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You should be able to solve this.

Name: im superior to you !SHjzCSZsqE 2007-08-21 16:48 ID:bo+yV2W9

x - 7 = 19 + x

A famous meme on /b/, well, I just solved it and want the credit.

x-x=26

(x^2)^0.5= a + and - number
For example. -3*-3=9, 3*3=9, therefore 9^0.5 = -3 and 3

+x--x=2x
x-x=26
2x=26
x=+13 and -13

Name: Anonymous 2007-08-21 17:26 ID:Heaven

great now go away

Name: Anonymous 2007-08-21 20:17 ID:2QGm8Q+F

>>1
>>1
>>1
What. Wrong on so many levels. peace out.

Name: Anonymous 2007-08-21 20:34 ID:EgTEfmWQ

If 13 = -13, then it's easy to see that every complex number is equal to every other complex number.  Are you really willing to admit this conclusion?

Name: Anonymous 2007-08-21 20:45 ID:+IbXFhXM

>>4
-13 < 0 < 13 = -|13| = |13| = 13 ?

Name: Anonymous 2007-08-22 12:01 ID:J/JAardS

Protip: x-x != 2x

Now go away faggot

Name: im superior to you !UdXMqJ7ILI 2007-08-22 12:14 ID:ff1l3w3d

>>4
13^2=(-13)^2
>>5
That's about right.
>>6
+x--x=2x

Name: Anonymous 2007-08-24 1:53 ID:Heaven

In during troll thread

Name: Anonymous 2007-08-24 16:19 ID:NnjSJ1OO

NOT THIS SHIT AGAIN

WE JUST DID THIS IN /a/ YESTERDAY

Name: RedCream 2007-08-24 18:05 ID:lgz2zZK3

Since:

x-7=19+x

... results in a contradiction ("-7=19"), then there is no solution for x.

Wow, gosh.  How fucking hard was THAT?

Name: sage 2007-08-24 18:22 ID:Heaven

Sage goes in every field.

Name: Anonymous 2007-08-24 21:28 ID:Heaven

In after RedCream.

Name: Anonymous 2007-08-25 14:02 ID:2fHbpfuo

x - 7 = 19 + x
x - 7 = x + 19

So you take a number and subtract seven.  That's equal to that same number plus nineteen!

So subtracting 7 and adding 19 are the same?  Only modular 26 can that have a hope of being true.

So x is anything modulo 26 (or 13 for that matter).

Q. E. D.

~Master's Student Anon with B. S. in Mathematics

Name: Sage 2007-08-25 15:15 ID:Z70nv0yP

OMG SUCH A FAILURE!

This simple mathethic equation for middle schooler.... -.-

And you faggots discuss this for so long......

FAIL!

Name: Anonymous 2007-08-25 15:33 ID:Heaven

>>14
sage does not go there

Name: 4tran 2007-08-25 18:09 ID:Heaven

>>13
Right, as was pointed out on /b/.  It's also true mod 2 and 1 (the latter being trivial, since everything is 0 mod 1).

Name: im superior to you !UdXMqJ7ILI 2007-08-25 18:57 ID:o9hwONLs

>>8-16
x - 7 = 19 + x
First we take the supposition X^2=169
This would mean X=+13 or -13

13-7=19-13=6

Name: RedCream 2007-08-25 21:58 ID:mwST66DF

>>17
FAIL.  x cannot take on TWO DIFFERENT VALUES at once to satisfy the equation.

At any rate, taking powers of a variable introduces extraneous roots.  Don't introduce x as anything other than x, fucko.

Name: Anonymous 2007-08-25 23:16 ID:Heaven

don't listen to redcream, he believes in alien dinnerware

Name: Anonymous 2007-08-26 2:56 ID:Heaven

>>18

x = |13|

Name: RedCream 2007-08-26 9:43 ID:1aP7WHQd

>>20
FAIL.  That equation cannot be used (i.e. built up) to satisfy the original equation, since the o.e. still requires 1 value for x.

Name: Anonymous 2007-08-26 12:03 ID:Heaven

>>21

Yes it can.

Name: im superior to you !UdXMqJ7ILI 2007-08-26 19:15 ID:6/ILCLaq

I can't believe you idiots cannot understand such a simple concept. Ok fine I'll do the thinking for you lazy ass retards.

x=169^0.5

169^0.5 - 7 = 19 + 169^0.5

I want all of you to line up, admit you are fucking stupid and kill yourselves.

Name: RedCream 2007-08-26 23:02 ID:ZXy/vx6G

>>23
FAIL.  Invoking higher powers results in creating EXTRANEOUS roots.  EXTRANEOUS means UNNECESSARY, motherfucker!  THERE IS NO S*I*N*G*L*E VALUE OF X THAT SATISFIES THE GIVEN EQUATION!  ±x is NOT a SINGLE value.  You can't ask for ±x change in the fucking grocery store!

You twentysomething collegefreaks just about take the cake.  You learn a mathematical tool, and then spend the next 20 years sticking it into the wrong spots like you're fucking Euler.

Name: Ikerous 2007-08-27 1:51 ID:uH1N1Dcu

The square root of 169 is a function not a value.  This square root function returns two values.  X can only equal a value not a function.


When using variables, "one important thing we are assuming is that the value of each occurrence of x is the same—that x does not get a new value between the first x and the second x." (wiki)

Name: 4tran 2007-08-27 4:00 ID:LGi8jXB3

RedCream is right.  x has no real or complex solution.

>>24
I don't think "twentysomething collegefreaks" are this stupid.  Some /b/tards are this stupid, while the trolls who pretend to believe this are in it for the lulz.  In short, >>17,21 phails unless s/he is a troll.

>>25
By strict mathematical definition, a function can only return a single value.  The term you're looking for is relation.

Name: im superior to you !UdXMqJ7ILI 2007-08-27 4:32 ID:l8noVEa6

>>24
Yes there is.

x = 169^0.5 = + or - 13

>>25
What's wrong with 2 values? They exist and comply with mathematical laws so they are TOTALLY legitimate. Just because you are mentally incapable of comprehending this fact it doesn't mean it doesn'texist or will just go away.

>>26
You can call me a troll all you want, but the truth is you keep coming back for more as you know you are in the presence of true genius.

Name: 4tran 2007-08-27 4:48 ID:LGi8jXB3

>>27
By that definition, there are an infinite number of solutions.

x = a or a-26...  GJ trivializing all of mathematics.  We bow before your holy creation of total uselessness.

Name: Anonymous 2007-08-27 8:52 ID:cBbVesY5

You stupid trolls.

Go back to MIDDLE SCHOOL!

Name: Anonymous 2007-08-27 12:18 ID:skmRDR2b

>>27

did you even read >>24 >>25 or >>26 ? if you did you would realise the obviousness of your ginormous fuckup.

Name: RedCream 2007-08-27 14:43 ID:8nkTRKLe

>>27
You're a fucking MORON.  You started with:

x - 7 = 19 + x

... and you're treating it like:

a - 7 = 19 + b

... where a=-b.  Note that "a=-b" is not the same as "x=x".

You can't really be this stupid, can you?

Name: Anonymous 2007-08-27 15:47 ID:EdPmnhPS

>>31

Set a = -b = x and it is. It's ok, I understand that completely elementary algebra can be difficult. To a special needs child.

Name: Anonymous 2007-08-27 15:54 ID:y2WuiWzx

Actually, they have a point.

OP didn't so much as specify the algebraic structure to which x belongs, much less "x real" or even "x complex"

Hell, x could come from a set defined as positive real on the left hand side of an equals sign and negative real on the right.

But if x is real anyone insisting on anything other than "no solution" is a faggot.

Name: RedCream 2007-08-28 1:15 ID:TtkRQAZP

>>32
You're still a fucking MORON.  There are two different equations, and can't make different things the same.  By invoking x, you're back to the 1 equation where there is no real solution, since ONE VARIABLE can't take on TWO VALUES.

>>33
An equation doesn't work that way.  If you're going to arbitrarily define how an equation works, then stop calling it an equation and stop using the equals sign.  Instead, call it something else of your own invention (like "efagtion") and use another symbol (like a tiny assface).  For efagtions, a variable xx has one value on one side, and the same variable x has another value on the other side.

Name: 4tran 2007-08-28 3:12 ID:KDKWo2OW

>>33
That breaks a simple requirement of equivalence relations: a ~ a, for all a in some set.  I guess that defines some new comparison of sorts, but I'm not sure what it would do.  Interesting nevertheless.

In this case, 1 = -1 -> 1 =/= 1
Ownd?

Name: Anonymous 2007-08-28 9:44 ID:Heaven

You know, if you just define
"x - 7 = 19 + x

A famous meme on /b/, well, I just solved it and want the credit.

x-x=26

(x^2)^0.5= a + and - number
For example. -3*-3=9, 3*3=9, therefore 9^0.5 = -3 and 3

+x--x=2x
x-x=26
2x=26
x=+13 and -13"
to stand for
"I am the weakest, most transparent troll on the planet, and yet I somehow succeed.", >>1 makes perfect sense.

Well done, >>1, well done. You certainly proved something.

>>33
You can't define a set as 'positive on the left and negative on the right'. The set is an abstract entity, the left/right/equal sign thing is just notation. You'd get stuff like x = 5, but 5 != x.
And basic operations like moving terms from one side of the equation to the other assumes that "x + 1 = x + 1"
And if you redefine x = x to mean x = -x, then x = x -> x = 0, which as >>35 says works out poorly for integer constants not 0.

Name: Anonymous 2007-08-29 9:08 ID:3Mdmsram

>>36


Er, yes you can. There are many relations that you can define on a set, say the integers, which are not reflexive (I think that's the right one), ie. x=y does not imply y=x.

And the second example you give is true in modular arithmatic in certain cases.

In fact if the original example were stated

x - 7 = 19 + x mod (y)

makes sense if y is not a multiple of x

Name: Anonymous 2007-08-29 10:53 ID:Heaven

i can't believe this has reached 37 posts.

Name: Anonymous 2007-08-29 12:50 ID:qIfKAuaG

>>37
Yes, you can have nonsymmetric relations, but that would be redefining '=', not having "a set defined as positive real on the left hand side of an equals sign". That makes as much sense as defining a constant to be 1 when you calculate it by hand and 2 on a calculator, except on Sundays, when it is always 5.
Sets and other mathematical entities are not defined in terms of their notation, and they don't have behaviour outside of the operations you perform on them.

Yes, I noticed the modular thing.
x = -y (mod 2), for example, simply checks for parity, so x = -x (mod 2) for all integers.

I don't see how
x - 7 = 19 + x mod (y)
makes any more sense, though?
x is still superfluous, you just get -7 = 19 (mod y). (True for y in {1, 2, 13, 26}))


>>38
You think that's bad, I can't believe it's reached 39.

Name: 4tran 2007-08-29 22:30 ID:agm2UH/d

>>39
At least we're getting a real mathematical discussion going.  It's better than Gilgamesh.
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