Question

Can anyone tell me the Q28 plzz​

Answer 1

Answer:

Let 2-√3 be rational no.

So,2-√3=p/q

-√3=p/q-2

Since,a and b are integers,a/b-2 is a rational number and

therefore,-√3 is a rational no.

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

but this contradicts the fact that it is a irrational number.

This contradiction arises as we have assumed 2-√3 a rational number.

So,2-√3 is a irrational no.