   SEARCH HOME Math Central Quandaries & Queries  Question from Taylor, a student: when doing a proof, how do i figure out the steps in which i find the statements? i find the reasons pretty easily but i do not understand how to get the proving part. that would be great if you can help me! Thanks Taylor,

Mathematics is often much more of an art than a science. The harsh truth is that there is no way to figure out what steps to use, or what order they should be in.

That being said, there are some good guidelines. One of them is to keep asking "what does that mean?" after every step in a proof. This includes the situation when you're looking at what you need to prove. Another one is that proofs don't always start at the beginning (only the write up does). If you look at where you need to get to, and can say "if I could get THIS, then I would be done", then you might have reduced your problem by one step. The reduced problem might be easier to solve. A third helpful suggestion is to ask "what do we know that might help?". A fourth suggestion is to ask "how do I get from here to there?"

Here is an example. Let's prove that the product of two odd integers is odd.

Suppose we are given two odd integers m and n. (What does that mean?) Then we know that m = 2k + 1 and n = 2t + 1 for some integers k and t.

We need to show that mn is odd. (What does that mean?) We need to be able to write mn = 2r + 1 for some integer r. (If I could get THIS, then I would be done.)

The product mn = (2k+1)(2t+1) = (2k)(2t) + 2k + 2t + 1 = 2(2kt + k + t)+1. Since (2kt + k + t) is an integer, we are done.

Here is another example. Let's prove that 1 + tan2(x) = sec2(x).

(What does that mean?) We know that tan(x) = sin(x)/cos(x), and sec(x) = 1/cos(x). So, we need to prove that 1 + (sin(x)/cos(x))2 = 1/cos(x)2. (If I could get THIS, then I would be done.)

(What do we know that might help?) We know that sin(x)2 + cos(x)2 = 1.

(How do I get from here to there?) If we divide both sides by sin(x)2, we get 1 + tan(x)2 = sec(x)2, as desired.

Best of luck.
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