Question

Prove that √2 √3 √5 √7 √11 are irrational numbers.

​

Answer 1

Mark me as A brainlist......

Answer 2

Answer: let us assume that √2 is a rational number

√2=p/q. (p and q are co-prime)

q√2=p. (Square both sides)

2q²=p²

So, p² is a multiple of 2

And , p is also a multiple of 2

p =2k

p²=2k²

(2k)²=2q²

4k²=2q²

2k²=q²

So, q² is multiple of 2

And, q is also a multiple of 2

But p and q are co-prime

This is because of our incorrect assumption of √2 as rational

So, √2 is a irrational number

-------------------------

Let us assume that√3 is rational number

√3=p/q (p and q are co-prime)

q√3=p. (Squaring both sides)

3q²=p²

So, p² is a multiple of 3

And, p is also a multiple of 3

p=3k

p²=3k²

(3k)²=3q²

9k²=3q²

3k²=q²

3k=q

So, q² is a multiple of 3

And, q is also a multiple of 3

But p and q are co-prime

This is because of our incorrect assumption of √3 as rational number

So, √3 is a irrational number

------------------

Let us assume that √5 is a rational number

√5=p/q. (p and q are co-prime)

q√5=p. (Square both sides)

5q²=p²

So, P² is a multiple of 5

And, P is also a multiple of 5

p=5

p²=5k²

(5k)²=5q²

25k²=5q²

5k²=q²

5k=q

So, q is a multiple of 5

And, q² is also a multiple of 5

But p and q are co-prime

This is because of our incorrect assumption that √5 is rational number

So,√5 is a irrational number

--------------------

Let us assume that √7 is rational number

√7=p/q. (p and q are co-prime)

q√7=p. (square both side)

7q²=p²

So, p² is a multiple of 7

And, p is also a multiple of 7

P=7k

P²=7q²

(7k)²=7q²

49k²=7q²

q²=7k²

q=7k

So, q² is a multiple of 7

And, q is also a multiple of 7

But p and q are co-prime

This is because of our incorrect assumption of √7 as a rational number

So, √7 is a irrational number

------------------------

Let us assume that √11 is a rational number

√11=p/q (p and q are co-prime)

q√11=p. (square both sides)

11q²=p²

So, p² is a multiple of 11

And, p is also a multiple of 11

P=11k

P²=11q²

(11k)²=11q²

121k²=11q²

11k²=q²

11k=q

So, q² is multiple of 11

And, q is also a multiple of 11

But p and q are co-prime

This is because of our incorrect assumption of √11 as a rational number

So,√11 is a irrational number

Step-by-step explanation: