danmac0710
  • danmac0710
How can I prove that: abs(x+y) less than or equal to abs(x) + abs (y) please?
Mathematics
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
think of some negative values ...
amistre64
  • amistre64
|x+y| <= |x|+|y| let x=3, and let y = -5 |3-5| <= |3|+|-5| |-2| <= 3+5
danmac0710
  • danmac0710
I have tried saying that if x>y>0, then abs(x+y) = abs(x)+abs(y) If x>0>y, then abs(x+y)x>y, abs(x+y) = abs(x)+abs(y) Does not feel like a proof though

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amistre64
  • amistre64
it wold prolly be fine with cases. let x,y be >= 0; then |x+y| = |x|+|y| = x+y let x,y be < 0; then |-x-y| = |-x|+|-y| = x+y the distance between same signed values is the same either way ... its when we get into different signs that we run into troubles let x < 0 and let y > 0; then |-x+y| < |-x|+|y| = x+y but trying to iterate it in another manner eludes me at the time
danmac0710
  • danmac0710
Hey I gotta dash... what do you think though. Is that enough? Thanks for your help though Amistre. I'll buy you a beer tonight, but I will have to drink it for you too!
terenzreignz
  • terenzreignz
Or just two cases... either same sign or different signs :D
terenzreignz
  • terenzreignz
oh... you already said that... nvm :D
amistre64
  • amistre64
|-5+3| = |-2| = 2, which is less than 8 |5-3| = |2| = 2, which is less than 8
danmac0710
  • danmac0710
Really gotta dash but many thanks for the ideas.
amistre64
  • amistre64
like signs increase distances; an additive effect different signs decreas distances; an "subtractive" effect
amistre64
  • amistre64
maybe some vector notations :)
anonymous
  • anonymous
note that \(xy\le |xy|\)\[|x+y|^2=(x+y)^2=x^2+2xy+y^2=|x|^2+2\color\red{xy}+|y|^2 \]\[|x|^2+2\color\red{xy}+|y|^2 \le |x|^2+2\color\red{|x||y|}+|y|^2=(|x|+|y|)^2\]it gives\[|x+y|^2 \le (|x|+|y|)^2\]take square root\[|x+y|\le |x|+|y|\]
amistre64
  • amistre64
are we allowed to just note that xy <= |xy| ? or would we have to prove that as well?
danmac0710
  • danmac0710
Thanks Mukushla, that's a nice way to look at it... Have you got a source for that proof?
danmac0710
  • danmac0710
I agree with Amistre too - proof needed for previous result. Thanks guys!
anonymous
  • anonymous
No source. Your welcome.

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