Q.E.D.-
Don’t chuckle with self-satisifed chummagury just yet. That isn’t the step at which an error occurs. Allow me to make that step more explicit.
A. -1 = -1
B. We know that -1/1 = -1, so we can substitute that in.
C. -1/1 = -1
D. We also know that 1/-1 = -1, so we can substitute that in.
E. -1/1 = 1/-1
I certainly need not “do the same thing to each side.” That’s a silly idea that gets ground into the heads of babes in high school, and then they get trampled by more complicated equations using trig functions and suchlike when you CAN’T do this. You can do whatever you like to either side of an equation as long as you preserve the relationship between left and right sides; that is, you ensure they stay equal.
Most brilliant math funny ever, from Calvin and Hobbes:
(Picture Calvin in a trenchcoat and fedora, on a rainy night in some gritty urban setting - he raises his collar against the downpour. Imagine Bogart or Mitchum as five-year-olds)
“I stepped out into the rainy streets and reviewed the facts. There weren’t many.”
(Calvin lights a cigarette under the shelter of his hat brim - for a moment, his features are clearly lit)
“Two saps, Jack and Joe, drive toward each other at 60 and 30 mph. After 10 minutes, they pass. I’m supposed to find out how far apart they started.”
(Calvin walks, trailing smoke, his tiny fedora’ed figure in silhouette)
“Questions pour down like the rain. Who are these mugs? What were they trying to accomplish? Why was Jack in such a hurry? And what difference does it make where they started from?”
(Long shot, Calvin walking away down a rain-slick, shiny street)
“I had a hunch that, before this was over, I’d be sorry I asked.”
There is no missing dollar. The bellboy has three dollars, each of the three men has nine dollars. 9 + 9 + 9 + 3 = 30. Perhaps you misled by the way the question confuses actual dollars with dollars spent.
In an alternative version of this riddle, there is still no missing dollar. 25 dollars are hanging out in the safe at the hotel, 2 are in the hands of the naughty bellboy, and 3 dollars are back at home in the pockets of the customers. That still adds up to 30 dollars.
And why CAN’T you apply trig functions to both sides? Give me an example where this can’t be done. I’d assume that if the two sides are equal, it must be able to be done. Guess I’m one of those babes who had this ground into his skull.
Personally, I like this this one better:
Sorry, cmosdes. You aren’t differentiating the (x times) thingie. You know as well as I do that, after differentiation, the value of X can (and usually does) change. You keep it constant, which is not couth.
x^2=4
d(x^2)=d(4)
2x=0
At the beginning, x=2. At the end, x=0. What gives?
For the love of god, someone start talking about sex or poop or how much they hate the president! We’ll all snap if this mathematical nonsense goes on much longer!
“Obviously” is a great word to use, especially when you’re wrong. The error is not introduced at that step. (At least, not for the first time.)
The square root can be distributed across a fraction. That is, sqrt(A/B) = sqrt(A)/sqrt(B). I don’t feel like looking the proof up right now, but test it out for yourself. Try any A and B you like.
I never said you couldn’t apply a trig function to both sides, if by that you mean that the operation that takes A = B to cos(A) = cos(B) preserves equality. It does. But it is just one of many things you can do.
I stated that the general rule for manipulating equations is NOT simply “doing the same things to both sides”, though in the sense you mean it this does preserve equality. The trig function equations I was referring to were those involving the trig substitions; where you need to solve equations by lining up the trig functions differently and simplifying them away.
There are contexts where “doing the same thing to both sides” gets you in trouble. When dealing with inequalities, for instance.
Yes, this one is clever too. A variant on the “subtly dividing by zero” school; sneaking in an equation that forbids certain operations. I wont point out the answer, though- for the sake of other’s enjoyment.
Uh… I began failing calculus when differentiation was introduced (chapter 2). You math bastards are standing between me and a degree, move before I run your pimply asses over with a rusty steamroller, k?
Actually, cmosdes is exactly right, and that is, in fact, where the error is introduced.
It’s odd that you would say this–I certainly can pick A and B where this equation is not true; in fact, I will use your A and B:
The left hand side of the first equation is sqrt(1/-1) = sqrt(-1) = i, while the left hand side of the second equation (obtained by distributing the square root over the division) is sqrt(1)/sqrt(-1) = 1/i = -i. And so sqrt(1/-1) is not equal to sqrt(1)/sqrt(-1).
