Question

prove that √5 is an irrational number​

Answer 1

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Answer 2

Step-by-step explanation:

Let root 5 be a rational number

Root 5=p/q where p and q are integers, q not equal to 0

Now, root 5=a/b,where a and b are co primes

5b=a,squaring on both sides,

5b^2=a^2

B^2=a^2/5,therefore 5 divides a^2,thus 5 divides a (theorem)

Let a/5=c

A=5c

A^2=25c^2

Substituting the value of a^2 as 5b^2 we get

5b^2=25c^2

5b^2/25=c^2

B^2/5=c^2,thus 5 divides b^2

Therefore 5 divides b(theorem)

Here 5 is a common factor of a and b which contradicts the fact that they are co primes. This contradiction has arisen from our wrong assumption that root 5 is a rational number. Therefore it is an irrational number