Proof that n^3 - n is divisible by 3 using Mathematical Induction

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In this video, I demonstrate how to use mathematical induction to prove that n^3 - n is divisible by 3 for all integers, n, that are greater than or equal to 2.

Mathematical induction is usually a mechanical process. There are really only 4 steps involved:

1. Test if the expression holds true for a particular value of n. In this case n = 2

2. Assume that it holds true for n = k

3. Use the above to prove that it is valid for n = k + 1

4. Conclude that the expression holds true for all values of n of a given condition.

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n^3 - n = n (n^2-1) = n (n-1)(n+1) = (n-1)(n)(n+1) = product of three consecutive integers
Product of three consecutive integers is always divisible by 3. In fact it is also divisible by 6. The result is also true for n =1 since 3 divides 0

cipherunity

My tutor taught me to do something very similar but for my induction hypothesis instead of keeping it has k^3 - k = 3j, he would normally take the - k and get it on the other side of the equation, so it would be k^3 = 3j + k . How would that have worked in this problem, because I can't seem to get it to make sense when I attempt to do it like that. I end up with..

3j + k + 3k^2 + 3k

Which would be.. 3(J + k^2 + k) + K .. right?

kzuik

Thank you so very much for this step by step explanation. It most certainly helped

valeriereid

thank some much for ur wonderful lecture 🙏🙏🙏🙏

mayieljrgobroh

very good but also very elementary. at first look its divisible by 3 because its a product of consecutive numbers and also by n. therefore it must be that n^3-n with > 1 is divisible by 3n.

greatkingkay

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