Question

Prove that √(7) is an irrational number

Answer 1

HEYA MATEY ..^_^❤✌️✌️

THERE'S YOUR ANSWER ...

LET, √7 BE A RATIONAL NUMBER .

THEREFORE,

$\sqrt{7} = \frac{p}{q }$

WHERE P, AND Q BELONGS TO Z, P, Q ARE COPRIME AND Q IS NOT EQUAL TO 0.

=> (q√7)^2 = p^2
=> 7q^2 = p^2

THEREFORE,
7 | p^2
=> 7. | p. ........1

let, p = 7x , ( where , x is any integer )

BY PUTTING THE VALUE FOR p IN EQUATION 1 , WE GET .....

7q^2 = ( 7 ) ^2
=> q^2 = 7x^2

THEREFORE,
7. | q^2
=> 7. | q

THEREFORE, 7 IS THE COMMON FACTOR OF p AND q BUT IT CONTRADICTS OUR ASSUMPTION , p AND q ARE COPRIME .

THEREFORE, √7 IS AN IRRATIONAL NUMBER ..

HOPE IT HELPS YOU ..^_^☝☝

Answer 2

hey buddy here is your answer

let √7 be a rational no.
therefore

√7 = a÷b ( a and b are co-prime no.)
√7b= a
spuaring both side
so,

${ \sqrt{7b} }^{2} = a ^{2}$
$a ^{2} is \: divisible \: by7$
a is also divisible by 7
so,
a can be expressed as

a = 7c
again spuaring ..

a^2 = 7c^2

a^2 = 49c^2. a^2 = 7b^2

7b^2 = 49c^2

b^2 = 7c^2

b^2 is also divisible by 7
so b is also divisible by 7

so this contradiction is just that a and b are co-prime no.

our supposition is wrong
√7 is an irrational number

I hope it helps you