- #1

mscbuck

- 18

- 0

## Homework Statement

"There is a very useful way of describing the points of the closed interval [a,b] (where we assume, as usual, that a < b)

a. Consider the interval [0,b]. Prove that if x is in [0,b] then x = tb for some t with 0 <= t <= 1. What is the significance of the number t? What is the mid-point of the interval [0,b]

b. Now prove that if x is in [a,b], then x = (1-t)a + tb for some t with 0 <= t <= 1. Hint: Can also be written as a + t(b-a). What is the midpoint of the interval [a,b]? What is the point 1/3 of the way from a to b?

c. Prove, conversely, that if 0 <= t <= 1, then (1-t)a + tb is in [a,b]

## Homework Equations

N/A

## The Attempt at a Solution

I think I'm really close and I'm just missing something extremely stupid. I solved part a like so

0 <= x <= b

0/b <= x/b <= b/b

0 <= x/b <= 1

Let x/b = t

0 <= t <= 1

I'm stuck on part b however. No matter how many times I did the algebra I just never could come up with the right equation. I assume to start out like this:

a <= x <= b

a/b <= x/b <= b/b

Let x/b = t

a/b <= t <= 1

a <= tb <= b

From here I'm somewhat stuck. I see that "tb" is part of the equation I"m looking for, I just need a (1-t)a from there.

I also think I have solved c:

0 <= t <= 1

0 * (b-a) <= t * (b-a) <= 1 (b-a)

0 <= t(b-a) <= b -a

a <= a + t(b-a) <= b

Thanks for all of your help in advance!

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