function $g: Q\times Q \to Q\times Q$ such that $g(x,y) = (2x + 3y, 3x + 2y)$ if $ x,y\in Q$, need to prove that g is reversible function

Question

I know that a reversible function is a function that is onto and injective, how can I prove that?

How about finding its inverse? – Angina Seng Dec 06 '19 at 20:24 — Angina Seng, Dec 06 '19 at 20:24

Mohammad Riazi-Kermani · Answer 1 · 2019-12-06T21:09:23.960

1

Solve $$ 2x+3y=a, 3x+2y=b$$ for $x$ and $y$ in terms of $a$ and $b$

You get $$ x=\frac {3b-2a}{5}, y=\frac {3a-2b}{5}$$

Thus for every pair $(a,b)$ of rational numbers you have $$g(\frac {3b-2a}{5},\frac {3a-2b}{5})=(a,b)$$

Thus we have $$g^{-1} (a,b) = (\frac {3b-2a}{5},\frac {3a-2b}{5})$$

That is the function $g$ is reversible.

edited Dec 06 '19 at 21:09

answered Dec 06 '19 at 20:29

Does this solution show that g is onto and injective? because I need to prove she's reversible and only then find g^-1 – Elior Dadon Dec 07 '19 at 07:38
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Yes, this is a proof for one-to one and onto property of $g$ by showing the unique pre- image for every (a,b). – Mohammad Riazi-Kermani Dec 07 '19 at 09:00

score 0 · Answer 2 · answered Dec 06 '19 at 20:17

0

Hint:

This function is linear so just check if $G$ takes only zero vector to zero vector, i.e. $\ker (G)=\{0\}$

answered Dec 06 '19 at 20:17

nonuser

2 Answers2