Question

prove √5 that is is irational number​

Answer 1

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hey mate here is your answer...

ANSWER

Let 5 be a rational number.

then it must be in form of  qp  where,  q=0     ( p and q are co-prime)

5=qp

5×q=p

Suaring on both sides,

5q2=p2           --------------(1)

p2 is divisible by 5.

So, p is divisible by 5.

p=5c

Suaring on both sides,

p2=25c2         --------------(2)

Put p2 in eqn.(1)

5q2=25(c)2

q2=5c2

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....

hope this answer helps you..❤❤✌✌

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Answer 2

Answer:

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

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