#### Prove that √3 is a rational number. Hence show that 5 – √3 is an irrational number.

**Solution:**

If √3 is a rational number

Consider √3 = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By squaring both sides

3 = p^{2}/q^{2}

So we get

P^{2} = 3q^{2}

We know that

3q^{2} is divisible by 3

p^{2} is divisible by 3

p is divisible by 3

Consider p = 3 where k is an integer

By squaring on both sides

P^{2} = 9k^{2}

9k^{2} is divisible by 3

p^{2} is divisible by 3

3q^{2} is divisible by 3

q^{2} is divisible by 3

q is divisible by 3

Here p and q are divisible by 3

So our supposition is wrong

Therefore, √3 is an irrational number.

In 5 – √3

5 is a rational number

√3 is an irrational number (proved)

We know that

Difference of a rational number and irrational number is also an irrational number

So 5 – √3 is an irrational number.

Therefore, it is proved.

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