Question

prove that root7 is irrational ​

Answer 1

Answer:

$let \: \sqrt{7} \: be \: a \: rational \: no \\ so \: \sqrt{7 } = \frac{a}{b} (where \: a \: and \: b \: are \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: co \: prime \: integers \: and \: b \: not \: equals \: to \: 0) \\ squaring \: on \: both \: sides \\ 7 {b}^{2} = {a}^{2} \\ a \: is \: divisible \: by \: 7 \\ a = 7f \\ squaring \: on \: both \: sides \\ {a}^{2} = 49 {f}^{2} \\ putting \: value \: of \: {a}^{2} \\ 7 {b}^{2} = 49 {f }^{2} \\ {b}^{2} = 7 {f}^{2} \\ b \: is \: divisible \: by \: 7 \\ \: this \: contradicts \: the \: fact \: that \\ \: a \: and \: b \: are \: co \: prime \: integers \\ hence \: \sqrt{7} is \: a \: irrational \: no$

hope it helps you friend.

Answer 2

Answer:

√7 is irrational

Step-by-step explanation:

let as assume that √7 is rational , ∴ it is of the form  p/q ; where q≠0, p and q are co-primes.       √7=p/q , ∴√7q=p , squre on both the sides then, 7q²= p² , when p²divides 7 ; then p divides 7. p=7r ; r is some integer , square both the side, p²= 49r²; then 7q²= 49r², ∴  q²=7r² , when q² divides 7 , then q divides 7 ∴ this contradict that our supposition is wroung . ∴p and q are not co-primes, ∴ √7 is irrational