How do you prove |a-b| = |b-a|?

Question

Let a and b be eal numbers.

Prove that |a-b| = |b-a|. (absolute values)

I proved it by cases (four), but I'm not sure whether I did it correctly or not. Please prove it if you can, I want to compare my answer and learn. Thanks.

Thanks guys, but those aren't proofs.

And it's "real numbers," not "eal." Typo.

Answer 1

|(a - b)| = |(-1) x (b - a)|= |(-1)| x |(b - a)| = |(b - a)|

Answer 2

a + b = b + a (a+b) - (b+a) = (b+a) - (b+a) Since b and a are part of the group, then the element (b+a) must also be an element of the group. We add the element -(b+a) which is the additive inverse in the group. Therefore, the right side is 0. However, we do not know yet for the left side. (a+b) - (b + a) = 0 a + b - b - a = 0 a + (b-b) - a = 0 b-b = 0, leaving us with a - a = 0 0 = 0 Does this work (or have I indirectly assumed commutativity in order to prove commutativity?)

Answer 3

if a is 2 and b is 3, 2-3 is -1.. 3-2 is 1 and tha absolute value of both is one .. proof right there

Answer 4

|a - b| = |b - a|

this becomes

a - b = b - a or a - b = -(b - a)
2a = -2b or a - b = -b + a
a = -b or 2b = 0a
a = -b, or -a = b or b = 0 and a = undefined

So proof is that a = -b or -a = b, or b = 0 and a = undefined, so i guess the true answer would have to be

Proof ANS : a = -b or -a = b

Answer 5

it cant...how can 3-4 be 4-3??

Answer 6

They are absolute values. There are no negatives in absoulte values. (unless the neg. is outside the sign)

Answer 7

1) |b-a|

2) multiply by 1 (or in this case, |-1|): |(-1)*(b-a)|

3) simplify: |a-b|

Therefore: |a-b| = |b-a|

Answer 8

|a-b| = |-1(a-b)|

since (b-a) = -1(a-b)

|a-b| = |b-a|

Answer 9

I can do it...

a=1
b=1

1-1=1-1

see that.... its done!

Answer 10

let {a-b} = x
then {-1}*{a-b} = {-1}*x
and -{a-b} = -x
so {b-a} = -x

By definition | x | = | -x |

and hence by substitution {a-b} = {b-a}