# Equivalent Statements

1. Jul 9, 2004

### Andante

I've been having a little "problem" with the concept of equivalent mathematical statements. As far as I can tell, normally in algebra any given number multiplied by 1 (1n) and the given number (n) are equal, and therefore anything function you perform on (1n) can also be done to (n) with the exact same results. However, I have stumbled across a case where this does not seem to be true, and my algebra teacher won't even deal with the question because she says "we haven't begun to deal with imaginary numbers yet" (apparently implying that because the rest of the class doesn't even know they exist, I shouldn't even be thinking about them until she introduces them to the class :grumpy:)

Okay, so the issue she won't address is this:

1. If you multiply a number by 1, it has the exact same value as the number itself.

--So, 25 and 25(1) are the exact same thing.

2. Therefore, if you perform any given operation on these two equivalent statements, the answer should be the same.

--So, the square root of 25 and the square root of 25(1) are both 5.

3. Therefore, in theory you should be able to multiply a number by something that is EQUIVALENT to 1, and the number should still equal (1n) and (n).

--So, 25(i^4) and 25(1) and (25) all have the exact same value. [Where i is an imaginary number and i^4= (i^2)(i^2)=(-1)(-1)=1]

4. Therefore #2 should still hold true. But it doesn't seem to in this case:

--The square roots of 25 and 25(1) are both 5, but the square root of 25(i^4) is 5(i^2) or 5(-1) or -5.

My problem: As far as I understand, (or have been led to understand), the basic rules of mathematics are supposed to apply universally...at least within algebra. Identity statements should always remain identity statements, no matter what you do to the two equivalent statements. That is what we have always been taught in school, ever since kindergarten. So what I'm wondering is this: do the rules really NOT always apply, and we've been lied to all along, or do they actually still apply in this case and I'm just not seeing how?

And if the rules DON'T apply, then why is mathematics dealt with as "truth" when the truths it relies upon are not universal?

Last edited: Jul 9, 2004
2. Jul 9, 2004

### master_coda

The square roots of 25 are 5 and -5...I'm not sure where you're getting the idea the the square root of 25 is only 5.

Perhaps you're confused by the fact that $\sqrt{25}=5$ and $\sqrt{25}\neq 5$, but that's just because $\sqrt{x}$ is defined as the positive square root. That doesn't mean -5 is not a square root; it's just the negative one.

3. Jul 10, 2004

### eJavier

Your problem seems to be

$$i^4 =i^0 \quad /()^{\frac{1}{2}}$$
$$i^2 \neq i^0$$

Well, the thing is that one has to be more careful when dealing with complex numbers, and specially with complex functions. In any case
$$(i^4) ^{\frac{1}{2}}^=1$$

That's because when dealing with complex numbers, $$a^b$$ is defined in terms of complex exponential and logarithm.

Anyway, rest assured, for when you have

$$A=B$$
and you apply a function f() to the above equation, since f is a function you can be sure that:
$$f(A)=f(B)$$

Provided that you know f is a well defined function.

Last edited: Jul 10, 2004
4. Jul 10, 2004

### Galileo

The problem you face arises from a property of the field of complex numbers which differs from the real numbers.

The number $$\sqrt{x}$$ is a number whose square is equal to $$x.$$ However, there are two such numbers: one positive and one negative. If you want $$f(x)=\sqrt x$$ to be a function you have to choose which square root to take. When dealing with real numbers we always take the positive one, but that's just a convention.

When dealing with complex numbers, you cannot order them like real numbers and so there is no such thing as a positive or a negative complex number. Therefore we cannot choose which solution to take when taking the square root of a complex number, there are always 2 solutions. (It is sometimes said that $$f(z)=\sqrt z$$ is a 2-valued function).

So there is no such thing as the square root. And a square root of $$25i^4$$ is indeed -5. The other one of course is 5.