1=-1? really?

[cameron marical]

ok cool thanks man.

[iNow]

No problem. You're welcome.

[chitrangda]

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

 

click on the [math]\sum[/math] sign while you type the reply."Latex reference" will open. do what so ever from there.

[lakmilis]

the one providing the modulus thing is right..

 

When starting off with a = b (it is more pertinent to state +-a = +-b)

 

Else, you will need to use the law of modulus which will lead to the only positive values of a n b and hence one will get 1= 1. If one doesn't use modulus, then one starts with the premise as stated above (+-a = +- b) hence 1 = -1 is a valid answer as the preise states (+- 1) = (+-) (-1)

 

For the poster who mentioned division.. yes , it is not* valid to divide by (a-b) with the premise a = b.

 

And about the square roots LOLOL.

 

same thing: [MATH] +-\sqrt{4} = +-2 [/MATH]

 

principal square root implies the modulus of the principla root, hence the positive value.. thus one can really suppress the +- ahead of the square root sign.

 

Ok my tex is rusty lol: cameron : to use math in symbolic form, there is a syntactical 'language' called (la)tex.. (look up tex).

 

if you wish to use here for example, you would type : "MATH] equation(s) according to tex syntax [/MATH"

 

(PS. YOu would have to add the square bracket in front of MATH anf after /MATH but if I would do it, the line would be parsed as a MATH segment. i.e. : [MATH] equation(s) - according - to - tex - syntax [/MATH] )

[max.yevs]

the one providing the modulus thing is right..

[MATH] +-\sqrt{4} = +-2 [/MATH]

 

for someone who knows all about latex, lets if you can do this...

 

[math]\pm\sqrt{4} = \pm2[/math]

[lakmilis]

for someone who knows all about latex, lets if you can do this...

 

[math]\pm\sqrt{4} = \pm2[/math]

 

Hehe.. the postds above weren't showing when I replied.. but still .. don't I just hover now

 

anyway they are in the quote

 

cameron: Using Laplace Transform to Solve ODE's with Discontinuous Inputs

 

oops wrong copy paste

 

here is the link for latex : http://www.scienceforums.net/forum/showthread.php?t=4236

[triclino]

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?

 

 

HERE you have :

 

1=1 OR -1 = 1 ,which is a true statement. The statement :

 

1 = 1 AND -1 = 1 IS a wrong statement

 

Other statements that are correct are:

 

 

1=1 or 3>9

 

-3 = 7 or ln 1 = 0

 

[math]\sqrt{x^2} = |x|[/math] or 1<0

 

1=1 = 2 or 2 + 3 =5

 

But the statements:

 

-3 =7 or 7<0

 

1=1 and 1<0 are wrong

[Fuzzwood]

You can, of course, throw i in there

[the tree]

Why would that be relevant?

[samtheflash82]

Why would that be relevant?

 

because you need i to find the square roots of negative numbers.

[the tree]

yeah, we know what i is for, jeez.