Wow. My algebra skills are rusty. I’m taking a special relativity class, and I’m in the middle of a proof, but I’m stuck. I know (from plugging in values) that the following two algebraic expressions are equivalent, yet I haven’t the foggiest idea how to get from the left side to the right side:

(1+v) / sqrt(1-v^2) = sqrt[(1+v) / (1-v)]

I derived the left hand side, and the book tells me that it should equal the right hand side. After plugging in various values (v=0.25, v=0.75, etc.), I’ve concluded that my solution is the same as the book’s solution, but I don’t know how to manipulate the left side to equal the right side. Any help would be appreciated!

Square both sides:
(1+v)^2 / (1-v^2) = (1+v) / (1-v)

Since (1-v^2) = (1+v)*(1-v). that’s always true.

Since (by convention) square roots are always non-negative, the original equation is truue as long at (1+v) > 0. (The equation is undefined for v = 1 or -1)

This is an incorrect method of proving that two sides of an equation are equal, as it assumes that they are in fact equal. Consider this “proof” that uses the same methodology:

-2x ?= 2x

Square both sides

4x^2 ?= 4x^2

You cannot assume that two expressions are equal in order to prove that they are equal. You have to prove their equality through some other method.

Dividing sqrt (1+ v) into itself leaves 1, which gives us:

sqrt (1 + v) / sqrt (1 - v)

or

sqrt ((1 + v)/(1 - v))

But, the point that I feel is most important here is that you cannot make the first step of “prove a = b” be “assume a = b”. This is as much an error in logic as it is in mathematics.

It’s perfectly valid to square both sides of the equation if we know both sides of the equation are positive (or negative) to begin with. Since this was a relativity (i.e., physics) question to begin with, it’s not unreasonable, that we might know the quantities to be positive.

In addition, Humanist, your prove suffers from the same flaw. If 1+v is negative, then you cannot equate 1+v with sqrt((1+v)^2) as you did in your first step

Nope. The method Giles used boils down to a = b [symbol]®[/symbol] a = a, which is true regardless of the truth of the hypothesis. It’s a pretty common mistake.

They are not the same at all. I made a mistake, yes, and I’ll get to that, but that’s entirely different from assuming that an equation is true to prove its truth. It is indeed valid to square both sides of the equation (or, indeed, perform any number of equivalent operations on both sides of the equation) IF we know that it is, in fact, a true equation.

Yes, you are correct that squaring two known equal quantities will result in another pair of equal quantities. However, my point is that such an equal result does not prove equality of the original quantities. It is the mathematical equivalent of “begging the question”, or assuming that a statement is true in order to prove a statement’s truth. To prove that a = b, there has to be a set of steps that shows that quantity a and quantity b are fundamentally equal. Showing that certain transformations of these quantities are equal does not constitute proof. When we’re trying to prove a = b, there’s a much more limited set of operations that we can perform, because, at every step, we have to know that our quantities are equal to the original quantities. We can, for instance, multiply by some form of 1, or add some form of zero, or raise a quantity to the first power, or perform some operation that in aggregate has one of these effects (and there are of course other legal operations).

As to your objection, let’s examine that in detail. You are correct that I did make a slight error that shows through in certain circumstances. It is possible that there is some reason to limit v to -1 <= v <= 1, in which case my earlier method works. However, I was given no reason to limit v in such a way, other than perhaps considering the course the material came from. However, we can, in fact, work this out a very slightly different way, if our domain is given no such restriction. It only requires a very slight correction:

This one, you’ll find, properly squares the imaginary result of sqrt (1 + v), where 1 + v < 0, resulting in the negative number 1 + v. My earlier mistake was entirely due to my rushing through this in order to get to work on time. Sorry about that.

So, I’ll make my important point again. You cannot act as though a = b until you know that a = b. Proof that “a = b” cannot simply be “Well, some derivative quantity we arrived at from operations performed on a equals some derivative quantity we arrived at from those same operations performed on b.” Rather, you have to show specifically that a is just another way of writing b, or vice versa, to prove equality.

In more basic terms, there is an underlying assumption made in saying that a=b in the OP. That assumption is what he is trying to prove, i.e., that a = b. So you can’t square “both sides” because you don’t know yet that you have an equivalency. You have to reduce the formulas to the point where you can declare them equivalent.

Of course, anytime you can manipulate the two formulas as equalities to get them to mirror themselves, you can manipulate one of the two formulas to where it will be the same as the other, at which point you have an equivalency.

Let me argue that Giles was correct: In order to prove that a = b, he first showed that a[sup]2[/sup] = b[sup]2[/sup]. He did not claim that that was sufficient by itself, but that it was once we knew that a and b were nonnegative.

Although not the best style, it is perfectly valid to prove the truth of an equation by manipulating it to get a known equality, if you know that each step is reversible (and, preferably, say so).

I guess I should clarify, as I seem to have sparked quite a discussion. I misspoke in the OP; I was really trying to reduce the left hand side of the equation to the right side of the equation. I had a brain-fart and forgot that (1-v^2) = (1-v)(1+v). Looking back on the problem again, I should have moved the (1+v) inside the square root symbol, leaving:

sqrt((1+v)(1+v) / sqrt(1-v^2)

then, moving the square root symbol around both quantities:

sqrt((1+v)(1+v) / (1-v)(1+v))

and, finally, dividing away the (1+v) from top and bottom.

sqrt((1+v)/(1-v))

If (1+v) = 0, this isn’t true, but given the assumption that 0 <= v < 1 (it’s a spaceship moving at constant velocity away from an observer, and v is actually the velocity of the spaceship divided by the speed of light).