Tricky, Interval Proof, is my Proof correct?

Question

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 considered only $0,b$ and the midpoint, while you were supposed to show this for all points in the interval. What about, say $\frac b4$ etc.? — Sarvesh Ravichandran Iyer, Nov 04 '17 at 11:55
yep, your right, can you provide maybe a hint, on how to prove for all points? — SFD, Nov 04 '17 at 12:00

score 1 · Answer 1 · answered Nov 04 '17 at 12:10

You're given that $0\le x\le b$ Suppose $b>0$. 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 $b>0$ there are infinitely many points in $[0,b]$, so you can't do it by cases.

score -2 · Accepted Answer · answered Nov 04 '17 at 12:00

-2

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 $t\le 1$

answered Nov 04 '17 at 12:00

In which way does it proof the statement? You assume that IF $x=tb$ THEN $t \in [0,1]$. But it shall be proven that if $x \in [0,b]$ THEN $x=tb$ for some $t\in [0,1]$… that's a huge difference. So nice proof but not for the given statement. – Gono Nov 04 '17 at 13:45

2 Answers2