Proving that Both m and n are Odd using Indirect Proof

  • Thread starter Thread starter blimkie
  • Start date Start date
  • Tags Tags
    Proof
AI Thread Summary
To prove that both integers m and n are odd given that their product mn is odd, the method of indirect proof is employed. The assumption is made that either m or n is even, which leads to the conclusion that the product mn must also be even. This contradiction arises because an even number multiplied by any integer results in an even product. Therefore, if mn is odd, both m and n cannot be even, confirming that they must both be odd. The discussion emphasizes the importance of correctly negating the statement to apply the indirect proof effectively.
blimkie
Messages
111
Reaction score
0
use the method of indirect proof. if m and n are integars and there product mn is odd, prove that both m and n are odd

so i wnat to prove that (m)(n) odd (mn)

so i assume that (mn) is even to go about the indirect method

now i need a jump start i wrote some equations and some algebra but dindt come up with a contradication

help is appreciative that's a lot guys
 
Physics news on Phys.org
Start off by

m = 2k+1

n = 2c + 1

where k and c is any natural number.
 
Using contradiction (indirect proof), you can assume either m or n to be even (that is, a multiple of two) and show that mn must also be even.
 
Cyclovenom said:
Start off by
m = 2k+1
n = 2c + 1
where k and c is any natural number.

No, that's exactly the wrong thing to do.

Since blimkie wants to use indirect proof to prove that "both m and n must be odd, he should negate that: "either m or n is even".

So assume m= 2p, which is even for any integer p, multiply by n and see what happens!
 
Last edited by a moderator:
A number to be even it should be divisible by 2. thus for the product mn to be divisible by 2 any of the integers m or n must be divisible by prime 2. thus the divisibility testdiectlygives you the answer that then anyone of them must be an even number.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top