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Malhar Rajpal Mathematicians pride themselves on being extremely exact and reliable with their findings. Almost always, their papers and discoveries are fully proved with no assumptions whatsoever. This contrasts most other STEM subjects, where findings are generally based on experimentation and thus have some degree of uncertainty and assumptions. You may wonder, ‘what methods of proof do mathematicians use to be absolutely confident on the validity of their findings?’. In this article, I will discuss what I think is the most beautiful and pure method of proof.
Direct proofs, or proofs by deduction, are often considered the purest form of proof since they use already proved statements or axioms to work forward and directly prove a new conjecture (a mathematical hypothesis). Let’s explore this through an example: Conjecture (statement to prove): n^3 - n is divisible by 6 for all integers n. Lemma 1 (already proven argument): if an integer is divisible by both 2 and 3, it must be divisible by 6 Thus we have to show that n^3 - n is divisible by both 2 and 3. We can begin by simply rearranging n^3 - n = n(n^2 - 1) by factoring out the n. Lemma 2: The difference of squares rule states that for some real numbers a and b, a^2 - b^2 = (a-b)(a+b). This can be shown by simple expansion of (a-b)(a+b). Thus by lemma 2, n^2 - 1 = (n-1)(n+1) and n(n^2-1) = n(n-1)(n+1). By rearranging, n(n-1)(n+1) = (n-1)n(n+1), which is the product of the three consecutive integers n-1, n, and n+1. Helper Conjecture: The product of three consecutive integers is divisible by 3. Let us prove this conjecture. In any set of three consecutive integers, one will be divisible by 3, as multiples of 3 appear every 3 consecutive integers. Since one of the three integers contributing to the product is divisible by 3, the product will have a factor of 3, and therefore be divisible by 3 itself. Helper Conjecture 2: The product of three consecutive integers is divisible by 2 Any set of two consecutive integers will have one number that is divisible by 2, by the same logic as above. Therefore, in a set of three consecutive integers, there will be at least one number that is even (divisible by 2). So their product will have at least one factor of 2, and will thus itself by divisible by 2. We have proved that n^3 - n = (n-1)(n)(n+1), which is the product of the three consecutive integers, n-1, n, and n+1. We have also shown that the product of three consecutive integers is divisible by both 2 and 3. Thus by using lemma 1 which states that any number n that is divisible by 2 and 3 is also divisible by 6, it is thus proven that the product of three consecutive integers is divisible by 6 and thus (n-1)(n)(n+1) must also be divisible by 6, for all integers n. Since n^3 - n = (n-1)(n)(n+1), and (n-1)n(n+1) is divisible by 6, our initial conjecture stating that n^3 - n is divisible by 6 for all integers n is proven. And since mathematicians like to end proofs with this sign, I will do it here too: Q.E.D.
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