Direct Proof With Odd Integers

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SUMMARY

The discussion centers on proving that if m is an odd integer and n divides m, then n must also be an odd integer. The proof begins with the representation of odd integers in the form m = 2k + 1. By assuming n is even and deriving a contradiction, the conclusion is reached that n must indeed be odd. This method effectively utilizes the properties of odd and even integers to establish the proof.

PREREQUISITES
  • Understanding of odd and even integers
  • Basic knowledge of divisibility
  • Familiarity with algebraic manipulation
  • Experience with proof techniques, particularly proof by contradiction
NEXT STEPS
  • Study the properties of odd and even integers in depth
  • Learn about proof by contradiction techniques
  • Explore integer divisibility rules and their implications
  • Investigate other proofs involving odd integers and their properties
USEFUL FOR

Mathematics students, educators, and anyone interested in number theory or proof techniques will benefit from this discussion.

smiles988
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Homework Statement


If m is an odd integer and n divides m, then n is an odd integer.


Homework Equations


Odd integers can be written in the form m=2k+1.
Since n divides m, there exists an integer p such that m=np


The Attempt at a Solution


We will assume that m is an odd integer and that n divides m. We will show that n is an odd integer. Since m is an odd integer, there exists an integer k such that m=2k+1. Since n divides m, there exists an integer p such that m=np.

I don't know where to go from here to arrive at n is equal to 2*an integer +1. Do I need to use cases?
 
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so you have m = np and m is odd. you want to show n is odd. so suppose it's even, then look what happens, you should get a contradiction.
 

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