Seven weeks ago, my wife Wendryn and I brought home our daughter, Katja, from the hospital. Since Katja was born, Wendryn has been going to work in the morning, and I have been going to work in the afternoon. Every once in a while, I try to get some work done while I am at home in the morning. It typically doesn’t go that well.
Suppose that I am trying to prove a theorem of some kind. For the sake of simplicity, I’ll use an elementary example—suppose that I am trying to show that the number \(\sqrt{2}\) is irrational (i.e. that \(\sqrt{2}\) cannot be written as a fraction):
Theorem: \(\sqrt{2}\) is irrational.
Proof: For contradiction, suppose that \(\sqrt{2}\) is rational. That is, suppose that \(\sqrt{2} = p/q\), where—
Oh! Katja! What’s the matter? We just changed your diaper, so that can’t be the problem. Let’s see… you last ate about two hours ago—maybe we should try feeding you again. Ah! Yes, there it is.
Man, you are a messy eater. And a burp!
Of course… now that you’ve eaten, you’re awake. How about some tummy time? Look! Baby push-ups! Man, you’re going to be crawling in no time! Oh, okay. I’ll stop tapping your nose. And maybe I’ll put you in the swing for a while, since you seem to be bored with the floor.
Right. Back to work.
Okay, where was I? Let’s see, I’ve got a \(p\) and a \(q\). What are they? Oh, right. Integers. Oh, of course. I’m looking for a contradiction. If we assume that \(\sqrt{2}\) is rational, it should pop out. Okay.
That is, suppose that \(\sqrt{2} = p/q\), where \(p,q\in\mathbb{Z}\). Then, squaring both—
Uh oh! Crying again, eh child? Well, it has been a few hours since we changed your diaper. Let’s take care of that now. Better? Excellent.
Oh, you don’t want to settle down? Hrm. Swing? Nope. Bouncy chair. Nyet. Rocking chair? No. Okay, I’ll just hang on to you for a while. Maybe a pacifier will help…
…or maybe it will just make you angrier. Got it. We’re just going to pace for the next hour. Groovy. I can dig it.
Yay! Sleeping baby. Let’s put you down, and see if I can finish that proof.
Now then, \(\sqrt{2}\) is rational, and expressed as a fraction. Where’s the problem? Oh, of course!
Then, squaring both sides of the equation, we have \(2 = p^2/q^2\). Moving the \(q^2\) term to the other side, we have \(2q^2 = p^2\). By the fundamental theorem of—
Arg! What now? Oh, man! It’s 10:30! You need to eat again. Where has the morning gone? Bottle, bib, rocking chair. Mmmm… rocking chair. That’s comfy.
This time, I think I’ll change your diaper now. Then I might have a whole hour to finish up. You know, Snurfles, this would go a lot faster if you relaxed, and let me get you changed. There it is. Great. Fed baby, clean diaper. Maybe I can finish up now.
So… “by the fundamental theorem of…?” What? Where was I going with that? Oh! Right! Prime factorizations. The problem is with uniqueness. Let’s follow through on that.
By the fundamental theorem of arithmetic, \(q = 2^{d_1}\alpha\) and \(p = 2^{d_2}\beta\), where \(d_1,d_2\in\mathbb{N}\) and 2 does not divide \(\alpha\) or \(\beta\). But then \(2q^2 = p^2\) implies that \(2^{2d_1+1}\alpha^2 = 2^{2d_2}\beta^2\). But then, as—
Son of a—! I was so close to being done! Can you let me finish? Please? No, of course not. What can I do for you now, darlin’? Your diaper is dry, you at half an hour ago… Maybe if we suction out your nose?
Where could the SnotSucker 2000TM be? Ah, here it is. Heh. Buzz-buzz, snort-snort. Hold still, this’ll be over in a minute. I know, it sucks (heh). There! I got a booger! Better? Yes!
What the? \(\alpha\)? Where the heck did that come from? Um… oh, I see. It is the remainder term when all powers of 2 are factored out of \(q\). And \(\beta\) plays the same role for \(p\). I think I see where this was going. Let’s just finish it up…
But then, as prime factorizations are unique, it follows that \(2d_1+1 = 2d_2\). This is a contradiction, as the left-hand side is odd, and the right-hand side is even. Thus there are—
Oh, crap! It’s 11:45! Time to go pick up Wendryn at work. I guess I’ll finish the proof later…
So it goes. :\
She’ll be helping you out with those proofs in no time. Just you wait.
She’s in the office with me right now, “helping” with my topology. ;)
xander
Xander, You are very funny and of course I noticed that Katja always comes first and that is the way it is. I am so happy for your new family. How is the doggy by the way?
Auntie, Suzanne
It’s always nice to have a reminder of why I did not procreate (even though I admit to going through the motions several times). Me + offspring would = a severely irrational me. Keep up the good work (somebody’s gotta do it). With love to you both, great-Uncle Tom (cabin optional).
Suzanne: The dog is fine. She and Katja are getting along very well.
xander