# Two Questions: (1) Is My Proof Method Legitimate, and (2) How Might Proof by Cases be Accomplished?

Prove 5x-4 is even iff 3x+1 is odd.

I have two questions: First, for example, could we assume that 5x-4 is even such that

5x-4 = 2k, for some integer k

and then manipulate the above equation to produce 5x-4, which would be conducted as follows

5x-4 = (3x + 2x) + (-5 + 1) = 2k

3x + 1 = 2k - 2x + 5 = 2(k-x+2)+1,

which shows that 3x+1 is odd, since (k-x+2) is an integer.

Would the above sequence of steps be acceptable?

My second question is, How might this proof be demonstrated by cases? I understand that there are easier methods of proof, but we were asked in class to do a proof by cases.

The difference $(5x-4)-(3x+1)=2x-5$ is odd.

It's sickening to hear that you are asked "to do a proof by cases", instead of invoking the overriding principle: Find the invariant! There are enough sad problems (e.g., the four color theorem) where chasing cases is indispensable so far.

• Duh on the rest of us....... – Mathemagician1234 Feb 13 '15 at 19:53
• I would go further: asking for a particular method of proof is in general likely to be a mark of incompetence on the part of the instructor. If you want to test students’s command of a particular technique of proof, devise a problem for which it’s the most natural approach — and don’t penalize the students who come up with a different approach anyway. – Brian M. Scott Feb 13 '15 at 20:45

Proof by cases might look something like this:

• Suppose $x$ is even. Then, $5x-4$ is even. Similarly, if $5x-4$ is even, then $x$ is even. Not then that $3x+1$ is odd.
• Suppose $x$ is odd. Then, $5x-4$ is odd. Similarly, if $5x-4$ is odd, then $x$ is odd. Note then that $3x+1$ is even.

These arguments can be formalized by writing $x = 2k$ for $x$ even, and $x = 2k+1$ if $x$ odd, but suffice it to say that we may permit the argument "an odd times an even is even; an odd times an odd is odd."

We can write these statements as follows: $$x\ \textrm{even}\ \Longleftrightarrow 5x-4\ \textrm{even} \implies 3x+1\ \textrm{odd} \\ x\ \textrm{odd}\ \Longleftrightarrow 5x-4\ \textrm{odd} \implies 3x+1\ \textrm{even}$$

Apply the law of the excluded middle to these two cases to establish the bidirectional implication

$$x\ \textrm{even}\ \Longleftrightarrow 5x-4\ \textrm{even} \Longleftrightarrow 3x+1\ \textrm{odd}$$

Your steps look good for proving "$5x-4$ is even $\Rightarrow$ $3x + 1$ is odd", now you have to prove the other direction, that "$3x + 1$ is odd $\Rightarrow$ $5x-4$ is even". This may be what is meant by "cases*". (equivalently, you can prove the contrapositive, and show "$5x-4$ is not even (i.e. odd) $\Rightarrow$ $3x + 1$ is not odd (i.e. even)")

*: i.e. your case structure is something like:

Case 1: Assume $5x-4$ is even $\dots$

Case 2: Assume $5x - 4$ is odd $\dots$

This looks ok,but an easier way would be to set up a system of linear equations and try and solve the system. If it has a unique solution,then it should be the case that the solution of 3y+1 is odd and the solution of 5x-4 is even.Consider the following system: 5x-4 = m 3y+1 = n Now try and solve the system,preferably as a matrix. You should get an odd integer as the solution for 3y+1 and an even one for 5x-4.

Hint $\,\ 3x\!+\!1\,$ odd $\iff 3x\,$ even $\iff \overbrace{3x+ \color{#c00}2(x\!-\!2)}^{\large 5x-4}\,$ even, $\$ by $\$ even $\pm$ even $\,=\,$ even.