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Answer 1

hi friend!
1)Let us assume that 5 + root 3 is rational. Then it can be written in the form

5 + root3 = p/q

or 5 +p/q = root3

It implies root3 is a rational number [Since 5 +p/q are rationals]

But this contradicts to the fact that root 3 is irrational. Hence our supposition was wrong. Therefore 5 + root 3 is irrational.

2)Let √2 + √3 = (a/b) is a rational no.
On squaring both sides , we get
2 + 3  + 2√6 = (a2/b2)
So,5 + 2√6 = (a2/b2) a rational no.
So, 2√6 = (a2/b2) – 5
Since, 2√6 is an irrational no. and (a2/b2) – 5 is a rational no.
So, my contradiction is wrong.
So, (√2 + √3) is an irrational no
3)3 root2 can be written as a÷b where a and b,b is not equal to zero are co prime.
hence 3 root 2=a÷b
root2=1÷3×a÷b
root2 =a÷3b
here root 2 is irrational and a ÷3b is rational.rational and irrational are not equal so 3 root 2 is irrational
hope my answer helps you
keep smiling

Answer 2

Hey!!!

As promised I am here to help you

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Chapter 1 = Real Numbers

To Prove : The given numbers are Irrational

Proof :

(i) 5 + √3

By the Method of Contradiction,

let 5 + √3 be a rational number

Then let 5 + √3 = x (where x is rational)

Then √3 = x - 5

Here we know √3 is irrational but x is rational

Thus our assumption is false

Thus 5 + √3 is an Irrational number

(ii) 2 + √3

By the Method of Contradiction,

let 5 + √3 be a rational number

Then let 2 + √3 = x (where x is rational)

Then √3 = x - 2

Here we know √3 is irrational but x is rational

Thus our assumption is false

Thus 2 + √3 is an Irrational number.

(iii) 3√2

By the Method of Contradiction

let 3√2 be a rational number

Then

$= > 3 \sqrt{2} = \frac{p}{q}$

where p,q not equal to 0 and HCF(p,q) = 1

=> p = 3√2q

Square both sides

=> p² = 18q² ----------(1)

This means p² is divisible by 18

Thus then p is also divisible by 18.

Let p = 18m

Square both sides

=> p² = 324m²

=> 18q² = 324m² (from 1 )

=> q² = 18m²

This means q² is divisible by 18

Thus then q is also divisible by 18.

Here HCF(p,q) = 18

but it should be HCF(p,q) = 1

Thus our assumption is wrong

Thus 3√2 is an irrational number

HENCE PROVED

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