Question

Prove that√3 is an irrational number hence prove 3-√5 is also an irrational number?

Answer 1

(I) let, √3 be the rational number .
√3=p/q
√3q=p
squaring both side,
(√3q)^2=p^2
3q^2=p^2
therefore,
p^2 is the factor of 3
so , p is also the factor of 3
now, let p=3m
put p =3m
3q^2=9m^2
q^2=3m^2
therefore,
q^2 is also a factor of 3
so q is also a factor of 3.

therefore p&q are not co prime numbers

so our assumption is wrong
and √3 is irrational number .
hence proved ..


(ii) let , 3-√5 be the rational number .
therefore,
3-√5=a
-√5=a-3
√5=a+3

as we know a+3 is an integer . so , it is rational number .
therefore √5 is also an integer but we know that √5 cannot be a integer .

therefore, our assumption is wrong .
and 3-√5 is irrational number .
hence proved...

Answer 2

Step-by-step explanation:

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number