Subset Math Help: Showing R-{0} is Subset of F's Image

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SUMMARY

The discussion centers on proving that the set R-{0} is a subset of the image of the function F defined as F(x) = 1/(3x-1) for x in R-{1/3}. The user demonstrates that for any y in R-{0}, the corresponding x can be expressed as x = (1+y)/3y, which confirms that F((1+y)/3y) equals y. This establishes that every element y in R-{0} can be obtained from the function F, thereby proving the subset relationship.

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  • Understanding of real number sets, specifically R-{0} and R-{1/3}
  • Knowledge of function images and how to determine them
  • Ability to manipulate algebraic expressions
  • Familiarity with basic calculus concepts related to functions
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Homework Statement



F : R-{1/3} ---> R-{0}
F(x) = 1/(3x-1)

to show R-{0} [tex]\subseteq[/tex] image of F

for any y in R-{0}, if we try to solve y=1/(3x-1), we find x=(1+y)/3y,

F((1+y)/3y)=y

Homework Equations



R is real number

The Attempt at a Solution



so, if (1+y)/3y are the values for x in R-{0} such that F((1+y)/3y)=y

but i don't understand, how is that showing R-{0} [tex]\subseteq[/tex] image of F

explain to me please
 
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You have shown that if y is in R\{0}, then it is in the image of F, too, since for such a number y there exists a number x = (1 + y) / 3y whose image under F is y. (btw the image of F is defined as {F(x) : x is in R\{1/3}})
 


radou said:
since for such a number y there exists a number x = (1 + y) / 3y whose image under F is y.

thanks, i get it now ^^
 

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