# Let a and b be real numbers with a < b.

## Homework Statement

Let a and b be real numbers with a < b.

a. Derive a formula for the distance from a to b. Hint: Use 3 cases and a visual argument on the number line.

b. Use your work in part (a) to derive a formula for the distance between (a,c) and (b,c) in a plane.

c. Use the Pythagorean theorem to derive a formula for the distance between the points(a,c) and (b,d) in the adjacent sketch. The sketch is a line on x-y coordinate with two endpoints (a,c) and (b,d) where b > a and d > c.

d. Generalize the distance formula to R^n.

## The Attempt at a Solution

I tried doing (a):

Derive a formula for the distance from a to b. Hint: Use 3 cases and a visual argument on the number line.

Case 1: Assume that 0 ≤ a < b. Since a ≥ 0, the distance from a to 0 is a. Since b ≥ 0, the distance from b to 0 is b. The distance from a to b is the distance from b to 0 minus the distance from a to 0 which is b-a.

Case2: Assume that 0 ≥ b > a. Since 0 ≥ a, the distance from 0 to a is -a. Since 0 ≥ b, the distance from 0 to b is -b. The distance from a to b is the distance from 0 to a minus the distance from 0 to b which is b-a.

Case3: Assume that b ≥ 0 ≥ a. Since b ≥ 0, the distance from b to 0 is b. Since 0 ≥ a, the distance from 0 to a is -a. The distance from a to b is the distance from b to 0 plus the distance from 0 to a which is b-a.

Does it look correct?

Thanks.

Office_Shredder
Staff Emeritus
Gold Member
You're missing a ≥ 0 ≥ b for sure, and also what happens if a = b? I believe the three cases they are referring to is when a>b, when b>a, and when a=b.

You're missing a ≥ 0 ≥ b for sure, and also what happens if a = b? I believe the three cases they are referring to is when a>b, when b>a, and when a=b.

When a ≥ 0 ≥ b. since a ≥ 0, the distance between a and 0 is a and the distance between 0 and b is -b because 0 ≥ b. The distance between a and b is the distance between a and 0 plus the distance between 0 and b: a - b.

If a = b and 0 ≥ a, 0 ≥ b, the distance from 0 to a is -a. The same is true for b. The distance from a to b is -a -(-b)= b -a. Since a = b, b - a = a - a = 0.

If a = b and a ≥ 0, b ≥ 0, the distance from 0 to a is a. The same is true for b. The distance from a to b is a - b. Since a = b, a - b = a - a = 0.

Does it look correct?

Thanks.

Office_Shredder
Staff Emeritus
Gold Member
I think that all of your formulas are correct, but you should notice that you never really needed to know whether a and b are positive or negative, just which one is larger.

Thanks. (a) says I should derive a formula, but there are 3 cases so- 3 formulas. Honestly, i have no idea what to do with these formulas. Do they all magically converge into one :) Can you elaborate a little on this, please?

Office_Shredder
Staff Emeritus
Gold Member
I mentioned it in my previous post: consider the cases a<b, a=b and a>b. It doesn't matter whether a or b are positive or negative (you can check that all your cases where a<b agree, where a=b agree, and where a>b agree).

I mentioned it in my previous post: consider the cases a<b, a=b and a>b. It doesn't matter whether a or b are positive or negative (you can check that all your cases where a<b agree, where a=b agree, and where a>b agree).

It still sounds cryptic to me. Does it pertain to case 1, too? I just copied that from the back of my textbook. The rest of the formulas I modeled after that. Thanks.

Ray Vickson
Homework Helper
Dearly Missed
You're missing a ≥ 0 ≥ b for sure, and also what happens if a = b? I believe the three cases they are referring to is when a>b, when b>a, and when a=b.

He was given the hypothesis that a < b.

He was given the hypothesis that a < b.

Can you give some clues and tips for (a)? Thanks.

Ray Vickson
Homework Helper
Dearly Missed
Can you give some clues and tips for (a)? Thanks.

I am not sure how (b) is drastically different from (a). Would the distance from (a,c) to (b,c) be b - a? The only thing that changed is, while y was equal to 0 in (a), y = c in (b). Maybe there's a more formal way of writing it down, taking into account y = c? If there's, I am not seeing it :)

Does (c) look like this: d = sqrt((b - a)^2 + (d - c)^2) ?

Thanks.

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Ray Vickson
Homework Helper
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I am not sure how (b) is drastically different from (a). Would the distance from (a,c) to (b,c) be b - a? The only thing that changed is, while y was equal to 0 in (a), y = c in (b). Maybe there's a more formal way of writing it down, taking into account y = c? If there's, I am not seeing it :)

Does (c) look like this: d = sqrt((b - a)^2 + (d - c)^2) ?

Thanks.

(b) is basically the same as (a); the only difference is that the line at elevation c instead of at elevation 0. And yes, you have done (c) correctly.

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(b) is basically the same as (a); the only difference is that the line at elevation c instead of at elevation 0. And yes, you have done (c) correctly.

Seeing as how much work went into defining the distance in (a), maybe, something to that effect needs to be done in (b) and (c). I mean it, probably, wouldnt be acceptable to just write down "the distance between (a,c) and (b,c) is a - b by analogy with the case 1 in (a) since y = c doesnt effect the distance between (a, c) and (b, c) because of bla bla" for (b). So how would you stylize (b) in math speak?

Could I write "According to the case 1 of (a) the distance between (a,c) and (b,c) is a - b. Likewise, the distance between (b,d) and (b,c) is d - c. Then by Pythagorean, the distance between (a,c) and (b,d) is d = sqrt((b - a)^2 + (d - c)^2)" for (c)?

Thanks.

Ray Vickson
Homework Helper
Dearly Missed
Seeing as how much work went into defining the distance in (a), maybe, something to that effect needs to be done in (b) and (c). I mean it, probably, wouldnt be acceptable to just write down "the distance between (a,c) and (b,c) is a - b by analogy with the case 1 in (a) since y = c doesnt effect the distance between (a, c) and (b, c) because of bla bla" for (b). So how would you stylize (b) in math speak?

Could I write "According to the case 1 of (a) the distance between (a,c) and (b,c) is a - b. Likewise, the distance between (b,d) and (b,c) is d - c. Then by Pythagorean, the distance between (a,c) and (b,d) is d = sqrt((b - a)^2 + (d - c)^2)" for (c)?

Thanks.

I hope you mean distance = |a-b| in question (b), since distance is always ≥ 0 and (b) does not tell us whether a < b or b < a. However, if we understand that a < b still applies, then you have written the negative of the distance: the distance is b-a, not a-b. And, yes: just saying that setting y = c does not affect the distance is good enough; that follows from some elementary geometry (but, unfortunately geometry is largely untaught now in the modern curriculum, or so I have been told).

I hope you mean distance = |a-b| in question (b), since distance is always ≥ 0 and (b) does not tell us whether a < b or b < a. However, if we understand that a < b still applies, then you have written the negative of the distance: the distance is b-a, not a-b. And, yes: just saying that setting y = c does not affect the distance is good enough; that follows from some elementary geometry (but, unfortunately geometry is largely untaught now in the modern curriculum, or so I have been told).

Thanks. Does putting it for (c) the way I did look acceptable?