Math Problem Statement

If a number n is odd, then n squared is even.

Solution

The statement "If a number nn is odd, then n2n^2 is even" is false.

In fact, if nn is odd, then n2n^2 will also be odd.

Here’s why:

Proof:

  1. An odd number can be expressed in the form n=2k+1n = 2k + 1, where kk is an integer.
  2. Now, square nn: n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1
  3. This shows that n2n^2 can be written as 2×(an integer)+12 \times \text{(an integer)} + 1, which is the definition of an odd number.

Thus, if nn is odd, then n2n^2 will also be odd, not even.

Correct statement:

If nn is odd, then n2n^2 is odd.

Would you like further details or clarifications on this concept?


Here are 5 related questions to explore:

  1. What happens if nn is even? Is n2n^2 even or odd?
  2. Can you prove that the sum of two odd numbers is always even?
  3. How does squaring affect other types of numbers, such as rational or irrational numbers?
  4. How can we extend the idea of odd and even numbers to modular arithmetic?
  5. Is there a pattern in the powers of odd numbers (e.g., n3n^3, n4n^4)?

Tip: When working with odd and even numbers, expressing them in the form 2k2k (even) or 2k+12k + 1 (odd) helps simplify proofs.

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Math Problem Analysis

Mathematical Concepts

Algebra
Odd and Even Numbers
Number Theory

Formulas

n = 2k + 1
n^2 = (2k + 1)^2 = 2(2k^2 + 2k) + 1

Theorems

Odd Numbers Theorem
Properties of Odd and Even Numbers

Suitable Grade Level

Grades 7-9