Re: Sign conventions

From: Jesse F. Hughes (jesse_at_phiwumbda.org)
Date: 02/21/05 

Date: Mon, 21 Feb 2005 07:56:03 +0100

jstevh@msn.com writes:

> Jesse F. Hughes wrote:
>> I want to try to follow your reasoning here. You start with an
>> equation, square both sides and solve. And then you conclude that
> the
>> solutions for the squared equation are also solutions for the
> original
>> equation, right?
>>
>> Let x + 4 = 1. Squaring both sides gives x^2 + 8x + 16 = 1. Thus,
>> (x + 5)(x + 3) = 0 and hence x = -5 or x = -3.
>>
>> We may therefore conclude that x = -5 is a solution to x + 4 = 1.
>> Fascinating reasoning there.
>>
>
> Sigh. If you have x+4 = 1, then x=-3, right?
>
> So how can you later claim that it doesn't?
>
> That's a basic contradiction.

Sigh. If z = sqrt(1) - sqrt(1) and sqrt(1) = 1 by definition, then
z = 0, right?

So how can you later claim that it doesn't?

That's a basic contradiction.

> Notice that (x+5)(x+3) = 0 IS in fact correct, if x=-3.
>
> You seem to be missing the significance of letters versus numbers,
> which is the basis for algebra.

Oh. *That's* the basis for algebra. Damn. I knew the category
theorists were off track, but I didn't know how far. Maps FX -> X
indeed. Geez.

> There is a difference between, say, x=3, and
>
> x+y=z
>
> and if you can't understand that, then you'll get lost.
>
> So, if
>
> z = sqrt(x) - sqrt(y)
>
> then it MUST be true that
>
> (z^2 + y - x)^2 = 4z^2 y
>
> or do you disagree?
>
> NOW I say, let x=y=1, and you get three solutions for z, right?

Oh. I get it. Let me try again.

Let x + y = z. Then x^2 + 2xy + y^2 = z^2. NOW I say, let y = 4 and
z = 1. Then you get two solutions for x, right (-3 and -5)?

Therefore, both -3 and -5 are solutions to x + y = z when y = 4 and
z = 1.

Thanks! Now I get it.

> Now do you understand?
>
> But if z = 1 - 1 = 0, then it equals 0.
>
> End of story at that point.
>
> In one case you have letters representing *possibility* while in the
> other you have certainty, a letter with a set value, like x=-3.
>
> Now then, there's a simple way for you to show you are rational on this
> subject:
>
> Given z = sqrt(x) - sqrt(y) is it or is it not true that
>
> (z^2 + y - x)^2 = 4z^2 y?
>
> Answer that one question, yes or no.

Yes, it's true, but it's also true that there are solutions to the
second equation that are not solutions to the first. Just like if
x + y = z, then x^2 + 2xy + y^2 = z^2 but not every solution of
x^2 + 2xy + y^2 = z^2 is a solution of x + y = z.
>
> I dare you. 

-- 
Jesse F. Hughes
"There's a thrill that's gone that I'll probably not have in quite the
same way again.  After all, FLT was a unique animal, and we had a
great dance."  -J.S. Harris on "proving" Fermat's last theorem

Relevant Pages
Quantcast