anonymous
  • anonymous
prove that any linear function is bijective
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Prettygirl_shynice
  • Prettygirl_shynice
okay im srry what the bijective mean
xapproachesinfinity
  • xapproachesinfinity
bijective means it is surjective and injective
anonymous
  • anonymous
I'm not quite sure how to structure the proof so that it shows that it's one-to-one and onto

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xapproachesinfinity
  • xapproachesinfinity
continuous *
xapproachesinfinity
  • xapproachesinfinity
oh structuring the proof is the task at hand lol
xapproachesinfinity
  • xapproachesinfinity
i guess we need an example f(x)=ax+b
xapproachesinfinity
  • xapproachesinfinity
formal definition of injection for any d,g in R f(d)=f(g) means d=g let's show this for f(x)=ax+b
xapproachesinfinity
  • xapproachesinfinity
start: f(d)=ad+b , f(g)=ag+b f(d)=f(g) implies ad+b=ag+b ==> ad=ag then it is clear that d=g f(x)=ax+b is one to one for any x in the domain
xapproachesinfinity
  • xapproachesinfinity
now comes onto
xapproachesinfinity
  • xapproachesinfinity
f is onto if there is an e in R such that f(e)=l l has to be in the range of f but for linear function it is just R
xapproachesinfinity
  • xapproachesinfinity
so start: f(e)=ae+b <== we assign this a value l since this maps reals to reals ( there is not range restrictions), so f must be onto
xapproachesinfinity
  • xapproachesinfinity
it is then that f(x)=ax+b is one to one correspondence this is for any linear equation
xapproachesinfinity
  • xapproachesinfinity
do you get it?
anonymous
  • anonymous
Hmm, I understand the one-to-one but I'm still having trouble with the onto.
xapproachesinfinity
  • xapproachesinfinity
yeh kinda sloppy argument the onto hinges on the fact that the elements of range (codomain) must not bet left out they all have to be mapped to elements of domain
xapproachesinfinity
  • xapproachesinfinity
|dw:1450231944686:dw|
xapproachesinfinity
  • xapproachesinfinity
think of the domain of linear function R is there any values for the range that are left out?
anonymous
  • anonymous
No, so that means the range = codomain
anonymous
  • anonymous
So all of the values in the codomain are mapped to some value in the domain
xapproachesinfinity
  • xapproachesinfinity
yeah codmain is just another name for range
xapproachesinfinity
  • xapproachesinfinity
yeah there is not single values that left out
anonymous
  • anonymous
Alright thank you!
xapproachesinfinity
  • xapproachesinfinity
agree onto argument was not that good enough
xapproachesinfinity
  • xapproachesinfinity
@ParthKohli
ParthKohli
  • ParthKohli
Yeah, we just want to prove that for a number, \(k\), in the codomain, there is some \(k'\) in the domain which maps to \(k\). We can easily show this:\[k = ak' + b\]\[k' = \frac{k-b}{a}\]Thus, we've shown that the real number\[\frac{k-b}{a}\]maps to \(k\)
xapproachesinfinity
  • xapproachesinfinity
good proof :)
xapproachesinfinity
  • xapproachesinfinity
that is the proof dear :) the point for proof of onto is to prove that some real k maps to k'

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