Let a, b, c $\in \mathbb{Q}$. Prove if $a + c < b + c$, then $a < b$.

Question

I'm lost on how to prove this statement.

Let $a, b, c \in \mathbb{Q}$. Prove if $a + c < b + c$, then $a < b$.

So far, I have

There exists $x_1, x_2, x_3 \in \mathbb{Z}$ and $y_1, y_2, y_3 \in \mathbb{Z}-{0}$ such that $a = x_{1} // y_{1}$, $b = x_{2} // y_{2}$, and $c = x_{3} // y_{3}$. Suppose $a + c < b + c$. Then, we have

$a + c < b + c \iff (x_1//y_1) + (x_3//y_3) < (x_2//y_2) + (x_3//y_3)$....

How do I proceed next? I don't even know if my idea is correct or not. Thank you.

Answer 1

It depends on your axioms.

You probably have an axiom that if $\alpha < \beta$ then $\alpha + \gamma < \beta + \gamma$ for all $\gamma$.

So you can either prove this by contradiction:

If $a=b$ then $a+c = b+c$ and we know that isn't true; and if $b < a$ then $b+c < a+c$ and we know that isn't true. So the only option left is $a< b$.

Or we can simply use the $-c$.

$a + c < b+c$ so $(a+c) + (-c) < (b+c)+(-c)$ so $a+0 < b+0$ and $a < b$.

(More thorough details:

$a +c < b+c$ was a given

$-c$ is a number that exists (by definition and existence of additive inverses)

$(a+c) + (-c) < (b+c) + (-c)$ by axiom

$a+(c+(-c)) < b +(c + (-c))$ by associativity

$a+ 0 < b + 0$ by definition of additive inverse

$a < b$ by definition of additive identity.)

.... but ... that's assuming that $a<b \implies a+c < b+c$ is your axiom. If you introduced you math system by different methods, you'd prove this in a different manner.

Answer 2

$a+c < b + c \implies (a+c) + (-c) < (b+c) + (-c) $

$\implies a+(c+(-c) < b+(c+(-c)) \implies a + 0 < b + 0 \implies a < b.$

Since $-c$ is the element $x$ such that $c+x = 0$.