Tricky, Interval Proof, is my Proof correct? Prove that if  x is in $[0,b]$, then $x=tb$ for some $t$ where  $0 \le t \le 1$

My Attempt at a proof: 

Interval Midpoint: $ x = \frac{b}{2} $

$  \frac{b}{2} = tb $
$  \frac{1}{2} = t $

This is implies that value of $t$ is equal to  $ \frac{1}{2} $  when $x$ is the mid point of  $[0,b]$.

Thus for the first element in the interval: $  t = \frac{1}{2}- \frac{1}{2} = 0  \rightarrow x= 0b = 0 $ 

Thus for the last element in the interval: : $  t = \frac{1}{2}+ \frac{1}{2} = 1 \rightarrow x= 1b = b $ 

Q.E.D

Is this a correct proof? is my reasoning correct? I am self learning maths but not sure if this rigorous enough.
You're given that $0\le x\le b$ Suppose $b&gt;0$. Then
$$
0\le\frac{x}{b}\le1
$$
and, setting $t=\frac{x}{b}$, you have $x=tb$ as required.

If $b=0$, then $x=0$ and $t=1$ is good.



Your proof is not enough: that's just for three points, but when $b&gt;0$ there are infinitely many points in $[0,b]$, so you can't do it by cases.
HINT: solve the inequality $$0\le tb\le b$$
since $b\geq 0$ we get $t\geq 0$ and $tb\le b$ gives $$b(1-t)\geq 0$$ and since $b\geq 0$ we get $t\le 1$