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Let R+ denote the set of positive real numbers. Let f : R × R+ → R be given by f(x, y) = x/y. (a) Is f an injective function? Prove your answer. (b) Is f a surjective function? Again, prove your answer. (c) Is f a bijection? Prove your answer.

Respuesta :

LammettHash

Let [tex]f:\Bbb R\times\Bbb R^+\to\Bbb R[/tex].

a. [tex]f[/tex] is injective if [tex]f(x_1,y_1)=f(x_2,y_2)[/tex] for any two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the two points must be identical with [tex]x_1=x_2[/tex] and [tex]y_1=y_2[/tex].

[tex]f[/tex] is not injective because we can pick two points for which [tex]f[/tex] gives the same value:

[tex]f(2,1)=\dfrac21=2[/tex]

[tex]f(4,2)=\dfrac42=2[/tex]

b. [tex]f[/tex] is surjective if there exists [tex](x,y)[/tex] for which every point in the codomain [tex]\Bbb R[/tex] is obtained by [tex]f(x,y)[/tex].

This should be somewhat obvious if you consider 3 different cases:

  1. [tex]f(x,y)<0[/tex] and can take on any negative real number for any choice of [tex]x<0[/tex]
  2. [tex]f(x,y)=0[/tex] if and only if [tex]x=0[/tex]
  3. [tex]f(x,y)>0[/tex] for any choice of [tex]x>0[/tex]

So [tex]f[/tex] is surjective.

c. [tex]f[/tex] is bijective if it is both injective and surjective. The conclusions above show [tex]f[/tex] is not bijective.

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