Question

Show that √5 is irrational.

Answer 1

We will prove whether √5 is irrational by contradiction method.
Let √5 be rational 
It can be expressed as √5 = a/b ( where a, b are integers and co-primes. 
√5 = a/b
5= a²/b² 
5b² = a²
5 divides a²
By the Fundamental theorem of Arithmetic
so, 5 divides a .

a = 5k (for some integer) 

a² = 25k² 
5b² = 25k² 
b² = 5k² 

5 divides b²
5 divides b. 

Now 5 divides both a & b this contradicts the fact that they are co primes. 
this happened due to faulty assumption that √5 is rational. Hence, √5 is irrational. 

One more handy question :- https://brainly.in/question/2642933 .

Answer 2

let us assume on of contrary that under √5 is a rational number then there exist co- prime positive integer A and B such that.

$\\ \sqrt{5} = a \div b \\ \\ 5b ^{2} = a^{2} \\ \\ 5 \div a ^{2} \\ \\ 5 \div a \\ \\ \\ let \: \: a = 5c \\ \\ a ^{2} = 25c ^{2} \\ \\ b {}^{2} = 5c {}^{2} \\ \\ 5 \div b {}^{2} \\ \\ 5 \div b$

here we find that a and b have at least 5 as a common factor this contradict the fact that a and b are coprime

hence ,

√5 is irrational Number.

hope it helps!!!!

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