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Let $\mathbb{R}$ denote the set of real numbers.
Consider the function $f$ on $R \times R$ defined by $f(x,y) = (x+y, 2x-3y)$.

Is $f$ onto? justify your answer.

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Again, I would request that you not post questions by merely quoting your homework assignment; some of us find posts in the imperative at least somewhat rude. I can suggest using a quotebox (preface the material you are quotient with a left-flushed > to make it appear in a quotebox). In addition, it's best if you at least say what you have tried, where you are stuck, or whether you are having some trouble even getting started. Thank you. –  Arturo Magidin Jan 11 '12 at 18:06

3 Answers 3

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If $f(x,y) = (x+y, 2x-3y)$ has an inverse $\left(\dfrac{3a+b}{5},\dfrac{2a-b}{5}\right)=g(a,b)$ then not only is $g$ the inverse of $f$ but $f$ is the inverse of $g$. So $f$ is bijective: both one-to-one and onto.

We know $f$ is onto because we can obtain any $a$ and $b$ using $f\left(\dfrac{3a+b}{5},\dfrac{2a-b}{5}\right)=(a,b). $

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Hint: does the system of equations $$ \eqalign{x+y&=a\cr 2x-3y&=b} $$have a solution for every $(a,b)$?

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x = 1/5(3a+b), y = 1/5(2a-b) ∴ it is one-to-one . But how onto? –  Q123 Jan 10 '12 at 13:15
    
@Q123: You've just calculated the inverse of $f$, which doesn't just show it's one-to-one, but also that it's onto. (A function has an inverse if and only if it is bijective.) –  Clive Newstead Jan 10 '12 at 13:18
    
all right. Now I know how to calculate the inverse of f. but not onto. somebody help me.. –  Q123 Jan 10 '12 at 14:00
    
@Q123: this is onto because for any a and b you can find an x and y. Sometimes when you "invert" a function you can't get a result for every element in the range-think of inverting $x^2$ to $\sqrt{x}$, but here you can. Please write your solution as x=(3a+b)/5 or (better) $x=(3a+b)/5$ or $x=\frac{3a+b}5$ so it is clear what is in the numerator. –  Ross Millikan Jan 10 '12 at 14:18
    
@RossMillikan: ok I will. But is there anyone here who is willing to tell me how I write x=(3a+b)/5 as RossMillikan did? Do I need to use something? –  Q123 Jan 10 '12 at 16:10

HINT $\ $ Since $\rm\:f\:$ is $\rm\:\mathbb R$-linear, i.e. $\rm\ f(a\ u + b\ v)\ =\ a\ f(u) + b\ f(v)\ $ for $\rm\ a,b\in \mathbb R,\ \ u,v\in \mathbb R^2\:,\:$ its image $\rm\:f(\mathbb R^2)\:$ spans $\rm\:\mathbb R^2\:$ if the image includes a basis, say $\rm\ (0,1),\ (1,0)\:.\:$ This is easy to verify, viz.

$\rm\qquad\qquad\quad\ v = (\:r,\:-r)\ \Rightarrow\ f(v)= (0,5r)\ $ so $\rm\ r = 1/5\ \Rightarrow\ f(v) = (0,1) $

$\rm\qquad\qquad\quad u = (3r,2r)\ \Rightarrow\ f(u)= (5r,0)\ $ so $\rm\ r = 1/5\ \Rightarrow\ f(u) = (1,0) $

Hence $\rm\ (a,b) = a\ (1,0) + b\ (0,1)\ =\ a\ f(u) + b\ f(v)\ =\ f(au)+f(bv)\ =\ f(au+bv)\:.$

For an analogous example look up the standard proof of the Chinese Remainder Theorem (CRT). It too uses linearity to construct the solution for $\rm\:(a,b)\:$ from the solutions for $\rm\:(1,0)\:$ and $\rm\:(0,1)\:.$

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