Question

prove that √3 is irrational also prove that 3+5√3​

Answer 1

Step-by-step explanation:

If possible , let

3

be a rational number and its simplest form be

b

a

then, a and b are integers having no common factor

other than 1 and b



=0.

Now,

3

=

b

a

⟹3=

b

2

a

2

(On squaring both sides )

or, 3b

2

=a

2

.......(i)

⟹3 divides a

2

(∵3 divides 3b

2

)

⟹3 divides a

Let a=3c for some integer c

Putting a=3c in (i), we get

or, 3b

2

=9c

2

⟹b

2

=3c

2

⟹3 divides b

2

(∵3 divides 3c

2

)

⟹3 divides a

Thus 3 is a common factor of a and b

This contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming

3

is a rational.

Hence,

3

is irrational.

2

nd

part

If possible, Let (7+2

3

) be a rational number.

⟹7−(7+2

3

) is a rational

∴ −2

3

is a rational.

This contradicts the fact that −2

3

is an irrational number.

Since, the contradiction arises by assuming 7+2

3

is a rational.

Hence, 7+2

3

is irrational.

Proved.

Answer 2

$\bold {\underline{To \: Prove :- }}$

prove that √3 + √5 is irrational number .

$\bold \red{\underline \blue{Solution :- }}$

Let us assume it to be a rational number.

Rational numbers are the ones that can be expressed in q/p form where p,q are integers and q isn't equal to zero.

⇒√3+√5 = p/q

⇒√3 = p/q–5

squaring on both sides,

⇒3 = p²/q² − 2.√5( p/q ) + 5

⇒(2√5p)/q = 5−3 + ( p²/q² )

⇒(2√5p)/q = 2q² − p²/q²

⇒√5 = 2q² − p²/q².q/2p

⇒√5 = (2q² − p²)/2pq

As p and q are integers RHS is also rational.

As RHS is rational LHS is also rational i.e √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction arose because of our false assumption.

so, √3 + √5 irrational.