Logic Behind a Proof: Injective Function G

[Mathematicsresear]

Homework Statement

suppose I have a function defined as:

G: ℚ--->ℚ
f(x)= { 2/ 3x if x does not equal to 0, 0 if x=0}

Homework Equations

Injective:if for all x,y in ℚ, f(x)=f(y) then x=y.
or if x does not equal to y then f(x) does not equal to f(y)

The Attempt at a Solution

I am confused as to the logic whilst proving that the above function is injective.

I understand that the contrapositive of the definition of injective can be used in the following case:

if x does not equal to 0 and y=0 then 2/ 3x =0 so f(x) does not equal to f(y).

However, why does this work for the following case:

if x does not equal to 0 and y does not equal to 0 then 2/ 3x = 2/ 3y so x=y.

Isn't this of the form p implies q implies r, and the definition of injective is not of that form?

 

[fresh_42]

Homework Statement

suppose I have a function defined as:

G: ℚ--->ℚ
f(x)= { 2/ 3x if x does not equal to 0, 0 if x=0}

Homework Equations

Injective:if for all x,y in ℚ, f(x)=f(y) then x=y.
or if x does not equal to y then f(x) does not equal to f(y)

The Attempt at a Solution

I am confused as to the logic whilst proving that the above function is injective.

I understand that the contrapositive of the definition of injective can be used in the following case:

if x does not equal to 0 and y=0 then 2/ 3x =0 so f(x) does not equal to f(y).

However, why does this work for the following case:

if x does not equal to 0 and y does not equal to 0 then 2/ 3x = 2/ 3y so x=y.

Isn't this of the form p implies q implies r, and the definition of injective is not of that form?

As you said, $x \longmapsto f(x)$ being injective means $f(x)=f(y) \Longrightarrow x=y$. This is equivalent to $x \neq y \Longrightarrow f(x)\neq f(y)$ with which we work here. So let us assume once and for all, that $x\neq y$. This is our premise, it is a given fact.

Now we have three cases to consider (the case $x=y$ being ruled out):

1. $x \neq 0 \, , \, y \neq 0$
2. $x \neq 0 \, , \, y = 0$
3. $x=0 \, , \, y\neq 0$

For symmetry reasons, the cases (2) and (3) are the same, just switch the roles of $x$ and $y$. Now what you wrote are the arguments in these two cases. Remember, $x\neq y$ being given.

Case (2) is what you said you understood: $0 \neq f(x) \neq f(y) = f(0) = 0$, which has to be shown.
Case (1) is then a contradiction: Assume $f(x) = f(y)$ for our $x \neq y$. Then we get from $\frac{2}{3x}=\frac{2}{3y}$ that $x=y$ which is a contradiction to our first assumption and so, $f(x) \neq f(y)$.

Each case is dealt with properly and all cases together give all possible constellations.

We can try and express case (1) without contradiction. Then we still have $x\neq y$ as our first requirement. Case (1) also means $x\cdot y \neq 0$.

Can you sow without indirect proof or contradiction, that $f(x) \neq f(y)$ in this case?
(Hint: Calculate $f(y) -f(x)$.)