**Linda in Northfield, ***Mensagenda* Editor

*Mensagenda*Editor

## About Mensagenda

Minnesota Mensa published Vol. I, No. 1 of our newsletter, then called the *Minnesota Mensa*, in June of 1965. Approaching six decades later and winning awards along the way, we continue to provide a monthly publication, now called *Mensagenda*.

As expected in a newsletter, we inform our local membership with organizational updates and provide details about our events. The real benefit is that, just like our events, *Mensagenda* is for our members, by our members.

The love of learning in Mensa is not just about supporting our scholarship but in enriching your own mind and sharing your knowledge, skills, and interests. Read articles and regular columns ranging from scientific explanations to humor in everyday life. Check out our members’ photography, drawing, painting, knitting and quilting, and crafting skills.

What would you like to share? Do you have expertise in a particular field of study or hobby? Want to express your opinion? Have you traveled recently? Do you write poetry? Can you create word games, numerical puzzles, or trivia questions? What could you say about…well, you get the picture.

*Mensagenda* is another way that Minnesota Mensa provides “a stimulating intellectual and social environment for its members.” What could you contribute if you joined Mensa?

#### There’s More to Read

Mensa membership provides access to the publications from other chapters, American Mensa, and Mensa International. Click here to learn more.

## Featured Cover Art

May/December Courtyard. Photo Montage by Linda in Northfield.

I get really tired of looking out of my window at nothing except a gray snowy courtyard for months on end. And my windows face east, so for three months or so around the solstice I get no direct sunlight at all, because the sun is so low my patio door is always in the shadow of the south wing (to my right).

This winter, I decided, I needed something different to look at. In May, I took a photo of the courtyard trees in full leaf. I found a company online that will print whatever you like on translucent film. Like a piece of kitchen cling wrap, if you smooth it onto a window, it just stays there—hence, “window cling.”

Businesses use these to post their hours or their daily specials in their windows, facing out so they can be read by those passing by.

This is what I see now, and the source of the cover photos (the “December” photo was actually taken Oct. 31, when the first snow fell). I find green trees and blue sky more cheering in these dark days.

## Featured Article

On a Whim by Mat in Vadnais Heights.

I seem to be hearing it more and more these days: “You can’t prove a negative.” Usually by folks who go on to use the statement as justification for “you can’t *prove *Bigfoot doesn’t exist” or “you can’t *prove *flying saucers don’t exist.” Sometimes I try to be polite, but in truth this absolutely sets my teeth on edge. Not because the specific examples cited above are silly, but because the assertion that “you can’t prove a negative” is totally, unequivocally, demonstrably *wrong*. And I will demonstrate this to you right now.

So, what am I going to do? Maybe prove that there *isn’t *a silver teapot full of mashed bananas orbiting a red dwarf star somewhere in the Andromeda galaxy? No, I shall leave that one alone. Instead, I will prove to you—big dramatic pause —that the square root of two is *not *a rational number.

A good number of you are tempted to roll your eyes and stop reading at this point, but really, It’s the same kind of question, isn’t it? What is a rational number? Simply a number that can be expressed as the ratio of two integers— like *2/3 *or *5/6 *or *22/7*. I claim that the square root of two (that is, the number that, multiplied by itself, equals two) *cannot *be represented by a ratio of *any *two integers. Stated that way, proving it sounds like a tall order. After all, there is an *infinite *number of integers. How can I say for certain that there does not exist a pair of them somewhere up in the squillions—let’s call them *p *and *q*—such that *p/q *is *exactly *equal to the square root of two (which I will henceforth refer to as “*root-two*” for brevity)? My extragalactic teapot problem seems almost preferable.

To spoil the ending, we do it with a tried-and-true style of mathematical proof called a *reductio ad absurdum*, or “reduction to absurdity.” We start by assuming the *opposite *of what we want to prove and show that doing so *always *leads to a logical contradiction. Since contradictions are the ultimate no-no in mathematics, any assumption that generates one *must be false*.

So, let’s begin. I promise this will involve nothing more advanced than arithmetic concepts you learned in high school or earlier … and an ability to focus for a few minutes even if you don’t like the subject matter.

As I stated above, we start by assuming the opposite of what we want to prove. That is, we assume that there *are *two integers *p *and *q *such that *p/q *equals *root-two*. Now we work through the implications.

Step one: If our *p *and *q *have any common factors, remove them before continuing. For example, say integers *p *= *9 *and *q *= *6 *make up our potential *p/q*, yielding *9/6*. Since *9 *and *6 *are both divisible by three, we divide both *p *and *q *by that common factor, redefining them as *p *= *3 *and *q *= *2*. We can get away with this because removing common factors always leaves a ratio with the same value (*9/6 *is precisely equal to *3/2*, in our example). That’s a long-winded explanation of common factors, but I want to avoid any cries of foul play later on.

Step 2: Square both sides of the equation: *p/q = root-two *becomes *p2/q2 = 2*.

Step 3: Multiply both sides by *q2*, in other words *p2/q2 = 2 *becomes *p2 = 2q2*. With me so far? Let’s pause here and take stock. *p2 = 2q2 *means that p2 is an *even *number (because it can be represented as some other integer—*q2 *in this case—multiplied by two). And if *p2 *is even, then *p *must also be even, because any even integer squared is *always *another even integer, and any odd integer squared is *always *odd. So … . Interim conclusion No. 1: *p *is an even number.

Step 4: Since *p *is even, we can define it as *2r*, where *r *is another integer (equal to half *p*’s value, duh). This seems obvious and rather pointless, but it’s important. Trust me. So… *p = 2r. *

Step 5: Square that last equation: *p = 2r *becomes *p2 = 22r2 *or, more simply, *p2 = 4r2*.

Step 6: Look back at step 3 above. From that we see that *p2 = 2q2*, right? But from step 5 we also know that *p2 = 4r2*. So, we know *two different ways *to define *p2*. Cutting out the *p2 *middleman, we can say *2q2 = 4r2*. Got that?

Step 7: Really simple. Divide both sides of that equation by two: *2q2 = 4r2 *becomes *q2 = 2r2*.

Similar to what we did after step 3 above … *q2 = 2r2 *means that *q2 *is an even number. And since *q2 *is even, we know that means *q *is even. So … .

Interim conclusion No. 2: *q *is an even number.

Well, so what? All we’ve shown is that if *p/q *= *root-two *then *both p *and *q *must be even. *But… *way back in step 1 we specified that *p *and *q *have been *reduced to have no common factors*. If they are both even, they have at least *one *common factor (namely *2*). We have arrived at contradictory statements: “*p *and *q *have *no *common factors” and also “*p *and *q *have *at least one *common factor.” These statements cannot both be true. That means that our initial assumption, that *root-two *can be represented as the ratio of two integers *p *and *q*, *must be false*.

Conclusion: *root-two cannot *be represented as the ratio of *any *two integers. QED.

There, I have successfully *proven a negative*. And it’s not like I invented this all by myself. Hippasus of Metapontum proved that *root-two *is not rational way back in the fifth century BCE. Legend has it that Pythagoras, who was a big believer in ratios, had Hippasus *sentenced to death by drowning *for his heretical discovery. Some people really can’t handle being wrong.