Any Math majors in the house? Or anyone that's taken Discrete and Combinatorial mathematics...

[FrogDog]

The last question on my assignment is asking me to prove something by using Proof by Contradiction. I have no idea how to go about finding the contradiction. I can set it up, but that's as far as I can go. Here's what I have so far:

Use proof by contradiction to prove the following: Assuming p is rational, and q is irrational, prove that if p > 0, then r = p*q is irrational.

So,

A:= (p is rational, q is irrational)
c:= (p > 0 --> r is irrational) where r = p*q

Assume A is true and c is false to find a contradiction. c is false when p > 0, but r (p*q) is not irrational.

p = a/b where 'a' and 'b' are integers and 'b' != 0
c = ?

Any help would be much appreciated.

 

[MadMark]

//Use proof by contradiction to prove the following: Assuming p is rational, and q is irrational, prove that if p > 0, then r = p*q is irrational.//

Rewrite:
p is rational. q ia irrational. prove p > 0 => pq irrational.

Proof:
Suppose pq is rational. Let pq = m/n where m,n are integers where their greatest common divisor is 1.
Since p is rational, let p = c/d (c and d have same same property as m,n above).
By properties of real numbers under multiplication, we have q = (m/n)(d/c)
=> q = md/nc.
=> q = a/b (where a = md, b = nc. a,b are integers)
=> q is rational!
-><- (contradiction!)


Hope this is legible. (as long as p and q are non-zero, this proof works. It can be written with more math and shorter if you want.)

 

[FrogDog]

Originally posted by: MadMark
//Use proof by contradiction to prove the following: Assuming p is rational, and q is irrational, prove that if p > 0, then r = p*q is irrational.//

Rewrite:
p is rational. q ia irrational. prove p > 0 => pq irrational.

Proof:
Suppose pq is rational. Let pq = m/n where m,n are integers where their greatest common divisor is 1.
Since p is rational, let p = c/d (c and d have same same property as m,n above).
By properties of real numbers under multiplication, we have q = (m/n)(d/c)
=> q = md/nc.
=> q = a/b (where a = md, b = nc. a,b are integers)
=> q is rational!
-><- (contradiction!)


Hope this is legible. (as long as p and q are non-zero, this proof works. It can be written with more math and shorter if you want.)

Thanks a lot, MadMark. That helps big time!