Question

prove that for any prime positive integer p, √p is a irrational number

Answer 1

If possible,let √p be a rational number.

also a and b is rational.

then,√p = a/b

on squaring both sides,we get,

(√p)²= a²/b²

→p = a²/b²

→b² = a²/p [p divides a² so,p divides a]

Let a= pr for some integer r

→b² = (pr)²/p

→b² = p²r²/p

→b² = pr²

→r² = b²/p [p divides b² so, p divides b]

Thus p is a common factor of a and b.

But this is a contradiction, since a and b have no common factor.

This contradiction arises by assuming √p a rational number.

Hence,√p is irrational.

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

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TO PROVE :- For any prime positive integer p, √p is an irrational number.

Let us assume that √p is a rational number.

Then, there exist positive co-primes a and b such that :-

√p = a/b

p = a²/b²

b²p = a²

p divide a²

p divides a.

a = pc ( positive integer c. )

Now, b²p = a²

b²p = p²c²

b² = pc²

p divide b²

p divides b

Therefore, p/a and p/b

This contradicts the fact that a and b are co-primes.

HENCE , √p IS IRRATIONAL NUMBER !

$\huge{\boxed{\sf{Hence\:Proved\:!}}}$

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