Question:
Prove that √5 is irrational.

Answer:

Let us consider √5 is a rational number.
Therefore, √5 = pq, where p and q are integers and q ≠ 0
If pa and q have any common factor ,then dividing by that common factor,
We have, √5 = ab, where a and b are co primes.
a = √5b
On squaring, a² = 5b² ........... (1)
Due to the presence of 5 on RHS, we say that 5 is a factor of a².

5 divides a²
Since, a is prime , therefore 5 divides a [ theorem ]
Therefore , a = 5k , where k is an integer.
Putting a = 5k in (1) , we have
25k² = 5b²

b² = 5k²
This shows that 5divides b². But b is a prime no and so 5 divides b also.
Thus 5 is a common factor of a and b . This is contradiction to the fact that
a and b have no common factor other than one.
Thus our consideration is wrong and so √ 5 is not a rational number.
Hence, proved.

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Question:
Prove that √5 is irrational.

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Question: Prove that √5 is irrational.

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Question:
Prove that √5 is irrational.