Question

Prove that root 7 is an irrational number by contradiction method

Answer 1

BY CONTRADICTION METHOD,

Let we suppose √7 is rational number so that it can be represented in the form of p/q where p and q are integers and q≠0.

hence we can let,

√7=p/q

squaring both sides

(√7)²=(p/q)²

7=p²/q²

7q²=p².......(i)

7 divides p²

therefore there exists an integer r

hence,

p=7r

squaring both sides

(p)²=(7r)²

p²=49r².........(ii)

From (i) and (ii)

7q²=49r²

q²=49r²/7

q²=7r²

7 divides by q²

7 divide q

But we already show that 7 divides p.

This implies that 7 is common factor of p and q.

but this contradict our assumption that p and q have no common factors.

hence our assumption is wrong that √7 is rational.

Therefore √7 is an irrational number.

hence proved :)

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Answer 2

HEY MATE HERE IS YOUR ANSWER--

Lets assume that √7 is rational number. ie √7=p/q.

suppose p/q have common factor then

we divide by the common factor to get √7 = a/b were a and b are co-prime number.

that is a and b have no common factor.

√7 =a/b co- prime number

√7= a/b

a=√7b

squaring

a²=7b² .......1

a² is divisible by 7

a=7c

substituting values in 1

(7c)²=7b²

49c²=7b²

7c²=b²

b²=7c²

b² is divisible by 7

that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.

√7 is irrational