1=-1? really?

[samtheflash82]

I am a bit confused by this. Here is the equation:

 

[math]

a=b[/math]

[math]a^{2}=b^{2}[/math]

[math]a^{2}-b^{2}=0[/math]

[math](a+b)(a-b)=0[/math]

[math]a=b,-b[/math]

let [math]a=1[/math]

[math]1=1,-1

[/math]

 

I understand that this cannot be true but why does it work algebraically? To my understanding, if [math] a=b [/math], than [math] a^{2}=b^{2}[/math] but if [math] a^{2}=b^{2} [/math], than [math] a\neq b [/math] What am I not understanding?

[max.yevs]

the flaw here is basically based on this-

 

[math]2^{2}[/math] = 4; [math]-2^{2}[/math] = 4

but

[math]\sqrt{4}[/math] = 2, not -2

 

so a and b could be 1 or -1 since both [math]1^{2}[/math] = 1 and [math]-1^{2}[/math] = 1

[iNow]

the flaw here is basically based on this-

 

[math]2^{2}[/math] = 4; [math]-2^{2}[/math] = 4

but

[math]\sqrt{4}[/math] = 2, not -2

 

Since when?

[Cap'n Refsmmat]

I think the trouble here arises from taking the shortcut.

 

[math](a+b)(a-b) = 0[/math]

[math]a+b = 0[/math] or [math]a-b=0[/math]

[math]2a = 0[/math] or [math]a - a=0[/math]

[math]a = 0[/math] or [math]0 = 0[/math]

 

You never get a chance to show that 1 = -1.

 

You can't jump ahead to the conclusions you made if you don't take the "shortcut" to finding solutions. The shortcut is invalid here.

[max.yevs]

well, when you square something negative it becomes positive, but when you take the square root you only take the positive

 

like you might think [math]\sqrt{(x)^2}[/math] = x

 

but for x= -2, [math]\sqrt{(-2)^2}[/math] = [math]\sqrt{4}[/math] = 2

 

as if -2 = 2

Edited March 25, 2009 by max.yevs

[Cap'n Refsmmat]

[math]\sqrt{2} = \pm 2[/math]

 

See http://en.wikipedia.org/wiki/Square_root

[moth]

I am a bit confused by this. Here is the equation:

 

[math]

a=b[/math]

[math]a^{2}=b^{2}[/math]

[math]a^{2}-b^{2}=0[/math]

[math](a+b)(a-b)=0[/math]

[math]a=b,-b[/math]

let [math]a=1[/math]

[math]1=1,-1

[/math]

 

I understand that this cannot be true but why does it work algebraically? To my understanding, if [math] a=b [/math], than [math] a^{2}=b^{2}[/math] but if [math] a^{2}=b^{2} [/math], than [math] a\neq b [/math] What am I not understanding?

 

when you get to [math](a+b)(a-b)=0[/math]

you could say a=1 or a = -1 but not both at the same time.

since b=a a can't equal -b unless they're 0.

i'm pretty sure if [math] a^2 = b^2 [/math] then [math] a = b[/math]

Edited March 25, 2009 by moth
case

[max.yevs]

i'm pretty sure if [math] a^2 = b^2 [/math] then [math] a = b[/math]

 

what if [math] a = 2 [/math], and [math] b = -2 [/math] ?

 

im pretty sure the flaw occurs because even though

if [math] a = b [/math] then [math] a^2 = b^2[/math]

it does not work backwards with

if [math] a^2 = b^2 [/math] then [math] a = b[/math]

 

because if [math] a^2 = b^2 [/math] then [math] a = b, -b[/math]

 

just like the equation he did, i can say

 

[math] a = b [/math]

[math] a^2 = b^2[/math]

[math] a = b, -b[/math]

let [math] a = 1 [/math]

[math] 1 = 1,-1 [/math]

 

of course this trick is a lot harder to find in his equation

and ignore my first two posts, both the positive and negative count as square roots, even if they're the reason for the flaw.

Edited March 25, 2009 by max.yevs

[moth]

what if [math] a = 2 [/math], and [math] b = -2 [/math]

 

im pretty sure the flaw occurs because even though

if [math] a = b [/math] then [math] a^2 = b^2[/math]

it does not work backwards with

if [math] a^2 = b^2 [/math] then [math] a = b[/math]

 

good point - i forgot negative numbers.my algebra has 20+ years of dust on it.

but a has to equal b or - b, not b and -b, right?

[max.yevs]

that's an interesting question, yeah i guess so

 

in fact its a good way of thinking why in the beginning, a equals b only, but when he somehow gets a equals both b and -b, there must be some mistake. only way i can think to replicate it is with a number and the negative of it both having the same square.

[gonelli]

I recall a maths lecturer taking a problem similar to this and going through it in front of the class. I'm pretty sure the initial premise of "a=b" was the problem, and wouldn't that suggest why you end up with a situation of the positive and negative answer both being equal to the same value?

[jian]

[math]a^2=b^2\Leftrightarrow|a|=|b|[/math]

[math]|a|=|b|\Leftrightarrow a=b or a=-b[/math]

[samtheflash82]

[math]\sqrt{2} = \pm 2[/math]

 

See http://en.wikipedia.org/wiki/Square_root

 

this is not true. at all.

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I was given a very similar equation to this again today. It was exactly the same as the one in the OP except at the step where you have [math](a+b)(a-b)=0[/math], the person divided the 0 by the [math]a-b[/math], and ended up with [math]a=-b[/math] therefore [math]1=-1[/math]. Am i right about the flaw being with the division? I mean, since [math]a=b[/math], wouldn't dividing by [math]a-b[/math] be the same as dividing by zero, which is "forbidden"?

[iNow]

this is not true. at all.

 

It's clear to me that Cap'n made a simple typo, and intended to show that [math]\sqrt 4 = \pm 2[/math], which is, in fact, QUITE true... and contrary to the assertion made by Max in post #2, which was challenged by me in post #3.

[samtheflash82]

It's clear to me that Cap'n made a simple typo, and intended to show that [math]\sqrt 4 = \pm 2[/math], which is, in fact, QUITE true... and contrary to the assertion made by Max in post #2, which was challenged by me in post #3.

 

alright, but [math]\sqrt 4 \neq \pm 2[/math]. [math]\sqrt{4}=2[/math] while [math]-\sqrt{4}=-2[/math]. the correct notation that you would need to use if you wanted a positive and negative answer would be [math]\pm\sqrt{4}=\pm2[/math]

[iNow]

Rubbish. Only if you restrict your definition of square root to the subset of positive roots. That's silly. Implied in the square root signage is a request for ALL roots. Unless otherwise specified, and unless a principal root is specifically requested, both positive and negative responses must be provided, not the other way around.

[Cap'n Refsmmat]

Oops. Yes, iNow is right, I meant [math]\sqrt{4} = \pm 2[/math], which is true.

 

Read the Wikipedia page I linked to.

[samtheflash82]

the radical sign [math]\sqrt{x}[/math] is a request for the principal square root of [math]x[/math].

 

Every non-negative real number x has a unique non-negative square root, called the principal square root, which is denoted with a radical symbol as [math]\sqrt{x}[/math], or, using exponent notation, as [math]x^{1/2}[/math]. For example, the principal square root of 9 is 3, denoted [math]\sqrt{9} = 3[/math], because [math]32 = 3 × 3 = 9[/math]. If otherwise unqualified, "the square root" of a number refers to the principal square root: the square root of 2 is approximately 1.4142.

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Consecutive posts merged

Oops. Yes, iNow is right, I meant [math]\sqrt{4} = \pm 2[/math], which is true.

 

Read the Wikipedia page I linked to.

 

I did, and I quoted it above.

[max.yevs]

nah, cap'n is right, i orignally mispoken.

 

[math]

\sqrt{4} = \pm 2

[/math]

 

but look at post #8, i think it captures the idea behind this whole thread.

[cameron marical]

wouldnt it be 1,-1=1,-1. not 1=1,-1. if a and b are the same thing, then they are, just we dont know what set they are. could be negatives, could be positives.

 

p.s. how do you guys write real math on here? nothing like that is on my keyboard. thanks.

[Cap'n Refsmmat]

I did, and I quoted it above.

Then you also saw:

 

Every positive number x has two square roots. One of them is [math]\sqrt{x}[/math], which is positive, and the other [math]-\sqrt{x}[/math], which is negative.

 

cameron: click on the fancy math to see how we did it. There should also be a link to a tutorial when you do.

[cameron marical]

where is it?

[samtheflash82]

whatever, the fact is that it's irrelevant.

[cameron marical]

whatever, the fact is that it's irrelevant.

 

...i dont know if this was directed at me, i was asking for the location of this fancy math thing.

[iNow]

It wasn't directed at you. Also, hover your mouse over the square root of 4 symbol in quoted text of Capn's post #21 above... Click it.

 

The window which opens shows the code which generates that, and also at the bottom right is available an additional link with more information (a tutorial) on how to do it. It's calle "LaTeX," and much like using quote tags, you use "math" tags and certain codes to make it.