Question

answer this for me ,,first who answered ,he would be the brainliest

Answer 1

Solution:

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

Answer:

For 5 points I can answer only one question and mark my answer as branliest