Question

prove that existence of √x for a given position real number x.​

Answer 1

For any positive real number x, prove that there exists an irrational number y such that 0< y< x.

Here is your perfect answer!

If x is rational number, then y/sqr2 is an irrational number such that 0 < y <x.

This is because square root 2 is more than 1.

(1.4.... > 1).

If x is irrational, then y = x/2 is also an irrational such that 0 < y <x.

Here x is positive real no

As we know square root of prime no is always irrational

So if x is prime no

Then √x will lie between 0 and x

So here there exists irrational

Now

If x is irrational

then its any fraction will also be irrational

So here y exists also

Now if x is not a prime and perfect square no

Then

As Between 0 and perfect square no except 1 there always exists a prime no ,also there exists square root of prime

Ex lets take x= 4

So 0 , 1 ,2,3,4

There exists 2,3 both prime

Also √2, √3 also lies between 0 and 4

So it would also give irrational no y