Math Problem Statement
Let n be an integer. Show that if n2 is a multiple of 3, then n must also be a multiple of 3.
Solution
We want to show that if is a multiple of 3, then must also be a multiple of 3. We'll use a proof by contrapositive and modular arithmetic.
Proof by Contrapositive:
We will prove the contrapositive statement:
"If is not a multiple of 3, then is not a multiple of 3."
The contrapositive is logically equivalent to the original statement. So, if we can prove this, the original statement will be true.
Step 1: Modular Arithmetic and Cases
If is not a multiple of 3, then can either be congruent to 1 or 2 modulo 3. In other words, we consider the following cases:
We will square each case and see if can be divisible by 3.
Case 1:
So, if , then , which means is not divisible by 3.
Case 2:
So, if , then , which again means is not divisible by 3.
Step 2: Conclusion
In both cases, when is not a multiple of 3, is also not divisible by 3. Hence, the contrapositive is true, and therefore, the original statement is also true:
This completes the proof.
Would you like a deeper explanation of any part? Here are 5 related questions to explore:
- How does modular arithmetic work for different numbers, not just 3?
- Can we generalize this result for other prime numbers like 5 or 7?
- What other proofs can be made using the contrapositive method?
- How does the original assumption help us solve problems in number theory?
- Can this proof method work for non-integer values like rational numbers?
Tip: When tackling divisibility problems, modular arithmetic is a powerful tool for simplifying and solving congruence relations.
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Math Problem Analysis
Mathematical Concepts
Number Theory
Modular Arithmetic
Proof by Contrapositive
Formulas
n^2 ≡ 1 (mod 3)
n^2 ≡ 4 (mod 3)
Theorems
Contrapositive Proof
Divisibility by 3
Suitable Grade Level
Grades 9-12
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