**Prove that the following numbers are irrational:**

**(i) 3 + √5**

**(ii) 15 – 2√7**

**Solution:**

(i) If 3 + √5 is a rational number say x

Consider 3 + √5 = x

It can be written as

√5 = x – 3

Here x – 3 is a rational number

√5 is also a rational number.

Consider √5 = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By squaring both sides

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

p^{2} = 5q^{2}

We know that

5q^{2} is divisible by 5

p^{2} is divisible by 5

p is divisible 5

Consider p = 5k where k is an integer

By squaring on both sides

p^{2} = 25k^{2}

So we get

5q^{2} = 25k^{2}

q^{2} = 5k^{2}

Here

5k^{2} is divisible by 5

q^{2} is divisible by 5

q is divisible by 5

Here p and q are divisible by 5

So our supposition is wrong

√5 is an irrational number

3 + √5 is also an irrational number.

Therefore, it is proved.

(ii) If 15 – 2√7 is a rational number say x

Consider 15 – 2√7 = x

It can be written as

2√7 = 15 – x

So we get

√7 = (15 – x)/ 2

Here

(15 – x)/ 2 is a rational number

√7 is a rational number

Consider √7 = p/q where p and q are integers

q ˃ 0 and p and q have no common factor

By squaring on both sides

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

p^{2} = 7q^{2}

Here

7q^{2} is divisible by 7

p^{2} is divisible by 7

p is divisible by 7

Consider p = 7k where k is an integer

By squaring on both sides

p^{2} = 49k^{2}

It can be written as

7q^{2} = 49k^{2}

q^{2} = 7k^{2}

Here

7k^{2} is divisible by 7

q^{2} is divisible by 7

q is divisible by 7

Here p and q are divisible by 7

So our supposition is wrong

√7 is an irrational number

15 – 2√7 is also an irrational number.

Therefore, it is proved.

Here

¾ is a rational number and √5/4 is an irrational number

We know that

Sum of a rational and an irrational number is an irrational number.

Therefore, it is proved.

**More Solutions:**

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