I was on some web site (Internet movie database of all places) and ran into this formula.
let a = b
a©÷ = ab
a©÷ + a©÷ - 2ab = ab + a©÷ - 2ab
2(a©÷ - ab) = a©÷ - ab
2 = 1
I think its wrong, obviously. I also think the error is in the 2nd to last line.
2(a©÷ - ab) ¡Á a©÷ + a©÷ - 2ab
I think it¡¯s a misuse of the substitution rule. Am I right?
CRAP!! damn the coding!!
let a = b
a ² = ab
a ² + a ² - 2ab = ab + a ² - 2ab
2(a ² - ab) = a ² - ab
2 = 1
Since a2 = ab, then a2 - ab = 0.
To get from the next-to-last line to the last line, you are dividing by (a2 - ab), which means you are dividing by zero.
Division by zero is undefined. Now you know why.
In the last step you are dividing both sides by zero, because:
a[sup]2[/sup]-ab = a(a-b) = a*0 = 0
Once you divide both sides by zero, everything beyond that point is invalid.
By the way, you can enter superscripts using the [ sup ][ /sup ] tag. This would result in fewer compatibility problems than using the “squared” special character. (e.g. I wouldn’t have to change the encoding setting on my browser to see it).
This is Why I am no longer an engineering Major. THanks Gary.
As an aside question: this looks like a pretty easy mistake to make in a proof. You’d have to make sure that every divisor in every expression is non-zero - I can imagine that would get really tough in a complex proof (e.g. Wiles’ proof of FLT). How do professional mathematicians avoid this kind of thing? Is it purely based on peer reviews? I know that in this example, the division by zero causes totally off-the-wall results, but is it possible that in another proof the results could look like something plausible (but incorrect) ?
Andrew Wiles did make a subtle mistake in his first version of the proof of Fermat’s Last Theorem, and someone caught it pretty quickly. Wiles and John Taylor soon figured out a slight modification of the proof that had no mistake. Mathematicians know how to read proofs to look for subtle errors (division by 0 is a pretty trivial error that no mathematician is going to make in a proof), and they’re competitive enough to want to find any errors in a paper. The real problem is not the subtlety of the errors; it’s that proofs are often so long that it takes a long time to read through them. Understanding the proof of some important theorems requires reading through dozens of papers that establish the proper subsidiary proofs before you can tackle the main proof.
As a former math teacher, I advocate that every algebra student should be able to explain the problem in a “proof” like this before he can pass the class. Here’s another one:
To show that all numbers are equal, let a and b be any two numbers. Define c = (a + b)/2. Then
2c = a + b
2ac - 2bc = a[sup]2[/sup] - b[sup]2[/sup] (multiply by a-b)
b[sup]2[/sup] - 2bc = a[sup]2[/sup] - 2ac
b[sup]2[/sup] - 2bc + c[sup]2[/sup] = a[sup]2[/sup] - 2ac + c[sup]2[/sup] (complete the square)
(b - c)[sup]2[/sup] = (a - c)[sup]2[/sup]
b - c = a - c
b = a
Since a and b were any two numbers, this proves that all numbers are equal.
Good one, Richard. Gotta watch the signs on those square roots.
Yes it’s an oldie, but it’s a goody, here’s another slightly more subtle one:
x[sup]2[/sup] = x*x
x[sup]2[/sup] = x(1 + 1 + 1 … + 1) (with x terms)
x[sup]2[/sup] = (x + x + x … + x)
taking the derivatives of both sides:
2x = (1 + 1 + 1 … + 1) (still with x terms in the brackets)
2x = x
2 = 1
Yep, that one’s one of my favorites.
Sometimes I think it’s more clear just why dividing by zero is such a big deal when actual numbers, rather than variables, are used in an example. Check this out:
100 = 50
Well, that’s true enough (0=0). However, it most obviously does not follow that:
10 = 5
Try performing the same manipulations as in the OP with an actual number (I think it’s more clear that way) - say X=Y=5:
You can’t just “cancel” anything in reality. That’s a shorthand for multiplying both sides by an inverse of something and reducing some factors to unity. In this case, that would require multiplying both sides of the equation by 1/0 to “reduce” (5-5)/(5-5) to unity. But 0/0 is not equal to 1; it’s undefined. I think this illustrates why that must be so.
BTW, I’ll bet everyone who debunked this knew exactly what to look for by just reading the thread title. This is an old, old trick. Lesson to learn: Whenever anyone pulls out something along these lines, the first thing to look for is “divide by zero”.
Yes, an old favorite of mine, too. In case anyone is wondering what the trick is, it is that differentiation of a function must be based on “continuity” of that function.
Expressing x squared that way is a “discrete” rather than a continuous approach. Hence you get fouled up.
True Blue Jack
For those following along at home, the “continuity” here is metric continuity. Basically, a metric space is metrically discrete iff there is a minimum distance between distinct points, and metrically continuous otherwise.
You can re-define derivative in such a way that you can differentiate a discrete function, but that doesn’t fix the problem. The real problem is that phrase (with x terms). If you’re differentiating with respect to x, you can’t ignore the x in that phrase. Take that into account, and you’ll find that d/dx( (x+x+x+…+x) (x times) ) = (1+1+1+…+1) (x times) + (x) (1 time) = x + x = 2x.
As for higher-level mathematical proofs, you don’t have to use division all that often (compared with addition, subtraction, or multiplication), but when you do, you always need to include a line showing that your denominator is nonzero. And since division is relatively uncommon and it can cause problems like this, division by zero is a fairly easy flaw to find.
Here’s another 2=1 puzzle, courtesy of The Derb:
(“Sqr” means square root, “*” means multiplication, and “i” is the square root of -1)
This one isn’t divide by zero.
This one is incredibly poor, and slightly not what we’re talking about, but I’m nevertheless fond of the following problem:
What is the following when expanded?
(a-n)(b-n)(c-n)…(z-n)
I’m not sure if the ‘…’ is properly used there, but hopefully it’s pretty clear what it’s supposed to mean.
Heheh - our maths teacher (this was the hardest year twelve maths subject) was pretty unimpressed when we revealed the answer.
~ Isaac
Not surprisingly, the use of the word “obvious” tips us off that this is where the problem lies 
If we let x=1 and y=0 in the above “identity”, then we get Sqrt(1)=i*Sqrt(-1), which is false.
All the best puzzles have the flaw in the premises, where nobody looks, rather than in the elaboration, where everyone stares and stares and figures and figures and …