SUNY Geneseo Department of Mathematics
Friday, March 12
Math 239 03
Spring 2021
Prof. Doug Baldwin
When you tell me about the contrapositive Canvas discussion’s proof, you don’t need to have it formally written up. I just want to hear your summary of it so that I know you followed and generally understood what was happening in the discussion.
A question came up in a recent meeting about how to make tables in LaTeX. Since I can’t find any examples in our LaTeX samples in Canvas, here’s one:
The key idea is the tabular environment, which defines data to be arranged in a table.
That environment has a second argument, in curly braces after \begin{tabular}, which specifies the columns and any separators between them. It’s a string of characters, the most common of which include “|” (a vertical bar) for a vertical line, typically used to separate columns, “l” (the letter “ell”) for a left-aligned column of data, “c” for a centered column of data, and “r” for a right-aligned column of data.
Write the data for the table inside the tabular environment, separating columns with an “&” character and rows with two backslashes (\\).
In a table, the \hline command is useful to insert a horizontal separator line.
You can center a table (or anything else) by placing it inside a center environment.
We looked at a complete example in LaTeX, whose source code and resulting output are available in Canvas. (This example is one that I originally made to demonstrate tables, but that then turned out to be convenient to also use for examples of a proof via the contrapositive, and a proof of a biconditional, so it has several things in it. The table is in the middle of the document.)
From section 3.2 in the textbook.
Prove that for all integers n, n is even if and only if n+1 is odd.
The key idea is to treat this as two conditionals, and prove each one separately. In other words, prove (1) n+1 is odd if n is even, and (2) n is even if n+1 is odd.
We wrote this as a formal proof in LaTeX. You can download the source code and PDF output from Canvas. (This is the document that has multiple examples in it; the biconditional proof is the last one.)
(This is my name for something that sometimes makes groups of biconditionals easier to prove; I don’t know if there is an “official” name for it.)
Sometimes you get a set of biconditional statements of the form “P if and only if Q, Q if and only if R, … if and only if Z.”
Rather than doing 2 proofs for each biconditional separately, you can do roughly half as much work if you do a series of related proofs that P implies Q, Q implies R, and so forth until finally you prove that Z implies P.
For example, prove that for all non-zero real numbers x,
1/x < 1 implies x > 1 because multiplying both sides by x (since x ≠ 0) gives 1 < x or x > 1.
x > 1 implies x3 > x2 because multiplying both sides by x2 gives x3 > x2.
x3 > x2 implies 1/x < 1 because dividing by x3 gives 1 > 1/x or 1/x < 1.
Most biconditionals need separate proofs for the “if” and “only if” directions. But sometimes you get a situation in which the logic works both forward and backwards, i.e., each implication in one direction is really an “if and only if,” and so you can do both directions of the proof in a single argument (but be careful that the logic really does work both ways).
For example, prove that for all real numbers x, x = 5 if and only if 2x + 1 = 11.
x = 5 if and only if 2x = 10 which holds if and only if 2x + 1 = 11.
Why can you not use a similar tactic to prove that for all real numbers x, x = 5 if and only if x2 = 25?
Because x = 5 implies x2 = 25, but that “implies” cannot be replaced with “if and only if.”
Another widely useful proof technique, proof by contradiction.
Please read section 3.3 of the textbook up to but not including “Rational and Irrational Numbers.”
Please also start practicing proofs by contradiction in this discussion.