Double Your Salary Using an Algebraic Loophole
Whoever it was that said you could get a number to say anything was absolutely right. What he didn't say, however, was how enormously therapeutic it is to get them to say what you want. So, enjoy the satisfying process of "proving" mathematically that you really make $200,000/year, have a 10000 square-foot home, and are as fit as a fiddle by following these steps.
Steps
- Roll up your sleeves. This accomplishes more than just the practical advantage of not getting your sleeves dirty and smudged with pencil lead. It lets onlookers know that you are serious about mathematically proving that you are really a 6' 6" person, just distorted by the light to look like a 4' 3" person. Also, it's good for letting them know that you don't have anything up your sleeves.
- Take out a sheet of paper and pencil. Note: Always use a pencil when employing Learn Algebraic Topology loopholes for any purpose. Sub-Note: Of course, you probably won't need to erase anything, but this will turn out to be because you used a pencil; if you use a pen, you will make a mistake and will wish that you had just used a pencil in the first place.
- Write at the top of the paper the real Do Number Magic and the "real" number. In the example, these will be 3 juggling balls and 11 Do a Top to Top Butterfly Pass in Contact Juggling balls in order to show that the person who juggles 3 balls is really juggling 11.
- Write 'Given a=b' on a new line.
- Perform the following mathematical operations:
3a = 3b (multiplying both sides by the real number) 11a = 11b (multiplying both sides by the "real" number) 3a2 = 3ab (multiplying by a on both sides) 11ab = 11b2 (multiplying by b on both sides) 3a2 - 11ab = 3ab - 11b2 (subtracting the above two equations to make one) 3a2 - 3ab = 11ab - 11b2 (subtracting 3ab and adding 11ab to both sides) 3a2 - 3ab + ab - b2 = 12ab - 12b2 (adding ab and subtracting b2 from both sides) 3a(a-b) + b(a-b) = 12b(a-b) (factoring out common factors ) 3a + b = 12b (removing common factors *Warning: this step divides by 0, which is algebraically undefined and therefore does not hold true in a proof) 3a = 11b (subtracting b from both sides) 3b = 11b (substituting a for b, remembering that they are equal) 3 = 11 (removing common terms) - Draw a little square at the end of your last line and fill it in. To a mathematician, this means "darn tootin" or "case closed" or "quod erat demonstrandum" or "this logic is as irrefutable as the Pied Piper's pipe after his son ran over it with the lawnmower" (well, actually, that last one is "irrefutable," but it's probably appropriate here anyway).
- Soak in the satisfaction of, for a moment, having the six-figure salary and the spacious summer home, amazingly enough, at the young age of 23! (Wow, you look great!) Or, of course, the satisfaction of being a master juggler, if that's what floats your boat.
Tips
- If you're feeling particularly nerdy, try proving that π = 3 or that e = 2. Now, that would really simpli-π life!
- Spoiler! Don't read if you want to figure out for yourself why this doesn't really work! This proof preys on an interesting mathematical no-no, hidden within the algebra of the proof: Division by zero. Notice that we decided in the beginning that a=b. So, when we factor out a-b, we are actually dividing all terms by a-b to do that. Well, a-b is 0 since a and b are equal, and division by 0 is illegal. (Sneaky mathematicians...) Anyway, there are lots of varieties of these loopholes that, while they may appear algebraically sound, they are in actuality false. Try hiding the square root of a negative in a variable or an imaginary number. What other things can you "prove"?
- Carefully consider the implications of the proof above; are there any problems with it? Read on, if you wish to know...
- There are countless numbers that would be nicer if they were just something different; age, GPA, credit card number, the amount that you owe the credit card company, and the number of times that you have been late to work this week, just to name a few.
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