Question

prove that root over 7 is irrational in leandy form​

Answer 1

Step-by-step explanation:

let's assume that√7 is rational let's find some integers like a\b

√7=a\b

=√7b =a

squareing ob both sides

(√7b)²=(a)²

7b²=a²

7 divide a²

7also divide a

put a =7c

7b²=(7c)²

7b²=49c²

7 divide b²

7 also divide b

this contradict the fact both a & b have same integer

so our assumptions is wrong √7 is irrigation

Answer 2

ǫᴜᴇsᴛɪᴏɴ →

prove that root over 7 is irrational in leandy form

ᴀɴsᴡᴇʀ →

Let us assume that 7 is rational. Then, there exist co-prime positive integers a and b such that

$\longmapsto\tt{ \sqrt{7} =\dfrac{a}{b}}$

Squaring on both sides, we get

$\longmapsto\tt{a² = 7b²}$

Therefore, a² is divisible by 7 and hence, a is also divisible by7

so, we can write a=7p, for some integer p.

Substituting for a, we get 49p² =7b² ⟹b² =7p².

This means, b²is also divisible by 7 and so, b is also divisible by 7.

Therefore, a and b have at least one common factor, i.e., 7.

But, this contradicts the fact that a and b are co-prime.

Thus, our supposition is wrong.

$\longmapsto\tt{Hence, \sqrt{7} \: is \: irrational.}$