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Let $$A=\{(x,y)\in\mathbb R^2:x>0, y>0\}$$ and define $f:A\to\mathbb R^2$ by $$f(a,b)=(a+b^2,2a^2+b).$$ Show that $f$ is one-to-one on $A$.

I know that a function is one-to-one if all values of the range are mapped to by at most one value in the domain, but do I show this for $f$?

A. Thomas Yerger
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jon
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  • I edited your question to make it more clear what you were asking; feel free to edit it back if this isn't exactly what you meant. – Math1000 Apr 02 '15 at 01:09
  • @jon Suppose on the contrary, and arrive at a contradiction. Suppose $f(a) = f(b)$. for $a \neq b$ OR prove it directly. – Anthony Peter Apr 02 '15 at 01:35

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you need to demonstrate that $f$ is injective and surjective. to show injectivity you need to prove that if: $$ f(a,b) = f(c,d) $$ then $$ a=c \\ b=d $$ this is a nice little exercise in algebra, so i will not spoil the fun by doing it for you!

to show surjectivity you need to prove that the equations: $$ x^2+y = s \\ x +2y^2 = t $$ with $s,t \gt 0$ always have a solution with $x, y \gt 0$. substitution will give you a degree-four equation in $x$ or $y$, which resolves to a quadratic in $x^2$ or $y^2$, so you need to show that this quadratic has a positive real root which is less than $s$ (in the $x$ case) or less than $\frac{t}{\sqrt{2}}$ (in the $y$ case)

David Holden
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