# If A=B, Then 2=1 WTF?!? **URL:** \ **Category:** Factual Questions\ **Created:** [November 25, 2003, 6:18am UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139 "2003-11-25T06:18:50Z")\ **Posts on this page:** 7\ **Page:** 2
**Author:** ![flex727](https://avatars.discourse-cdn.com/v4/letter/f/258eb7/32.png) [@flex727](https://boards.straightdope.com/u/flex727)\ **Post date:** [November 27, 2003, 2:58pm UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139/21 "2003-11-27T14:58:40Z")
> [@](#): > > _Originally posted by Orbifold \* > \*\*… then we get Sqrt(1)=i_Sqrt(-1), which is false. \*\* Hmmm, that equation isn’t really false, is it? 😕 But, then again, realizing why that equation isn’t false reveals the answer to the puzzle. :smack: ---
**Author:** ![MonkeyMensch](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@MonkeyMensch](https://boards.straightdope.com/u/MonkeyMensch)\ **Post date:** [November 27, 2003, 3:26pm UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139/22 "2003-11-27T15:26:12Z")
> [@](#): > > \*Originally posted by flex727 \* > \*\*Hmmm, that equation isn’t really false, is it? 😕 > > But, then again, realizing why that equation isn’t false reveals the answer to the puzzle. :smack: \*\* Am I being whooshed or is there some underlying truth to _1 = -1?_ ---
**Author:** ![MonkeyMensch](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@MonkeyMensch](https://boards.straightdope.com/u/MonkeyMensch)\ **Post date:** [November 27, 2003, 3:37pm UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139/23 "2003-11-27T15:37:58Z")
On further inspection of **flex727** ’s first line I assume it comes from the derivation Sqr(x-y) = Sqr(x-y) Sqr(x-y) = Sqr[(-1) \* (y-x)] Sqr(x-y) = Sqr(-1) \* Sqr(y-x) Sqr(x-y) = i \* Sqr(y-x). I can see that the last statement is false, as I saw it was at the beginning of **flex727** ’s problem. It’s still early out here but I can’t seem to find the error in that derivation. Math geeks? ---
**Author:** ![brad\_d](https://avatars.discourse-cdn.com/v4/letter/b/50afbb/32.png) [@brad\_d](https://boards.straightdope.com/u/brad_d)\ **Post date:** [November 27, 2003, 5:43pm UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139/24 "2003-11-27T17:43:24Z")
I think it has to do with the square root function being multi-valued. 1 (like all numbers) really has two “square roots”: 1 and -1: (-1)[sup]2[/sup]=1. It’s often convention for the “sqrt” function to imply the _positive_ root, but messing about too haphazardly with this results in ridiculousness like that above. Similarly, -1 has two square roots: _i_ and -_i_. So, 1[sup]1/2[/sup] = _i_ \* (-1)[sup]1/2[/sup] is extremely dangerous to write, as both sides of that equation have two values. 1[sup]1/2[/sup] has the values 1 and -1; _i_ \* (-1)[sup]1/2[/sup] has the values _i__i = -1 and i_-_i_ = 1. Pick and choose which values you want, and the relation can be either true or false. I’m pressed for time so I can’t flesh this out as thoroughly as I’d prefer to, but I think the whole “proof” makes use of this sort of logical nonsense. ---
**Author:** ![Mogle](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/mogle/32/12021_2.png) [@Mogle](https://boards.straightdope.com/u/Mogle)\ **Post date:** [November 27, 2003, 6:48pm UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139/25 "2003-11-27T18:48:35Z")
[Lurker mode off] ‘Sqr(x-y) = i \* Sqr(y — x)’ is not the problem. Let’s rename a couple of the variables in the puzzle: Start from this obvious truth: Sqr(x-y) = i \* Sqr(y — x). Since this is plainly the case for any numbers x and y, put x = a and y = b. Then Sqr(a — b) = i \* Sqr(b — a). On the other hand, since that obvious truth I started with is the case for any numbers x and y, I could equally well put x =\*\* c\*\* and y = **d** , to get another, equally true statement: Sqr( **c** — **d** ) = i \* Sqr( **d** — **c** ). Now, if P = Q and R = S, then obviously P \* R = Q \* S. So: Sqr(a — b) \* Sqr( **c** — **d** ) = i [sup]2[/sup] \* Sqr(b — a) \* Sqr( **d** — **c** ). Replace a,b,c,d with x,y and we get: x — y = i[sup]2[/sup] \* (y — x) \<=\> x — y = x — y [Lurker mode on] ---
**Author:** ![danceswithcats](https://avatars.discourse-cdn.com/v4/letter/d/9e8a1a/32.png) [@danceswithcats](https://boards.straightdope.com/u/danceswithcats)\ **Post date:** [November 27, 2003, 7:47pm UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139/26 "2003-11-27T19:47:00Z")
b4i4q ru/18 qt pi? 😉 ---
**Author:** ![MC\_Master\_of\_Ceremonies](https://avatars.discourse-cdn.com/v4/letter/m/bb73d2/32.png) [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)\ **Post date:** [November 27, 2003, 7:47pm UTC](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139/27 "2003-11-27T19:47:48Z")
Unfortunately I can’t see some of the binary operations (in flex’s or moigle’s post (i.e. they’re not in my character set), what is the binary operation, that I’ve replaced with the % “sqrt(x-y) = i\*sqrt(x % y)”? Am I right in assuming it’s a minus sign (though why can I see some minus signs and not others)? Though it’s hard to follow not seeing all the binary operations, it looks like a variation on the one that **Richard** posted. [Previous page](https://boards.straightdope.com/t/if-a-b-then-2-1-wtf/215139.md?page=1)