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``mcs'' --2015/5/18 --1:43 -- page 15 --\# 23
{\it 1.7. Proof by Cases 15}
Now since zero is the only number whose square root is zero, equation $()$ holds iff
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$(x_{1}-\mu)^{2}+(x_{2}-\mu)^{2}+\cdots+(x_{n}-\mu)^{2}=0$. (1.5)
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Squares of real numbers are always nonnegative, so every term on the left hand side of equation $()$ is nonnegative. This means that $()$ holds iff
Every term on the left hand side of $()$ is zero. (1.6) But aterm $(x_{i}-\mu)^{2}$ is zero iff $ x_{i}=\mu$, so $()$ is true iff
Every $x_{i}$ equals the mean.
$$
\blacksquare
$$
1.7 Proof by Cases
Breaking a complicated proof into cases and proving each case separately is a com- mon, useful proof strategy. Here's an amusing example.
Let's agree that given any two people, either they have met or not. If every pair of people in a group has met, we'll call the group a {\it club}. If every pair of people in a group has not met, we'll call it a group of {\it strangers}.
Theorem. {\it Every collection of 6 people includes a club of 3 people or a group of 3 strangers}.
{\it Proo}. The proof is by case analysis. Let $x$ denote one of the six people. There are two cases:
1. Among 5 other people besides $x$, at least 3 have met $x.$ 2. Among the 5 other people, at least 3 have not met $x.$
Now, we have to be sure that at least one of these two cases must hold, but that's easy: we've split the 5 people into two groups, those who have shaken hands with $x$ and those who have not, so one of the groups must have at least half the people. Case 1: Suppose that at least 3 people did meet $x.$
This case splits into two subcases:
5Describing your approach at the outset helps orient the reader.
6Part of a case analysis argument is showing that you've covered all the cases. This is often obvious, because the two cases are of the form $P$'' and ``not $P$ However, the situation above is not stated quite so simply.
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