# Math Question **URL:** \ **Category:** Factual Questions\ **Created:** [April 17, 2006, 8:26pm UTC](https://boards.straightdope.com/t/math-question/353054 "2006-04-17T20:26:35Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Fuzzy\_Wombats](https://avatars.discourse-cdn.com/v4/letter/f/d6d6ee/32.png) [@Fuzzy\_Wombats](https://boards.straightdope.com/u/Fuzzy_Wombats)\ **Post date:** [April 17, 2006, 8:26pm UTC](https://boards.straightdope.com/t/math-question/353054/1 "2006-04-17T20:26:35Z")
So I was cruising along fark and ran into this link: [http://archives.math.utk.edu/subjects/1equals2.txt](http://archives.math.utk.edu/subjects/1equals2.txt) Contained are three equations with rather interesting results… If you look at the first one they begin with a = b and end with 2 = 1 through some clever algebraic manipulations. Now, while I have some math background, I’m no expert. It seems to me that this is incorrect, what jumps out to me immediately is, for instance, on the first equation, the third line results in 0 = 0, before they perform the actions required to yield 2 = 1. Are these valid results? Or just a clever attempt to boggle the minds of those without a strong math background? ---
**Author:** ![KarlGauss](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/karlgauss/32/3713_2.png) [@KarlGauss](https://boards.straightdope.com/u/KarlGauss)\ **Post date:** [April 17, 2006, 8:39pm UTC](https://boards.straightdope.com/t/math-question/353054/2 "2006-04-17T20:39:13Z")
If a=b, the (a - b) = 0. Notice how they divide by (a- b), i.e. by 0 ---
**Author:** ![aerodave](https://avatars.discourse-cdn.com/v4/letter/a/f08c70/32.png) [@aerodave](https://boards.straightdope.com/u/aerodave)\ **Post date:** [April 17, 2006, 8:40pm UTC](https://boards.straightdope.com/t/math-question/353054/3 "2006-04-17T20:40:01Z")
Even more importantly than 0=0 appearing in the middle if it, is that the step between the fourth and fifth lines relies on dividing by 0. (Since a = b -\> a – b = 0). The fact that you can’t do that is what invalidates the “proof.” ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [April 17, 2006, 8:42pm UTC](https://boards.straightdope.com/t/math-question/353054/4 "2006-04-17T20:42:46Z")
In the first set, a=b, so a-b=0, and you cannot divide by a-b, which they did to get lin 5. In the last set, sqrt(1) = 1 or -1, not just 1. You can’t just select 1. And sqrt(-1) is i or -i, not just i. You can’t select just i. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 17, 2006, 8:43pm UTC](https://boards.straightdope.com/t/math-question/353054/5 "2006-04-17T20:43:58Z")
The bad step in the third equation is in going from -1/1 = 1/-1 to sqrt(-1/1) = sqrt(1/-1). For a, b \> 0, sqrt(ab) = sqrt(a)sqrt(b), but not if a or b is \< 0. The funky step in the second equation has to do with adding the x[sup]2[/sup] term to both sides. Strange things can happen when you increase the degree of an equation to find solutions. ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [April 17, 2006, 8:44pm UTC](https://boards.straightdope.com/t/math-question/353054/6 "2006-04-17T20:44:57Z")
The mistake in the first is division by zero (line 4 to line 5). In the second, the mistake is between lines 7 and 8. When they take the square root of both sides, it should be |3 - x| = |Pi - x|, which then becomes x - 3 = Pi - x. In the third, the mistake is between lines 4 and 5. There are different ways to explain the mistake. One is the explanation that what you really have is one square root should be i, while the other is -i (different “branches” of the square root). An “alternate” explanation is that, in general, sqrt(a/b) is not sqrt(a)/sqrt(b). In particular, it fails when a or b is negative. The reason it fails is the different “branches” explanation given earlier. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [April 17, 2006, 8:46pm UTC](https://boards.straightdope.com/t/math-question/353054/7 "2006-04-17T20:46:53Z")
> [@Fuzzy Wombats](#): > > for instance, on the first equation, the third line results in 0 = 0 More pointedly, to get from the 4th to the 5th line involves division by zero (often present in these sorts of things). ---
**Author:** ![Fuzzy\_Wombats](https://avatars.discourse-cdn.com/v4/letter/f/d6d6ee/32.png) [@Fuzzy\_Wombats](https://boards.straightdope.com/u/Fuzzy_Wombats)\ **Post date:** [April 17, 2006, 8:47pm UTC](https://boards.straightdope.com/t/math-question/353054/8 "2006-04-17T20:47:42Z")
I knew something wasn’t quite right… just wasn’t sure what it was. Thanks! ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 17, 2006, 8:57pm UTC](https://boards.straightdope.com/t/math-question/353054/9 "2006-04-17T20:57:00Z")
Others have answered while I typed, but maybe the same explanations in different words will help illuminate things. In the first sequence of equalities, the move from the 4th line to the 5th line is invalid. They divide both sides by a - b, but since a = b, we have a - b = 0, so they’re dividing by zero. That’s never allowed. In the second sequence, the error is in passing from the 7th to the 8th line. From (3-x)^2 = (Pi-x)^2 you can only conclude _Either_ 3-x = +(Pi-x) _or_ 3-x = -(Pi-x). That page erroneously assumes the first option, when in reality it happens to be the second option that’s true. (Solve for x in the second option and you’ll see that you get the correct value.) The third sequence is a little more difficult to explain precisely if you lack experience with imaginary numbers. Basically, when working with the sqrt function, you either need to treat it like a multivalued function, so that sqrt(1) = {1, -1}, or you need to consistently choose a particular so-called “branch” of the function. When passing from the 4th to the 5th line, that page uses different branches on the left and right-hand side of the equation. In short, they are not evaluating the same function on both sides, so it’s not surprising that they get different answers. ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [April 17, 2006, 9:00pm UTC](https://boards.straightdope.com/t/math-question/353054/10 "2006-04-17T21:00:58Z")
The middle one is a little tricker. There is nothing in the original equation that requiers “Pi” to equal 3.14… Therefore, “Pi” is just another variable. Let’s call it “y”. Then you have x = (y+3)/2 or y = 2x - 3 “y” has no unique solution, but is a function of x. ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 17, 2006, 9:07pm UTC](https://boards.straightdope.com/t/math-question/353054/11 "2006-04-17T21:07:19Z")
> [@ultrafilter](#): > > The funky step in the second equation has to do with adding the x[sup]2[/sup] term to both sides. Strange things can happen when you increase the degree of an equation to find solutions. That step is perfectly valid. There’s nothing funky about it. Adding x^2 to both sides will never change the truth-value. It’s also a completely reversible step, because you can subtract off the x^2 no matter what value x has. The problem arises in the next couple steps. ---
**Author:** ![JustAnotherGeek](https://avatars.discourse-cdn.com/v4/letter/j/58f4c7/32.png) [@JustAnotherGeek](https://boards.straightdope.com/u/JustAnotherGeek)\ **Post date:** [April 17, 2006, 9:24pm UTC](https://boards.straightdope.com/t/math-question/353054/12 "2006-04-17T21:24:32Z")
> [@John Mace](#): > > The middle one is a little tricker. There is nothing in the original equation that requiers “Pi” to equal 3.14… Therefore, “Pi” is just another variable. Let’s call it “y”. > > Then you have x = (y+3)/2 > or > y = 2x - 3 > > “y” has no unique solution, but is a function of x. Possibly, but I can still see an error in the math when “Pi” = 3.14159… In the step where they take the sqrt of both sides, remeber, there is always a + and a - choice. Since the original equation was a first order, only one of those is garunteed. As mentioned above, you need to take the negative root of the left hand side, and you will get: -(3-x) = Pi-x which, when rearranged, gives you the second line. Which basically means (on rereading), “what **Tyrrell McAllister** said three posts up.” ---
**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\ **Post date:** [April 18, 2006, 12:27am UTC](https://boards.straightdope.com/t/math-question/353054/13 "2006-04-18T00:27:44Z")
BTW, a while back, I came up with the following ```auto e[sup]iφ[/sup] = e[sup]i 2π ( φ/2π ) [/sup] = (e[sup]i 2π[/sup])[sup]( φ/2π ) [/sup] = 1[sup]( φ/2π ) [/sup] = 1 ==> e[sup]iφ[/sup] = 1 forall φ ``` Of course, it’s wrong, but the fun is in finding which particular line is wrong. ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [April 18, 2006, 1:18am UTC](https://boards.straightdope.com/t/math-question/353054/14 "2006-04-18T01:18:44Z")
> [@Polerius](#): > > BTW, a while back, I came up with the following > > ```auto > > e[sup]iφ[/sup] = e[sup]i 2π ( φ/2π ) [/sup] > = (e[sup]i 2π[/sup])[sup]( φ/2π ) [/sup] > = 1[sup]( φ/2π ) [/sup] > = 1 > > ==> e[sup]iφ[/sup] = 1 forall φ > > ``` > > Of course, it’s wrong, but the fun is in finding which particular line is wrong. Without even examining it, I can tell you that exponentiation is multi-valued. For example there are two values to 4[sup]1/2[/sup] can be 2 or -2. ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [April 18, 2006, 3:00am UTC](https://boards.straightdope.com/t/math-question/353054/15 "2006-04-18T03:00:01Z")
> [@Tyrrell McAllister](#): > > That step is perfectly valid. There’s nothing funky about it. Adding x^2 to both sides will never change the truth-value. It’s also a completely reversible step, because you can subtract off the x^2 no matter what value x has. The problem arises in the next couple steps. Actually, everyone’s missed it. **ultrafilter** is right that there’s tweakiness when you increase the degree of an equation, but adding x[sup]2[/sup] isn’t where the degree went up. **John Mace** is right that in this example “Pi” is just a variable, so let’s rename it y. x = (y+3)/2 All well and good. 2x = y+3 Multiply each side by 2. Perfectly valid. 2x(y-3) = (y+3)(y-3) AHA! We’ve multiplied by (y-3) which _adds solutions_ (namely y=3) to the equation. The rest is just cancelling off the part that depends on x to get only this solution at the end. y = 3 ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 18, 2006, 3:18am UTC](https://boards.straightdope.com/t/math-question/353054/16 "2006-04-18T03:18:58Z")
> [@](#): > > Actually, everyone’s missed it. > … > AHA! We’ve multiplied by (y-3) which adds solutions (namely y=3) to the equation. You didn’t catch anything that others missed. You are just giving a different interpretation to the (admittedly terse) “proofs”. The rest of us have been reading the lines as connected by “implies” (or inclusion of varieties) while you are reading them as connected by “if and only if” (or equality of varieties). Under our reading, there is nothing wrong with that step, though it is certainly the point at which carelessness can lead to problems. ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [April 18, 2006, 3:57am UTC](https://boards.straightdope.com/t/math-question/353054/17 "2006-04-18T03:57:31Z")
> [@Mathochist](#): > > John Mace is right that in this example “Pi” is just a variable, so let’s rename it y. I guess I see what you’re saying to a point, which I understand to be that since Pi is transcendental, it is “algebraically independent” of rationals/radicals, so in that sense it behaves just like the variable “x”. On the other hand, if we replace “Pi” in the proof with “4”, this seems equivalent to saying, “‘4’ is just a variable…” You might say that “4” has a well-defined value but, then again, so does Pi (it’s just not expressed decimally, of course). I still say the real mistake is where the square root is taken. ---
**Author:** ![Tyrrell\_McAllister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tyrrell_mcallister/32/16772_2.png) [@Tyrrell\_McAllister](https://boards.straightdope.com/u/Tyrrell_McAllister)\ **Post date:** [April 18, 2006, 4:20am UTC](https://boards.straightdope.com/t/math-question/353054/18 "2006-04-18T04:20:17Z")
> [@Cabbage](#): > > On the other hand, if we replace “Pi” in the proof with “4”, this seems equivalent to saying, “‘4’ is just a variable…” Indeed, in the given proof, the 3 may as well be considered a variable. Read this way, the proof “shows” that x = (y+z)/2 implies that z = y (or “is equivalent to”, under **Mathochist** ’s reading). ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [April 18, 2006, 5:27am UTC](https://boards.straightdope.com/t/math-question/353054/19 "2006-04-18T05:27:19Z")
> [@](#): > > Without even examining it [**Polerius** ’ “proof”], I can tell you that exponentiation is multi-valued. For example there are two values to 4[sup]1/2[/sup] can be 2 or -2. I think, though, that the real problem here is that logarithms are multi-valued. ---
**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\ **Post date:** [April 18, 2006, 3:16pm UTC](https://boards.straightdope.com/t/math-question/353054/20 "2006-04-18T15:16:00Z")
> [@Cabbage](#): > > I guess I see what you’re saying to a point, which I understand to be that since Pi is transcendental, it is “algebraically independent” of rationals/radicals, so in that sense it behaves just like the variable “x”. No, I mean there’s nowhere in the “proof” that any actual value of “Pi” is used. > [@](#): > > I still say the real mistake is where the square root is taken. The square root is problematic, but it’s just what separates the solutions of the original equation from the solutions introduced when you multiply by (Pi-3). If you made the other choice of signs you’d be back to the first equation again. [Next page](https://boards.straightdope.com/t/math-question/353054.md?page=2)