Question

4. Represent √10 and √20 on the number line.
5. Prove that the following numbers are irrational.
(1) 2+ √3
(2)3-√5
(3) √2+√5​

Answer 1

Step-by-step explanation:

5. Prove that the following numbers are irrational.

(1) 2+ √3

Solution -

Let's assume 2+ √3 is a rational number.

Rational number = if a and b are in the form of a/b where a and b are integers and b is is not equal to 0

2+ √3= a/b

√3=a/ b-2

Rational number =a/b =a/b-2

So, √3 is also rational number

But we know that √3 is an irrational number

So our contradiction is wrong

Hence 2+ √3 is an irrational number. (proved)

(2)3-√5

Solution -

Let's assume 3-√5 is a rational number.

Rational number = if a and b are in the form of a/b where a and b are integers and b is is not equal to 0

3-√5=a/b

-√5=a/b-3

Rational number =a/b =a/b-3

So, √5 is also rational number

But we know that √3 is an irrational number

So our contradiction is wrong

Hence 3- √5 is an irrational number. (proved)

(3) √2+√5

Solution -

Let us assume that √2+√5 is a rational number.

A rational number can be written in the form of p/q where p,q are integers and q≠0

√2+√5 = p/q

On squaring both sides we get,

(√2+√5)² = (p/q)²

√2²+√5²+2(√5)(√2) = p²/q²

2+5+2√10 = p²/q²

7+2√10 = p²/q²

2√10 = p²/q² – 7

√10 = (p²-7q²)/2q

p,q are integers then (p²-7q²)/2q is a rational number.

Then √10 is also a rational number.

But this contradicts the fact that √10 is an irrational number.

Our assumption is incorrect

√2+√5 is an irrational number.

Hence proved.