Question

prove that √7 is irrational
Give step-by-step explanation​

Answer 1

On the contrary,let us assume √7 be a rational.

So,

√7 = p/q [where p,q€N ,q≠0]

and p and q are co primes.

Squaring on both sides,

7 = p²/q²

→p²=7q²

This implies that p² is divided by 7

By Fundamental Theory of Arithmetic,

p is also divided by 7.

Now,

Let p = 7r,for some positive integer r

Squaring on both sides,

→p² = 49r²

→7q² = 49r²

→q² = 7r²

Implies,

7 also divides q²

Similarly,7 also divides q

As,p and q have more than one factor,they aren't co primes

This contradicts our assumption

Hence,√7 is an irrational

Answer 2

Let us assume, to the contrary, that √7 is rational.

$Let \: \: \sqrt{7} = \frac{a}{b} \: \: where \: \: a \: \: and \: \: b \: \: \\ are \: \: co-prime \: \: integers \: \: and \\ b \: \neq0$

b√7 = a.

Squaring on both sides:

7b² = a².

Since 7 divides a² [b² = a²/7], 7 also divides a[Theorem: If an integer p divides a², then it also divides a].

Let a = 7c, for some integer c. Then:

7b² = 49c².

b² = 7c².

Since 7 divides b² [c² = b²/7], 7 also divides b[Theorem: If an integer p divides a², then it also divides a].

But, we stated that a and b are co-primes[have no other common factors other than 1].

This contradiction arises because √7 is irrational.

Therefore, √7 is irrational[Proof by Contradiction].