Answer:

• Let's TAKE that ✓2 is a rational number where √2 = p/q and are in their LOWEST terms.

p^2= 2q^2

p^2 is a multiple of 2

hence, p is a multiple of 2

let P= 2m

(2m)^2 = (√2q)^2

4m^2 = 2q^2

2m^2= q^2

q^2 is a multiple of 2

q is a multiple of 2

it is against our assumptions and hence it is proved that √2 is irrational.

Let's take that ✓3 is a rational number where √3 = p/q and are in their lowest terms.

p^2= 3q^2

p^2 is a multiple of 3

hence, p is a multiple of 3

let P= 2m

(3m)^2 = (√3q)^2

9m^2 = 3q^2

3m^2= q^2

q^2 is a multiple of 3

q is a multiple of 3

it is against our assumptions and hence it is proved that √3 is irrational.

• Let's take that ✓5 is a rational number where √5 = p/q and are in their lowest terms.

p^2= 5q^2

p^2 is a multiple of 5

hence, p is a multiple of 5

let P= 5m

(5m)^2 = (√5q)^2

25m^2 = 5q^2

5m^2= q^2

q^2 is a multiple of 5

q is a multiple of 5

it is against our assumptions and hence it is proved that √5 is irrational.

• Let's take that ✓7 is a rational number where √7 = p/q and are in their lowest terms.

p^2= 7q^2

p^2 is a multiple of 7

hence, p is a multiple of 7

let P= 7m

(7m)^2 = (√7q)^2

49m^2 = 7q^2

7m^2= q^2

q^2 is a multiple of 7

q is a multiple of 7

it is against our assumptions and hence it is proved that √7 is irrational.

Above it is clearly proved that √2 , √3 , √5 and √7 are irrational.

"The SUM of irrational numbers will always be irrational. "

hence it is proved that the sum of √2 , √3 , √5 and √7 will be an irrational number.

hope it helps!

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