# Thread: False Proof: -1 = 2

1. ## False Proof: -1 = 2

Dearly enforummed,

Woe is me. Up is down, left is right, and apparently $\displaystyle -1 = 2$.

Here's the problem from my old textbook:

Discuss the following false "proof":

Suppose $\displaystyle a$ is a solution to $\displaystyle a = 1 + a^2$. Clearly $\displaystyle a$ cannot be 0; hence $\displaystyle 1 = (1 / a) + a$.

Use this expression to substitute for 1 in the first conditional equation:

$\displaystyle a = [(1 / a) + a ] + a^2$. Therefore,

$\displaystyle 0 = (1 / a) + a^2$, and

$\displaystyle -(1/a) = a^2$; hence

$\displaystyle a^3 = -1$.

We now infer that $\displaystyle a = \sqrt[3]{-1} = -1$.

Substituting this value into our original equation, we have

$\displaystyle -1 = 1 + (-1)^2 = 1 + 1 = 2.$
Here's the very (un)helpful little hint they give:

$\displaystyle (x^2 - x + 1)(x + 1) = x^3 + 1$

So, my questions are, in order of importance:

0. How do I make a cube root symbol?
1. What kind of sick person dreams up false proofs?
2. How is the hint supposed to help? If I transform the step $\displaystyle a^3 = -1$ to $\displaystyle a^3 + 1 = 0$, and from there to $\displaystyle (a^2 - a + 1)(a + 1) = 0$, I now have a more complicated expression -- one for which the solution is still -1 or +1, neither of which still play well in $\displaystyle a = 1 + a^2$.
3. If I encountered this kind of problem without this helpful hint, how would I know to use it? It seems like a singular rabbit to pull out of a hat...
4. This is the big one... Following normal algebraic processes, starting from a reasonable assumption, this horrific result, uh, resulted. Is this a failure of algebra, or of the idiot user attempting to make it work? Is there a tell-tale sign anywhere along the steps that something went wrong?

I vaguely remember this problem from as far back as high school, and I think it even made sense at the time -- but that was a rather long time ago from my perspective. May it please the court, I humbly beseech any assistance anyone can muster...

- Bill
Attempting to Relearn Calculus
For Masochism and Glory

2. I did not even read your post at all. And I already know the mistake! The mistake is division by zero.

3. Originally Posted by ThePerfectHacker
I did not even read your post at all. And I already know the mistake! The mistake is division by zero.
Er... it is stated in the problem that a can be assumed not to be zero, due to the structure of the original formula. It ain't that simple.

Can you maybe read the post some and just maybe give some insight?

Edit: Or *is* it that simple? I don't know. If I knew, and if I could unravel the answer from the rather cryptic "division by zero" statement, I wouldn't be asking.

I'll unpack the reasoning why I'm fairly sure it's not division by zero, as well:

Simple substitution. If you have:

$\displaystyle a = 1 + a^2$ and if you were to substitute $\displaystyle a = 0$, then you would have

$\displaystyle 0 = 1 + 0^2$.

This clearly ain't kosher, so you can, right across the top, divide by $\displaystyle a$ with impunity. No special cases around zero are necessary, because it simply can't be a solution. Since the only division I see in the above hokey "proof" is by $\displaystyle a$, I don't see this as a possible mistake. I could be wrong -- I'm no longer the math freak I was in high school and college, so doubtless I've left a few rabbits overlooked in my hat -- but if I am, please let me know how and why.

But it is still wrong, if $\displaystyle a^3 = -1$ we cannot conclude that $\displaystyle a=-1$. That is where the mistake happens.

5. Originally Posted by ThePerfectHacker

But it is still wrong, if $\displaystyle a^3 = -1$ we cannot conclude that $\displaystyle a=-1$. That is where the mistake happens.
I can accept that that is where the mistake is, but I don't grok in fullness. The proof is wrong, and the exercise is given to try to explain why it's wrong. Why is this a mistake? After all, $\displaystyle (-1)^3 = -1$.

- Bill
Attempting to Relearn Calculus
For Masochism and Glory

6. Originally Posted by relearning_calculus
I can accept that that is where the mistake is, but I don't grok in fullness. The proof is wrong, and the exercise is given to try to explain why it's wrong. Why is this a mistake? After all, $\displaystyle (-1)^3 = -1$.

- Bill
Attempting to Relearn Calculus
For Masochism and Glory
the solution to a = 1 + a^2 is complex. so we cannot assume a = -1, since that is a real solution.

besides this, a being complex can cause a whole host of problems. for instance, a^2 can be negative, etc. so be careful and look out for things like that

7. Originally Posted by Jhevon
the solution to a = 1 + a^2 is complex. so we cannot assume a = -1, since that is a real solution.

besides this, a being complex can cause a whole host of problems. for instance, a^2 can be negative, etc. so be careful and look out for things like that
I see it now! All the monkeybusiness manipulating the a in order to get it cubic obscured the original roots, which were complex. I'd quite overlooked that little tidbit somehow.

I work the roots out as being $\displaystyle \frac{1}{2} \pm \frac{\sqrt{3i}}{2}$, the cube of which is -1 for both possibilities.

8. Originally Posted by relearning_calculus
I see it now! All the monkeybusiness manipulating the a in order to get it cubic obscured the original roots, which were complex. I'd quite overlooked that little tidbit somehow.
correct.

this is one of the reasons why a neat and elegant solution (which more than often means a short one) is preferred in math. sometimes we get lost in the monkeybusiness.

- For Masochism and Glory! Power to you!

9. Originally Posted by relearning_calculus
0. How do I make a cube root symbol?
the code is \sqrt[3] {x} to yield $\displaystyle \sqrt[3] {x}$

1. What kind of sick person dreams up false proofs?
not necessarily sick. most of the time it is an extremely bored, smart person. but when it is a sick person, they tend to be the sickest, like beyond belief

2. How is the hint supposed to help? If I transform the step $\displaystyle a^3 = -1$ to $\displaystyle a^3 + 1 = 0$, and from there to $\displaystyle (a^2 - a + 1)(a + 1) = 0$, I now have a more complicated expression -- one for which the solution is still -1 or +1, neither of which still play well in $\displaystyle a = 1 + a^2$.

3. If I encountered this kind of problem without this helpful hint, how would I know to use it? It seems like a singular rabbit to pull out of a hat...
hehe, we didn't even use the hint. frankly, i don't see how it is helpful really.

4. This is the big one... Following normal algebraic processes, starting from a reasonable assumption, this horrific result, uh, resulted. Is this a failure of algebra, or of the idiot user attempting to make it work?
definitely the latter

10. Originally Posted by relearning_calculus
1. What kind of sick person dreams up false proofs?
I agree with Jhevon to a point: there are many bored people out there making up such "proofs" for fun.

However there is a serious teaching point here. Are you now likely to forget that there are other answers to $\displaystyle \sqrt[3]{-1}$ than just -1?

-Dan

11. Originally Posted by topsquark
I agree with Jhevon to a point: there are many bored people out there making up such "proofs" for fun.

However there is a serious teaching point here. Are you now likely to forget that there are other answers to $\displaystyle \sqrt[3]{-1}$ than just -1?

-Dan
Certainly I won't forget this. One of the deficiencies of my early education was that I never took high school algebra 1, and I think this is maybe why up until now, I never think about the possibilities of complex numbers mucking up the works. (I also never learned much of anything about using logarithms, I'm finding.)

On the other hand, this aspect of the question was made in pure jest. After all, I myself am the sick kind of person who is studying calculus, not because I need it for a grade or a degree -- these are long behind me -- but largely because I'm bored and have a hard time understanding what most other people consider "fun".

12. Originally Posted by topsquark
However there is a serious teaching point here. Are you now likely to forget that there are other answers to $\displaystyle \sqrt[3]{-1}$ than just -1?
indeed!

but you don't have to do crazy proofs to get such results. for example, the same lesson could be taught if they just posed a question like:

"Find all three roots of $\displaystyle a^3 - 1 = 0$"

i guess crazy proofs are more fun than mechanically solving equations though

13. Originally Posted by Jhevon
indeed!

but you don't have to do crazy proofs to get such results. for example, the same lesson could be taught if they just posed a question like:

"Find all three roots of $\displaystyle a^3 - 1 = 0$"

i guess crazy proofs are more fun than mechanically solving equations though
The crazier it is, the more likely it is to stick with them. For example, I taught as a GTA at Purdue for two years teaching a tutorial section for Intro Physics. When the students got to rotational motion an estimated 90% of them didn't label the unit for their angles. (Yes, I realize the unit radian is defined as a distance/distance making it effectively unitless, but you have to have some way to distinguish that the quantity is an angle after all.) The answer to this particular question was $\displaystyle \pi$ rad.

So I made up this complicated story involving you (the student) in a dark alleyway, water dripping from the smelly dumpsters, odd choking sounds coming from the darkness, Shaq walking toward you with a mean look in his eye, etc. (Fill in your own gory details. For fun I varied it.) At the end of the story I have me coming up behind you and shouting "Pi!" I then ask the student what I meant by that.

Some would answer 3.14, some would answer an angle, some would look at me blankly. I would then say something along the lines of "But I like apple!" Then they'd really look at me like I was some kind of crazy man (which I am!) and I'd point out that they couldn't know what I was talking about until I gave them some sort of reference. Then I'd point out their unitless angle.

It's long and involved (it took me the full hour to make it around the room once) but do you know that most of my students didn't forget to label their angles on the next exam?

Weird works as a teaching method.

-Dan

14. Originally Posted by Jhevon
indeed!

but you don't have to do crazy proofs to get such results. for example, the same lesson could be taught if they just posed a question like:

"Find all three roots of $\displaystyle a^3 - 1 = 0$"

i guess crazy proofs are more fun than mechanically solving equations though
This was actually taught to me, in this way, just before my 16th birthday, back in high school during Calc AB. I remember the problem being posed on an overhead projector, I remember about where I was sitting in the classroom, I remember my teacher's drawling, nasal voice talking about it. I remember her writing it down and solving it, in red ink. I remember the solution making perfect sense, and I remember I was bored and ready for the REAL calculus to start.

My memory is weird. I DON'T remember the actual solution. Probably because it was posed as "Find all three roots of $\displaystyle a^3 - 1 = 0$". I was bored by the lecture, and it didn't stick.

I haven't touched calculus since I was about 19 and switched from engineering to compsci. That itself was 11 years ago. What you used to think of as boring and easy gets hard, years later, if you don't have something to anchor it and its context securely in your memory, and especially you don't use it! I'll whine and gripe about silly problems like these, but Dan is absolutely right, *weird* is a great teacher, and I won't forget this now.

I'm struggling hard with another crazy proof, if I can't solve it by Monday, I'll have to go begging the community for more help...