Problem 1 — Listing Even Numbers
Question: List all even numbers strictly between \(-4\) and \(20\).
Solution:
\(-2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18\)
Definitions of Even and Odd Numbers
Even numbers are integers divisible by \( 2 \) with no remainder.
Examples: \(\dots, -4, -2, 0, 2, 4, \dots\)
Algebraically, every even number can be written in the form: \[ 2n \] where \( n \) is an integer.
Odd numbers are integers that are not divisible by \( 2 \).
Examples: \(\dots, -5, -3, -1, 1, 3, \dots\)
Algebraically, every odd number can be written in the form: \[ 2n + 1 \] where \( n \) is an integer.
In the proofs below, each property is demonstrated by writing the numbers in these algebraic forms — the standard method for showing that a statement holds for all even or odd integers.
Worked Problems
Problem 2 — Listing Odd Numbers
Question: List all odd numbers strictly between \(3\) and \(30\).
Solution:
\(5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\)
Problem 3 — The Sum of Two Even Numbers is Even
Question: Show that the sum of two even numbers is always even.
Solution: Let \(2n\) and \(2k\) be two even numbers. Their sum is:
\[ 2n + 2k = 2(n + k) \]Setting \( N = n + k \), the sum is \( 2N \), which is a multiple of \( 2 \) and therefore even.
Problem 4 — The Sum of an Even and an Odd Number is Odd
Question: Show that the sum of an even number and an odd number is always odd.
Solution: Let \(2n\) be an even number and \(2k + 1\) be an odd number. Their sum is:
\[ 2n + (2k + 1) = 2(n + k) + 1 \]Setting \( N = n + k \), the sum is \( 2N + 1 \), which is the algebraic form of an odd number.
Problem 5 — The Sum of Two Odd Numbers is Even
Question: Show that the sum of two odd numbers is always even.
Solution: Let \(2n + 1\) and \(2k + 1\) be two odd numbers. Their sum is:
\[ (2n + 1) + (2k + 1) = 2n + 2k + 2 = 2(n + k + 1) \]Setting \( N = n + k + 1 \), the sum is \( 2N \), which is a multiple of \( 2 \) and therefore even.
Problem 6 — The Sum of Three Odd Numbers is Odd
Question: Show that the sum of three odd numbers is always odd.
Solution: Let \(2m + 1\), \(2n + 1\), and \(2k + 1\) be three odd numbers. Their sum is:
\[ (2m + 1) + (2n + 1) + (2k + 1) = 2m + 2n + 2k + 3 \]Break the \( 3 \) into \( 2 + 1 \):
\[ 2m + 2n + 2k + 2 + 1 = 2(m + n + k + 1) + 1 \]Setting \( N = m + n + k + 1 \), the sum is \( 2N + 1 \) — an odd number.
Problem 7 — The Square of an Odd Number is Odd
Question: Show that the square of an odd number is always odd.
Solution: Let \( 2n + 1 \) be an odd number. Squaring it gives:
\[ (2n + 1)^2 = (2n + 1)(2n + 1) = 4n^2 + 4n + 1 \]Factor out a \( 2 \) from the first two terms:
\[ 2(2n^2 + 2n) + 1 \]Setting \( N = 2n^2 + 2n \), the expression is \( 2N + 1 \), which is odd.
Problem 8 — The Product of an Odd and an Even Number is Even
Question: Show that the product of an odd number and an even number is always even.
Solution: Let \( 2m + 1 \) be an odd number and \( 2n \) be an even number. Their product is:
\[ (2m + 1)(2n) = 4mn + 2n = 2(2mn + n) \]Since the product is a multiple of \( 2 \), it is even.
Summary: Parity of Operations
| Operation | Result | Reason |
|---|---|---|
| Even + Even | Even | \( 2(n+k) \) |
| Even + Odd | Odd | \( 2(n+k)+1 \) |
| Odd + Odd | Even | \( 2(n+k+1) \) |
| Even − Even | Even | \( 2(n-k) \) |
| Even × Even | Even | \( 4np \) |
| Even × Odd | Even | \( 2(2mn+n) \) |
| Odd × Odd | Odd | \( 2N+1 \) |
| Even² | Even | \( 4n^2 \) |
| Odd² | Odd | \( 2N+1 \) |
Interactive Practice Questions
Complete each sentence using Even or Odd. Expand View Solution to reveal the algebraic proof.
1. The product of two odd numbers is ______.
Answer: Odd
Proof: Let \( 2n + 1 \) and \( 2p + 1 \) be two odd numbers. Their product is:
\[ (2n + 1)(2p + 1) = 4np + 2n + 2p + 1 = 2(2np + n + p) + 1 \]This fits the form \( 2N + 1 \), which is odd.
2. The product of two even numbers is ______.
Answer: Even
Proof: Let \( 2n \) and \( 2p \) be two even numbers. Their product is:
\[ (2n)(2p) = 4np = 2(2np) \]Since it is a multiple of \( 2 \), it is even.
3. The difference of two even numbers is ______.
Answer: Even
Proof: Let \( 2n \) and \( 2p \) be two even numbers. Their difference is:
\[ 2n - 2p = 2(n - p) \]The result factors out a \( 2 \), so it is even regardless of which number is larger (the difference may be negative, but it is still even).
4. The sum of two even numbers and one odd number is ______.
Answer: Odd
Proof: Let \( 2n \) and \( 2p \) be even numbers and \( 2m + 1 \) be an odd number. Their sum is:
\[ 2n + 2p + (2m + 1) = 2(n + p + m) + 1 \]This fits the form \( 2N + 1 \), making it odd.
5. The square of an even number is ______.
Answer: Even
Proof: Let \( 2n \) be an even number. Its square is:
\[ (2n)^2 = (2n)(2n) = 4n^2 = 2(2n^2) \]Because the result is a multiple of \( 2 \), it is even.
Frequently Asked Questions
What is an even number?
An even number is any integer divisible by \( 2 \) without remainder. Every even number can be written as \( 2n \) where \( n \) is an integer. Examples include \(\dots, -4, -2, 0, 2, 4, \dots\)
What is an odd number?
An odd number is any integer that is not divisible by \( 2 \). Every odd number can be written as \( 2n + 1 \) where \( n \) is an integer. Examples include \(\dots, -5, -3, -1, 1, 3, \dots\)
Is the sum of two odd numbers even or odd?
Even. If \( 2n + 1 \) and \( 2k + 1 \) are two odd numbers, their sum is \( 2n + 2k + 2 = 2(n + k + 1) \), which is a multiple of \( 2 \).
Is the product of two odd numbers even or odd?
Odd. If \( 2n + 1 \) and \( 2p + 1 \) are two odd numbers, their product is \( 4np + 2n + 2p + 1 = 2(2np + n + p) + 1 \), which is of the form \( 2N + 1 \) and is therefore odd.
Is zero even or odd?
Zero is even. It is divisible by \( 2 \) with remainder \( 0 \), and it can be written as \( 2 \times 0 \).
Is the square of an even number even or odd?
Even. If \( 2n \) is even, then \( (2n)^2 = 4n^2 = 2(2n^2) \), which is a multiple of \( 2 \).
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Last updated: October 17, 2026