Algebraic Proof with Odd, Even and Consecutive Numbers
Prove a general claim about integers by naming them algebraically, expanding, and showing the required factor is always there.
What a learner can do afterwards
- Write any even number, any odd number and consecutive integers in terms of n
- Prove that the product of two consecutive integers is even, showing the factor explicitly
- Say why an algebraic argument settles a claim that no number of tested cases can
1 · Read
Proofs start by writing numbers in general form. Any even number is 2n, any odd number is 2n plus 1, and consecutive integers are n and n plus 1. These definitions turn wordy claims into algebra you can push around.
To prove the product of two consecutive integers is even, write n times n plus 1. Neighbours always split odd and even, so one factor is even. Pull the 2 out front, as in 7 times 8 equals 2 times 28. That visible 2 is the whole proof.
Sums work the same way. Two consecutive integers add to n plus n plus 1, which is 2n plus 1, the odd form. Equivalent expressions like 2n plus 2 and 2 times n plus 1 match for every n, so swapping them is always safe.
No count of tested cases proves a claim for all integers, since cases never cover infinity. Algebra over n settles every case at once. Try it: three consecutive integers sum to 3n plus 3, which is 3 times n plus 1, always a multiple of three.
Name numbers with n, display the factor the claim needs, and let the algebra cover every integer at once.
2 · Watch
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Where this leads
Jobs that lean on this skill. Follow one to see everything it is built on.
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.