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**Prove that the sum of any 3 consecutive integers is always a multiple of 3.**

The required insight is to use the way the three numbers are related to each other - ie. that they are consecutive. Since the three numbers are consecutive, the second is

*necessarily*one more than the first and the third

*necessarily*two more than the first.

This means starting with

*as few arbitrary elements as possible*.

*necessary*properties of algebraic equivalence - that repeated addition can

*necessarily*be written as multiplication (ie, terms can be collected) and that an expression with a common factor can

*necessarily*be factorised (ie. the distributivity law) thus proving that the sum is a multiple of 3.

The lens of

*arbitrary and necessary*thinking is important here: to successfully build a proof through a chain of deductive reasoning, the minimum number of

*arbitrary*elements should be used. Much more importantly, necessary, consequential,

*must-be*mathematical properties must be teased out and used to produce the argument. In the arbitrary there is doubt, but in the necessary there is certainty.

Thus it is to me, that to

*think mathematically*(and proof is surely the ultimate form of mathematical thinking for the pure mathematician) is to use

*arbitrary*elements sparingly and with caution, and always to look for the

*necessary, must-be*properties of the situation.

Agree with comments - and I always use numicon shapes as a great visual

ReplyDelete@NMPartnershipUK pip@nationalnumeracy.org.uk