Question

prove that √7 is irrational number​

Answer 1

Answer:

let us assume that √7 be rational. then it must in the form of p / q [q ≠ 0] [p and q are co-prime]

√7 = p / q => √7 x q = p

squaring on both sides

=> 7q2= p2 ------> (1)

p2 is divisible by 7 p is divisible by 7 p = 7c [c is a positive integer] [squaring on both sides ] p2 = 49 c2 --------- > (2)

subsitute p2 in equ (1) we get 7q2 = 49 c2 q2 = 7c2 => q is divisble by 7 thus q and p have a common factor 7.

there is a contradiction as our assumsion p & q are co prime but it has a common factor. so that √7 is an irrational.

HOPE IT HELPS U...

Answer 2

√7 is irrational number

__________ [TO PROVE]

Let us assume that, √7 is a rational number

√7 = $\dfrac{a}{b}$

Here, a and b are co-prime numbers.

• Squaring on both sides, we get

=> (√7)² = $( { \dfrac{a}{b}) }^{2}$

=> 7 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

=> 7b² = a² ________( eq 1)

Clearly;

a² is divisible by 7.

So, a is also divisible by 7.

Now, let integer be c.

=> a = 7c

• Squaring on both sides.

=> a² = 49c²

=> 7b² = 49c² [From (eq 1)]

=> b² = 7c² _______(eq 2)

This means that, 7 divides b², and so 7 divides b also.

7 divides b² and 7 divide a² also.

So, our assumption is wrong.

√7 is irrational number.

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