if p and q are prime positive integers then prove that a√p -b√q are irrational numbers About the author Sarah
Answer: First, we’ll assume that p and q is rational , where p and q are distinct primes p + q =x, where x is rational Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides. ( p + q ) 2 =x 2 p+2 pq +q=x 2 2 pq =x 2 −p−q pq = 2 (x 2 −p−q) Now, x, x 2 , p, q, & 2 are all rational, and rational numbers are closed under subtraction and division. So, 2 (x 2 −p−q) is rational. But since p and q are both primes, then pq is not a perfect square and therefore pq is not rational. But this is contradiction. Original assumption must be wrong. So, p and q is irrational, where p and q are distinct primes. Reply
Answer:
First, we’ll assume that
p
and
q
is rational , where p and q are distinct primes
p
+
q
=x, where x is rational
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.
(
p
+
q
)
2
=x
2
p+2
pq
+q=x
2
2
pq
=x
2
−p−q
pq
=
2
(x
2
−p−q)
Now, x, x
2
, p, q, & 2 are all rational, and rational numbers are closed under subtraction and division.
So,
2
(x
2
−p−q)
is rational.
But since p and q are both primes, then pq is not a perfect square and therefore
pq
is not rational. But this is contradiction. Original assumption must be wrong.
So,
p
and
q
is irrational, where p and q are distinct primes.