Verifying that f(x,y) is one to one.

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SUMMARY

The function f(x,y) = 2x + y is proven to be not one-to-one but onto the set of real numbers. A function is one-to-one if distinct inputs do not yield the same output; in this case, f(3, 1) and f(2, 3) both equal 7, demonstrating that f is not one-to-one. Conversely, f is onto because for any real number z, there exist multiple pairs (x, y) such that 2x + y = z, confirming that the function covers all real numbers.

PREREQUISITES
  • Understanding of functions and their properties, specifically one-to-one and onto functions.
  • Familiarity with linear equations and their graphical representations in two dimensions.
  • Basic knowledge of set theory, particularly the concepts of domain and range.
  • Experience with algebraic manipulation to solve equations involving two variables.
NEXT STEPS
  • Study the definitions and properties of one-to-one and onto functions in more depth.
  • Learn about linear transformations and their implications in higher dimensions.
  • Explore the concept of inverse functions and conditions for their existence.
  • Investigate the graphical representation of functions in R² and how it relates to their properties.
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Mathematics students, educators, and anyone interested in understanding the properties of functions in multivariable calculus and linear algebra.

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Homework Statement


How do you show that a function with two variables f(x,y) is one-to-one and onto?
example f(x,y) = 2x+y

Homework Equations



Do we have to use linear algebra?
 
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f(x,y) represents a 2D surface.

To have (x,y) -> f(x,y) be 1 to 1 and onto, I think that means that at every height (every value of f) you have a unique point (x,y).

Unless there is some additional constraint for x and y, so that f(x,y) represents a curve I'm not sure that this is possible.

In your example f(x,y) = 2x+y is a 2D plane. For each value of f, you get a line 2x +y = const.

Interesting to think about, though.

-James
 
Proving that f(x,y) is "one-to-one" and "onto" depends upon the range space!

Since f(x,y)= 2x+ y is, for numbers x and y, a single number, the "default" assumption would be that f maps R2 to R.

And to prove, one way or the other, use the definitions!

A function, f, from set U to set V, is said to be "one to one" if and only if two different values, p and q, in U cannot give the same point in V.

Here, U is R2 so we can write p= (x_1, y_1) and q= (x_2, y_2)
. V is R so any point in V is a single number, z. Now, if f= 2x+ y were not one to one, then there would exist p= (x_1, y_1) and q= (x_2, y_2) such that f(p)= 2x_1+ y_1= 2x_2+ y_2= f(q). That would mean 2(x_1- x_2)= -(y_1- y_2)= 0 Well, take x_1= 3, x_2= 2, y_1= 1, y_2= 3. Then x_1- x_2= 3- 2= 1 so 2(x_1- x_2)= 2 and y_1- y_2= -2 so -(y_1- y_2)= 2 also.

That is, f(3, 1)= 2(3)+ 1= 7 and f(2, 3)= 2(2)+ 3= 7. No, f(x, y)=2x+ y is NOT "one to one".

A function, f, from set U to set V, is "onto" (the set V) if and only if, for any point q in V, there exist at least one point p, in U so that f(p)= q. To show that f(x,y)= 2x+ y is "onto" R, let z be any real number (any point in R) and look for (x, y) (a point in R2) such that 2x+y= z. In fact (exactly because this function is not "one to one") there are many such points. Take, for example, y to be z- 2 and x to be 1.

Yes, f(x, y)= 2x+ y is "onto" the set of real numbers.
 
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thank you for all your replies would try it out :)
 

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