Exploring Number Parity

and an Introduction to Proofs

Number parity is the grown up term for talking about whether a number is even or odd. Today we’ll take this elementary concept, define it formally and use it as a launching point to prove some basic results.

Yes, the proof. I know this is a terrifying topic for many, but it is the true cornerstone of mathematics. Logical reasoning in action. The sooner you get comfortable with them, the easier they will be.

Definitions

First let’s define even and odd numbers. No need to get too technical at the moment, just tell me what you would tell someone who was learning about number parity for the first time.

You might begin by listing an some examples:

• Even Numbers: …, -6, -4, -2, 0, 2, 4, 6, 8, 10, …
• Odd Numbers: …, -3, -1, 1, 3, 5, 7, 9, 11, …

Or you might give instructions:

• Evens are found by counting by two’s in either direction from zero.
• Odds are found by counting by two’s in either direction from one.

Since we’ll be dealing with proofs, we’ll also need formal definitions.

Definition of Even

A number is even if it can be written as the product of two times an integer.

( Note: An integer is a number in the set: { …, -3, -2, -1, 0, 1, 2, 3, … } )

Definition of Odd

Odd numbers are always offset by one from an even number. So a number is odd if it can be written as the product of two times an integer plus 1.

Now let’s prove some basic results regarding number parity. If this is your first attempt at proofs, hang in there! They get easier with practice.

Proof 1: The Sum of Two Even Numbers is Even

Our game plan here is to define two anonymous even numbers and perform arithmetic with the goal of writing the sum in the form of an even number.

Proof:

Let m and n be two arbitrarily chosen even numbers such that m and n can be expressed as follows, where j and k are integers.

Take the sum of m and n.

Since both terms on the right are multiples of 2, rewrite the statement.

Both j and k are integers, so their sum is an integer. Let p = j + k, where p is an integer. Substitute p into the equation.

Since the sum of m and n is in the form of an even number, we conclude that the sum of any two even numbers is, in fact, an even number. QED

( Note: QED is latin for ‘quod erat demonstrandum’ that means ‘which is what had to be proven’. It is customary to end a proof with QED or a black box. )

Proof 2: The Sum of Two Odd Numbers is Even

Follow the same pattern as above to prove that the sum of two odds is even, but use the definitions of odd in place of even.

Proof:

Let m and n be two arbitrarily chosen odd numbers such that m and n are expressed as follows, where j and k are integers.

Take the sum of m and n.

Which simplifies to:

Since all the terms on the right hand side are multiples of 2, rewrite the above statement as follows:

Because j, k and 1 are all integers, their sum is an integer. Therefore, make the replacement: j + k + 1 = p, where p is an integer.

Since the sum of m and n is in the form of an even number, we conclude that the sum of any two odd numbers is, in fact, an even number. QED

Proof 3: Sum of an Even & Odd Number is Odd

Let m be even and n be odd numbers such that they are in the following forms, where j and k are integers.

Sum m and n.

Group 2j and 2k together.

Factor out 2 from the grouping.

Since j and k are integers, their sum is an integer. Make the replacement: p = j + k, where p is an integer.

Since the sum of m and n is in the form of an odd number, the sum of an even and odd number is, in fact, an odd number. QED

Proof 4: Product of Two Even Numbers is Even

Let m and n be two arbitrarily chosen even numbers such that m and n are expressed as follows, where j and k are integers.

Take the product of m and n.

Using commutativity, rewrite the product.

Since both j and k are integers, their product is an integer. Make the replacement: p = 2jk, where p is an integer.

Hence, the product of two even integers is even. QED

Proof 5: Product of Two Odd Numbers is Odd

Let m and n be two arbitrarily chosen odd numbers such that m and n are expressed as follows, where j and k are integers.

Take the product of m and n.

Multiply the binomials.

Group together all terms that are multiples of two.

Factor out 2.

Since j and k are integers, their product and sum remain integers. Therefore, make the replacement: p = 2jk + j + k, where p is an integer.

Since the product of m and n is in the form of an odd number, we conclude that the product of any two odd numbers is, in fact, an odd number. QED

More Proofs

Here’s a few similar-styled proofs to try for yourself. Let me know how they go!

• Prove the square of an odd number is odd.
• Prove the square of an even number is even.
• Prove the product of an odd and even number is even.

Next Lesson: Problems from New York’s Algebra 1 Exam

Thanks for reading!

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