# How to prove that $\frac{x}{2}$ is smaller than $x$ for positive $x$

Could someone provide me a valid proof that $\frac{x}{2}$ is smaller than $x$. It seems obvious but i cannot think of a proof. Or just prove that $x+x$ is larger than $x$ for positive $x$.

• $$\frac{x}{2} < \frac{x}{2} + \frac{x}{2}.$$ – Cameron Williams Oct 24 '15 at 4:59
• How do we know this is true? – Sorfosh Oct 24 '15 at 5:00
• You may want to look at the Peano axioms – Samrat Mukhopadhyay Oct 24 '15 at 5:01
• You would have to prove that if $c > 0$ and $a < b$, then $a+c < b+c$. – Cameron Williams Oct 24 '15 at 5:01
• So it is an axiom? That's great. Thanks – Sorfosh Oct 24 '15 at 5:02

\begin{aligned} x \phantom{\:+0} &= x \\ 0 &< \phantom{x+\:} x \\ x + 0&< x + x \\ x/2 &< x/2 + x/2 \\ x/2 &< x \end{aligned}

• Oh god, this was easy. Thanks a bunch :D – Sorfosh Oct 24 '15 at 5:10
• You're welcome. You could try just writing down Cameron's comment if this is homework or something and see what your teacher gives you. I doubt they'll require you to write down something as unnecessary as mine! – eyqs Oct 24 '15 at 5:11
• It's not homework, just something i thought of. – Sorfosh Oct 24 '15 at 5:12
• Alright. See if you can prove that $x/2 > x$ when $x < 0$, $x^2 < x$ when $|x| < 1$, $\sin x < x$ when $|x| < 1$, and even more! – eyqs Oct 24 '15 at 5:13
• These are easy, i don't know why i did not think of that. Brian fart i guess – Sorfosh Oct 24 '15 at 5:14

Note that if:

$$x>0$$

Then add $x$ to both sides:

$$x+x>0+x$$

Or

$$2x>x$$

Then we may divide by $2$

$$x>\frac{x}{2}$$