Proof by contrapositive involves proving that if the conclusion is false, then the premise must also be false.

The contrapositive of the statement "if a^3 is even, then a - 1 is odd" is "if a - 1 is not odd (i.e., even), then a^3 is not even (i.e., odd)".

Here's the proof:

1. Assume that a - 1 is even. This means that a - 1 can be written as 2k for some integer k (since all even numbers can be expressed as 2 times an integer).

2. Then, a = 2k + 1.

3. The cube of a is (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1.

4. The expression 2(4k^3 + 6k^2 + 3k) + 1 is of the form 2m + 1, where m = 4k^3 + 6k^2 + 3k is an integer.

5. Numbers of the form 2m + 1 are odd.

6. Therefore, if a - 1 is even, then a^3 is odd.

This proves the contrapositive, and hence the original statement "if a^3 is even, then a - 1 is odd" is also true.

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Proof by contrapositive involves proving that if the conclusion is false, then the premise must also be false.

The contrapositive of the statement "if a^3 is even, then a - 1 is odd" is "if a - 1 is not odd (i.e., even), then a^3 is not even (i.e., odd)".

Here's the proof:

1. Assume that a - 1 is even. This means that a - 1 can be written as 2k for some integer k (since all even numbers can be expressed as 2 times an integer).

2. Then, a = 2k + 1.

3. The cube of a is (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1.

4. The expression 2(4k^3 + 6k^2 + 3k) + 1 is of the form 2m + 1, where m = 4k^3 + 6k^2 + 3k is an integer.

5. Numbers of the form 2m + 1 are odd.

6. Therefore, if a - 1 is even, then a^3 is odd.

This proves the contrapositive, and hence the original statement "if a^3 is even, then a - 1 is odd" is also true.

Knowee AI · Accepted Answer

Proof by contrapositive involves proving that if the conclusion is false, then the premise must also be false.

The contrapositive of the statement "if a^3 is even, then a - 1 is odd" is "if a - 1 is not odd (i.e., even), then a^3 is not even (i.e., odd)".

Here's the proof:

1. Assume that a - 1 is even. This means that a - 1 can be written as 2k for some integer k (since all even numbers can be expressed as 2 times an integer).

2. Then, a = 2k + 1.

3. The cube of a is (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1.

4. The expression 2(4k^3 + 6k^2 + 3k) + 1 is of the form 2m + 1, where m = 4k^3 + 6k^2 + 3k is an integer.

5. Numbers of the form 2m + 1 are odd.

6. Therefore, if a - 1 is even, then a^3 is odd.

This proves the contrapositive, and hence the original statement "if a^3 is even, then a - 1 is odd" is also true.

Use a proof by contrapositive to show that for all integers a ∈ Z, if a3 is even, then a − 1 is odd. Justify all of yoursteps.

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