Ikou
  • Ikou
Can someone help me? I need to prove this Identity is true (x - y) (y - x) = (xy – x^2 - y^2 - yx)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
Just multiply it out... \[(x-y)(y-x) = x(y-x) -y(y-x)\]by distributive property Now distribute again on that and you should be done!
Ikou
  • Ikou
Oh, well any way I could plug numbers in? Like to get 35=35
Ikou
  • Ikou
Somebody please help me, this is taking much longer than it should

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UsukiDoll
  • UsukiDoll
no need... just expand the (x-y)(y-x) with the distributive property and it's there.
UsukiDoll
  • UsukiDoll
start with the left side of the equation.
UsukiDoll
  • UsukiDoll
it's like either use the distributive property on the left to get the right hand side of the equation or factor the right hand side of the equation to get the left side of the equation
UsukiDoll
  • UsukiDoll
ugh man I fell for that.. nice @whpalmer4 I saw what you did there :/ you used factor by grouping dude. >_< good one :)
Ikou
  • Ikou
Thank you for explaining, but if you, say, make x=# and y=#, what should I do to make them equal to eachother
UsukiDoll
  • UsukiDoll
hmmm we could try letting x and y be the same value
UsukiDoll
  • UsukiDoll
though I'm trying to do one in my head ... letting x = 1 and y = 1 the result would become (1-1)(1-1)=((1)(1)-1^2-1^2-(1)(1)) (0)(0)=(1-1-1-1) 0=(0-1-1) 0=(-1-1) and that's a false statement
Ikou
  • Ikou
I don't think that would work, because on the left it would be zero right off the bat
UsukiDoll
  • UsukiDoll
0 doesn't equal to -2. I know that.. so x and y needs to be different
Ikou
  • Ikou
in looking for something that gives the same number on the left to the right
Ikou
  • Ikou
I've been working on this for days and its really confusing
UsukiDoll
  • UsukiDoll
maybe x = 2 , and y = 3 ? (2-3)(3-2)= ((2)(3)-2^2-3^2-(3)(2)) (-1)(0) = (6-4-9-6) 0=(2-9-6) nooooooope I'm assuming that there is no x,y pair and we may need to do with what @whpalmer4 suggested.
Zarkon
  • Zarkon
\[(x - y) (y - x) = (xy – x^2 - y^2 \color{red}{+} yx)\]
Ikou
  • Ikou
@Zarkon Why would there be a +?
Zarkon
  • Zarkon
\[(-y)(-x)=yx\]
UsukiDoll
  • UsukiDoll
DOH! Sign error that makes a big difference.
UsukiDoll
  • UsukiDoll
-___________________-!
UsukiDoll
  • UsukiDoll
a negative times a negative is positive . Now let's see if we can find an x,y pair that makes the equation equal
Ikou
  • Ikou
If this is what was hanging me up for days I will go into a damn screaming fit
Zarkon
  • Zarkon
also, you cant prove it with examples. you need to multiply it out like @whpalmer4 was doing
UsukiDoll
  • UsukiDoll
which is either start from the left to get the result on the right hand side of the equation or vice versa right @Zarkon ?
UsukiDoll
  • UsukiDoll
(1 - 1) (1 - 1) = (1 – 1^2 - 1^2 +1) (0)(0) = (1-1-1+1) (0) = (0-1+1) 0=(-1+1) 0=0 ok I'm gonna be guilty for doing this because a proof shouldn't be plug x and y but I let x = 1 and y =1 and the equation holds true. It's the sign error that threw everything out of the mix.
Ikou
  • Ikou
Could this come out the same if it was a bigger number like 3?
whpalmer4
  • whpalmer4
Yeah, those pesky - signs are the work of the devil! I'm told that in days of old, mathematicians would often write their equations so they didn't need to use them! \[x^2 -3x + 6 = 0\]would be written \[x^2 + 6 = 3x\]and so on.
UsukiDoll
  • UsukiDoll
ALso I didn't eat dinner yet. Can't do math without energy. yeah it should be the same
Ikou
  • Ikou
heh, Im so happy this is finally put to rest. Sorry for my sign mistake, this damn thing kept me up till 3 am each night
whpalmer4
  • whpalmer4
Actually, I think the original problem is incorrectly stated. \[(x-y)(y - x) = x(y -x) - y(y - x) = x(y-x) - y(-1)(x-y) = x(y-x) + y(x-y) \]\[= xy - x^2 + yx - y^2 = 2xy - x^2 - y^2\]
Ikou
  • Ikou
Why have you dragged me back to hell
whpalmer4
  • whpalmer4
\[y-x = -1(x-y)\]so we just have \[-1(x-y)(x-y) = -1(x-y)^2 = -1(x^2 - 2xy +y^2) = -x^2 + 2xy- y^2\] Hey, who copied the problem?
UsukiDoll
  • UsukiDoll
wasn't me.
Ikou
  • Ikou
This is for an assignment where I have to create a new identity for polynomials and have to go through proving it
whpalmer4
  • whpalmer4
Possibly the problem was to prove that \[(x-y)(y+x)= (xy - x^2 - y^2 - yx)\]? Notice that the right side simplifies to just \(-x^2 -y^2\) because \(xy - yx = xy - xy = 0\)
UsukiDoll
  • UsukiDoll
omg I see it now. -_-
UsukiDoll
  • UsukiDoll
OMG! THIS IS TROLL LEVEL >:O
UsukiDoll
  • UsukiDoll
like xy and yx cancel out
whpalmer4
  • whpalmer4
Well, anyhow, to prove an identity, just manipulate both sides of the equation as necessary to meet in the middle with the same thing on both sides...
Ikou
  • Ikou
I need a result where (x - y) (y - x) = (xy – x^ - y^2 + yx) have a result of (example) 20=20
UsukiDoll
  • UsukiDoll
leaving (x-y)(y+x) = -x^2-y^2 and then (x-y)(y+x) = -(x+y)(x-y) ............ x+X
whpalmer4
  • whpalmer4
Well, you don't prove it works for all numbers by plugging in numbers, though you can prove that it DOESN'T work for all numbers by plugging in numbers.
Ikou
  • Ikou
This assignment is going to be the death of my grade in this class. I really should know how to do this, it really shouldn't be this hard, and theres no + in (y - x)
whpalmer4
  • whpalmer4
Well, I'm sorry, but the identity you posted simply is not an identity. Try plugging in \(x =1, y = 2\)
Ikou
  • Ikou
wait a sec, I worked it out and it was -1=-1
whpalmer4
  • whpalmer4
\[(x-y)(y-x) = xy - x^2 - y^2 - yx\]is your statement, yes? \[(1-2)(2-1) = (1)(2) - (1)^2 - (2)^2 - (2)(1)\]\[-1*2 = 2-1-4-2\]\[-2=-5\]not in my universe :-)
Ikou
  • Ikou
no it was + yx I screwed up in my signs and (2-1)=1 so -1*1=-1
whpalmer4
  • whpalmer4
so for the last time, what is the identity you are trying to prove?
Ikou
  • Ikou
(x - y) (y - x) it's a real pain in the retriceand confusing
whpalmer4
  • whpalmer4
no, what is the entire statement? You have no = sign there
Ikou
  • Ikou
ah, sorry im a bit tired, its is (x - y) (y - x) = (xy – x^ - y^2 + yx)
whpalmer4
  • whpalmer4
Okay, that's what I was saying earlier:\[(x-y)(y-x) = 2xy - x^2 - y^2\]your version just hasn't collected the \(xy\) terms together Here is a proof: \[(x-y)(y-x) = (xy - x^2 - y^2 + yx)\]we expand the left side twice with the distributive property: \[x(y-x) - y(y-x) = xy - x^2 - y^2 + yx\]\[xy - x^2 - y^2 -(-yx) = xy - x^2 - y^2 +yx\] \[xy - x^2 - y^2 + xy = xy - x^2 - y^2 + yx\]now we change the order of the product for the last term on the left hand side because multiplication is commutative: \[xy - x^2 - y^2 + yx = xy - x^2 - y^2 + yx\] Done.
whpalmer4
  • whpalmer4
actually, didn't even need that last step if I hadn't changed \(-(-yx)\) to \(+xy\) by habit of always going in ascending lexicographic order
Ikou
  • Ikou
Thank you for clearing things up! I love you all for helping ♥
Ikou
  • Ikou
Thank you @whpalmer4 senpai!
whpalmer4
  • whpalmer4
You're welcome!

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