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you can use the Euclidean Algorithm to find the gcd

Let's work through the problem step by step using the Euclidean Algorithm.

### Problem Breakdown
We are given:
1. An integer \( a \) is even if \( a = 2k \) for some integer \( k \).
2. An integer \( a \) is odd if \( a = 2k + 1 \) for some integer \( k \).

We need to prove:
- \( \gcd(a, a+2) = 1 \) if \( a \) is odd.
- \( \gcd(a, a+2) = 2 \) if \( a \) is even.

### Proof
We'll consider both cases for \( a \): odd and even.

#### Case 1: \( a \) is Odd
If \( a \) is odd, we can write \( a = 2k + 1 \) for some integer \( k \). Then:
\[
a + 2 = (2k + 1) + 2 = 2k + 3.
\]
Thus, we have:
- \( a = 2k + 1 \) (an odd number),
- \( a + 2 = 2k + 3 \) (also an odd number).

Now we want to find \( \gcd(a, a+2) = \gcd(2k+1, 2k+3) \).

Using the Euclidean Algorithm:
\[
\gcd(2k+1, 2k+3) = \gcd(2k+1, (2k+3) - (2k+1)) = \gcd(2k+1, 2).
\]
Since \( 2k+1 \) is odd, it is not divisible by 2, so the only common divisor is 1. Therefore:
\[
\gcd(a, a+2) = 1.
\]
This proves the first part: if \( a \) is odd, \( \gcd(a, a+2) = 1 \).

#### Case 2: \( a \) is Even
If \( a \) is even, we can write \( a = 2k \) for some integer \( k \). Then:
\[
a + 2 = 2k + 2 = 2(k + 1).
\]
Thus, we have:
- \( a = 2k \) (an even number),
- \( a + 2 = 2(k + 1) \) (also an even number).

Now we want to find \( \gcd(a, a+2) = \gcd(2k, 2k+2) \).

Using the Euclidean Algorithm:
\[
\gcd(2k, 2k+2) = \gcd(2k, (2k+2) - 2k) = \gcd(2k, 2).
\]
Since \( 2k \) is divisible by 2, we can factor out 2 from both terms:
\[
\gcd(2k, 2) = 2.
\]
Therefore:
\[
\gcd(a, a+2) = 2.
\]
This proves the second part: if \( a \) is even, \( \gcd(a, a+2) = 2 \).

### Conclusion
We have shown that:
- If \( a \) is odd, \( \gcd(a, a+2) = 1 \).
- If \( a \) is even, \( \gcd(a, a+2) = 2 \).

Would you like more details on any part of this solution, or have any questions?

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Here are five related questions that might extend your understanding:

1. What is the Euclidean Algorithm and how is it used in finding the greatest common divisor?
2. How does the parity (odd or even nature) of numbers affect the gcd calculations?
3. Can you find the gcd of numbers with a larger difference, such as \( a \) and \( a+4 \)?
4. What would the gcd of \( a \) and \( a+1 \) be in both odd and even cases?
5. How does modular arithmetic simplify the proof of gcd properties?

**Tip:** When proving properties involving gcd, consider using the Euclidean Algorithm, as it simplifies finding common factors between two numbers.

An integer a is even if a = 2k for some integer k. Likewise, a is odd if a = 2k + 1 for some integer k. Prove that gcd(a, a + 2) = 1 if a is odd and gcd(a, a + 2) = 2 if a is even.

Explore how to prove that gcd(a, a + 2) equals 1 when a is odd and equals 2 when a is even. This proof uses the Euclidean Algorithm and explores the parity of integers, suitable for students in Grades 11-12.

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