anonymous
  • anonymous
Show that the set T ={(w,x,y,z)∈R4 such that y=w and x^2 =z^4} is not the graph of any function of w and x.
Mathematics
jamiebookeater
  • jamiebookeater
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helder_edwin
  • helder_edwin
well u have the correspondence (i don't know if this is the word in english) \[ \large (w,x)\mapsto(y,z) \] such that \(y=w\) and \(z^4=x^2\).
helder_edwin
  • helder_edwin
do u remember the definition of function?
anonymous
  • anonymous
A function is a rule that assigns a unique element in Rn to Rm. It is one to one.

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helder_edwin
  • helder_edwin
the last part is something else.
helder_edwin
  • helder_edwin
a function assigns a UNIQUE element to EVERY element of its domain.
anonymous
  • anonymous
Ok I understand that.
anonymous
  • anonymous
So a certain w and x cannot have two outputs.
helder_edwin
  • helder_edwin
what does (w,x)=(1,-1) get assigned to?
helder_edwin
  • helder_edwin
yes. precisely.
anonymous
  • anonymous
(1,-1) is assigned to (1,1)?
helder_edwin
  • helder_edwin
just that?
helder_edwin
  • helder_edwin
let's see: \[ \large y=w=1 \] right?
anonymous
  • anonymous
right
helder_edwin
  • helder_edwin
BUT \[ \large z^4=x^2=(-1)^2=1\Rightarrow z=\pm\sqrt[4]{1}=\pm1 \]
helder_edwin
  • helder_edwin
so \[ \large (1,-1)\mapsto(1,1) \] and \[ \large (1,-1)\mapsto(1,-1) \]
anonymous
  • anonymous
So does that mean T is not a function?
helder_edwin
  • helder_edwin
yes. that's what u were asked to prove.
anonymous
  • anonymous
T is graph of something, but that something is not a function. right?
helder_edwin
  • helder_edwin
yes.
anonymous
  • anonymous
Thank you SO much!
helder_edwin
  • helder_edwin
u r welcome.

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