Given that the definition of an odd number is , whereas an even number is , we can prove that the sum of two odd integers is even.
Given some odd integer, , and another , the sum is which is of form .
Let be some rational number and be some irrational number.
Suppose is rational, this means that there exists some such that , where is rational.
You can isolate , . Since and are both rational, the difference of two rational numbers is also rational. This would imply that is rational, but that contradicts the original assumption.
Since the assumption that is a contradiction, must be irrational.
The sum of an irrational number and a rational number is irrational.
A rational number is defined as the ratio of two integers, such that , where and are integers, and .
Let and be two rational numbers, , and .
Multiplying and results in .
Since the product of two integers is also an integer, follows the form of , where both and are integers.
The product of two rational numbers is always rational.
A number is even if it follows the form , and odd if it follows the form .
Given that the product of is even, it is of the form .
Assuming both are odd, , and .
which is odd.
Since is odd if both and are odd, then at least one of or must be even for to be even.
The contrapositive is if is odd, then is even.
Assuming is odd, , , which is odd.
would therefore have to be even, since the sum of two odd numbers is even.
By contraposition, must be even give that is odd.
Suppose is odd, and is also odd.
Suppose is odd, , which is odd.
would be even, but the original statement was that is odd, which is a contradiction.
We can conclude that must be even given that is odd.