Expert: Paul Klarreich - 1/26/2007

Question
i'm studying concepts of math

can you please help me with the following 2 questions

1)if a and b are positive integers with (a,b)=1 and if ab is a square, prove that both a and b are squares.

2) prove that if m is a positive integer for which sqrt(m) is rational, then m is a perfect square. Conclude that m is not a perfect square, the sqrt(m) is irrational.  

Answer
Questioner:   lina
Category:  Advanced Math
 
Subject:  Primes
Question:  i'm studying concepts of math

can you please help me with the following 2 questions

1)if a and b are positive integers with (a,b)=1 and if ab is a square, prove that both a and b are squares.

2) prove that if m is a positive integer for which sqrt(m) is rational, then m is a perfect square. Conclude that m is not a perfect square, the sqrt(m) is irrational.  
....................................
Hi, Lina,  

For 1) Both a and b have a unique factorization into primes.  Since (a,b) = 1, no prime divides both a and b.

n also has a U.F.i.P.

Let n = q1^e1 q2^e2 .. qk^ek  be the factorization.  [According to the Fundamental Theorem of Arithmetic, this is how we write the factorization -- each prime appears raised to some power, which could, of course, be 1.]

Then

n^2 = q1^2e1 q2^2e2 .. qk^2ek

i.e. every prime divisor of n^2 appears as an even power (which could be 2)

A. Let p1 be one of the prime divisors of a.  Then p1 divides n^2, and therefore p1 is one of the q's.

B. Since that q appears as an even power on the right, it must appear as an even power on the left.

C. Since (a,b) = 1, none of those q's appear in b.

D. So every prime factor of a appears as an even power, and a is a square.

A similar argument applies to b.

.....................................................  

2) Suppose that sqrt(m).  Then sqrt(m) = a/b, and we can assume that a/b is in lowest terms, so that  (a,b) = 1.  Then:
      a^2
m^2 = -----
      b^2

And  a^2 = b^2 m^2  
Now then:

A. a, b, and m have a U.F.i.P.  

B. Every prime that divides a^2 must divide a and divide the RHS.

C. If p1 divides b^2 m^2 it must divide either b^2 or m^2.

D. But if p1 divides a it cannot divide b, therefore it divides m.

E. If every prime that divides a divides m, we can simply cancel all the factors of a^2 and the LHS reduces to 1.

F. Since  b^2 and m^2 were integers, b^2 = 1, and m^2 = a^2, thus a perfect square.

(Your required conclusion follows.  All you need to say is:

If sqrt(m) is rational, m is a square. (We just proved that.)
The contrapositive statement:
If m is not a square, sqrt(m) is not rational.
is equivalent.