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/sci/ - Science & Math

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6038759 No.6038759[DELETED]  [Reply] [Original]

ok, i am doing homework for real analysis, and there is one problem i cannot figure out. i need help, or hints.
show this to be true:
if x, y, z are in R, then x <= y <= z iff |x-y|+|y-z|=|x-z|

>> No.6038773

>>6038759
assume x <= y <= z, simplify |x-y|+|y-z|
assume y < x <= z, simplify |x-y|+|y-z|
assume x <= z < y, simplify |x-y|+|y-z|

>> No.6038774

sage for homework and being wrong.
Take x=y=1, z=0. Then LHS is false but RHS is true.

>> No.6038777 [DELETED] 

>>6038774
>0 + 1 = 1 is not true

you are a genuine retard

>> No.6038781

>>6038774
>1 - 1 + 1 - 0 != 1 - 0
wat

>> No.6038786

>>6038781
he said RHS is true, and you missed the modulus signs

>> No.6038792

>>6038781
He's saying the left side of the implication is false. It's supposed to be an if and only if statement.

>> No.6038816

>>6038774
>>6038792
>>6038786
its pretty obvious that this exercise has the condition x <= z that OP just didnt mention. have you never worked through an anal textbook?

>> No.6038822

>>6038816
>anal textbook
this isn't an engineering problem dipwad

>> No.6038829

>>6038816
?
it still fails with relaxing the condition to x<=z

it helps to draw a number line and actually see what's going on here.

>> No.6038837

>>6038816
i wrote down everything given written in the textbook. the only condition is that x,y,z exist in the real numbers. x<=y<=z is what i need to show, but it still implies x<=z
>>6038774
that is enough to disprove the statement. thank you!!

>> No.6038842

>>6038837
the other implication is true, though. draw a picture. you don't even need to split it into cases.

>> No.6038843

>>6038759
actually, the RHS just means y is in [x,z]
(equality case of triangular inequality)

>> No.6038847

>>6038842
you are right, that one part is true, especially if you think of it as distances. but the if and only if part doesn't hold true, so the statement as a whole isn't true.