \: p \: - - - - - - - (i) \\ p = 5 \times m \: is \: an \: intergers - - - - (ii) \\ (5m) {}^{2} = 5q {}^{2} \\ 25m {}^{2} = 5q {}^{2} \\ q {}^{2} = 5m \\ 5 \times divides \: q {}^{2} = 5 \times divides \: q \\ q = 5 \times n \: is \: an \: intergers - - - - (iii) \\ from \: (ii) \: and \: (iii) \: is \: 5 \: common \: factor \: of \: p \times q$

And, the contradictors of p& q is co-prime

Step-by-step explanation:

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