Prove that $t$ is a one-to-one correspondence.

Question

I am having problems with this discrete math proof. I have made it this far, but I do not understand how to go from here.

Problem:

Define a map $t: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}$ by $t(a, b) = (a + b, a - b)$. Prove that $t$ is a one-to-one correspondence.

Solution:

Proof: Let $z,x,c,v\epsilon\mathbb{R}$ such that $(z, x),(c, v) \epsilon \mathbb{R} \times \mathbb{R}$. Because $t(z, x) = t(c, v)$ \begin{align} \begin{split} (z + x, z - x) &= (c + v, c - v) \\ 2z &= 2c \\ z &= c \\ \end{split} \end{align}

I do not know how to prove that x = v from here.

score 2 · Answer 1 · answered Sep 19 '13 at 05:29

2

You've concluded $z = c$ and you know started with $x + z = c + v$, by definition of $=$ for ordered pairs. Combining the equations gives

$$x + c = c + v \implies x = v$$

answered Sep 19 '13 at 05:29

score 2 · Accepted Answer · answered Sep 19 '13 at 05:31

2

$z+x=c+v$ and $z=c$, so $c+x=c+v$.

Alternatively, your method of adding the equations could be applied again, only this time subtract instead to get $2x = 2v$.

Often "one-to-one correspondence" is used (unfortunately) to mean not just one-to-one, but bijective, both one-to-one and onto. This function is also onto, and it may be that you are supposed to prove it.

answered Sep 19 '13 at 05:31

Jonas Meyer

53,602

How did he get to 2z=2c ? – c3cris Oct 19 '17 at 11:49
@c3cris adding the entries of the ordered pairs on each side of the previous equation. – Jonas Meyer Oct 19 '17 at 12:23

Prove that $t$ is a one-to-one correspondence.

2 Answers2