# Why is this bad logic ?

1. Dec 10, 2005

### dimachka

$$-1 = \sqrt[3]{-1} = -1 ^ \frac{1}{3} = -1 ^ \frac{2}{6} = ((-1)^2)^\frac{1}{6} = 1^\frac{1}{6} = 1$$

2. Dec 10, 2005

### matt grime

because roots are not one to one functions, they require a choice of branch, you've just chosen wrongly. search these forums; there are sadly many instances of this question.

i really want that FAQ.

mirror question: what makes you think it is good logic?

3. Dec 10, 2005

### dimachka

i know about branches, hence the problem lies in the fact that i changed that 1/3 to a 2/6 because i introduced an extra solution, namely 1. But im wondering how i can explain something like this using elementary mathematics. Thanks a bunch for your input.

4. Dec 10, 2005

### matt grime

why would you need to explain it using elementary mathematics? rule: you cannot replace fractional powers with other (equivalent in Q) fractions. reason: cos it buggers up the maths. you don't need any more justification than that.

there are six numbers that raised to the sixth power give one. -1 is one of them, so it's as good as it can be, why even go to this length? square -1 then square root it, 1=-1 apparently, this is just the same.

Last edited: Dec 10, 2005
5. Dec 10, 2005

### dimachka

i would need to explain it using elementary mathematics and elementary concepts because not everybody has a collegiate mathematics education... :surprised

6. Dec 10, 2005

### matt grime

tell them that when we take (even) roots like 1/2 or 1/4 or 1/6 the map is not invertible, and obviously i trust you to put that in pictures nicely for them: squaring takes 2 to 1, so can't reverse the arrows unambiguously. Honestly the 1/6 thing in your post is merely over egging the pudding, the square root contains the same phenomenology. That requires no more than knowing how to think of functions in terms of pictures, which is after all they can possibly know anyway. Nothinh I've written takes any collegiate knowledge really (though the word branch is high-brow, the idea it contains is quite simple)

7. Dec 10, 2005

### nazgjunk

Yeah, like me. I am seventeen years old, I have had some maths (over here in holland, i am in fifth grade VWO), but I can't see a logical or mathematical error in this one, other than it obviously being false.

8. Dec 10, 2005

### matt grime

you are presuming that equality is preserved under all maps, and there is no reason to suppose that, that is all. there is nothing deep going on:

sqauring takes the reals to the positive reals, and we can square root only positive reals to obtain reals, and for convenience of the two numbers that sqaure to give x we take the positive square root for all positive x.

so it is no surprise that squaring and square rooting 1 and -1 gives the same answer. by design we cannot end up with a negative square root, and that is our choice.

the cuberooting and taking sixth roots is just the same 'trick'

it is not deep! it is not hiding some great mystery.

9. Dec 10, 2005

### dimachka

yes i feel you are correct matt and this question can just be explained by explaining why $-1 = (-1)^\frac{2}{2} = (-1^2)^2 = 1^2 = 1$ is wrong. But i dont exactly see how this can be explained through $\sqrt{1} = 1 & \mbox{ or } -1$ Care to explain?

Last edited: Dec 10, 2005
10. Dec 10, 2005

### matt grime

What? I thought you were supposed to be teaching this? non-bijective functions cannot be inverted. taking the power 1/2 is not the inverse function of the map R --> R, x--->x^2. That map has no inverse. End. Fin. Nuff said.

Last edited: Dec 10, 2005
11. Dec 10, 2005

### nazgjunk

Well, i still don't really get it. Guess I'll let my maths teacher have a look at it next wednesday, I hope he can explain it to me.

Probably I am just too stubborn too see it, as usual. For some reason I tend to forget what I am doing, and then I don't understand a thing anymore.

12. Dec 10, 2005

### dimachka

I think a better way to explain the problem someone would have understanding this is: "Why must every step in a series of algebraic manipulations be reversible to guarantee that the logic will be sound?"

13. Dec 10, 2005

### nazgjunk

I think i am finally starting to understand this. And as for "Why must every step in a series of algebraic manipulations be reversible to guarantee that the logic will be sound?", no one ever told me that, but it sounds good to me.

14. Dec 10, 2005

### matt grime

see what? do you understand the idea of bijective function? of inverse functions? that is the reason why this happens; you're looking far too hard for something that isn't there.
if you square to different numbers you can get the same answer thus there is no way to undo the operation of sqauring on all numbers: the square root cannot know, if you squared -1, that you want it to give the negative square root. Look at it this way. i have a number, its square is 1, what was the number? see, can't tell me. that is all that's going on.

Last edited: Dec 10, 2005
15. Dec 10, 2005

### matt grime

that would depend upon the person you're taking to obviously. however, that is for you to decide in teaching it since only you know to whom you are explaining it.

16. Dec 10, 2005

### matt grime

but that isn't accurate. it is prefectly possible to have none reversible implications and still have a sound argument, indeed that is the point of implication.

the difference here is that you are *claiming* that the steps are reversible implicitly, when they aren't. the point is yo'ure just making a false and unkustified claim that people don't notice is false.

17. Dec 10, 2005

### matt grime

i want to amend my opinion: that is false.

it is not the algebriac manipulations need to be reverisible, just that each logical step is actually sound to begin with. it so happens that becuase you cannot invert squaring that you have made a logically inconsistent deduction.

18. Dec 10, 2005

### nazgjunk

Well as far as i can see is every step fully logical, but that might be my mistake.

Geez, this is getting me depressed again.

19. Dec 10, 2005

### matt grime

the step you claim is logically consistent is: if x^2=y^2 then x=y. now, surely you can see that is false (we even have a counter example that shows this is false). that is what it is saying when we take square roots and think that the answers must be what we started with.

20. Dec 10, 2005

### dimachka

I think it is that each algebraic manipulation must be reversible, i think that is the crux of algebra. In this case, the reason the operation of squaring is not reversible is precisely because you cannot invert to get a unique answer, but instead have two answers.