Question

Why is it |a-b| ≤ |a|+|b| and not |a-b| = |a|+|b|

Answer 1

my teacher said it is ≤ and not = but why is this?

Answer 2

pls give a minute.

Answer 3

do you have any idea?

Answer 4

If we assume \( |a-b| = |a|+|b| \forall a,b \in \mathbb{R} \)

Lets choose a = 3 and b =5 does your equality holds ?

Answer 5

what your teacher wants to say is \( |a-b| ≤ |a|+|b| \forall a,b \in \mathbb{R} \) holds for any a,b within that restriction.

Answer 6

hello foolformath,

i dont understand the ≤

Answer 7

\( \le \) means less than or equal to here, substitute a and b and see how it works!

Answer 8

*"less than or equal to"

Answer 9

lets take an example:

a = 3
b = 5

|3-5| ≤ |3|+|5|?
this will be: |-2|≤|3|+|5| right?

Answer 10

well you can say from that: "The difference between two numbers is always less or equal that the sum of both"

Answer 11

Yes, let me continue after you \( |-2|≤|3|+|5| \Rightarrow 2 \le 8 \) and that holds!!

Answer 12

I don't know if i wrote that in a good way

Answer 13

There is a interesting inequality of this form which is well explained here: http://en.wikipedia.org/wiki/Triangle_inequality#Normed_vector_space

Answer 14

oke... i will take another one just to be sure:

a=-5
b = 2

|-5-2| ≤ |-5| + |2|
|-7| ≤ 5 + 2
7 ≤ 7

right?

Answer 15

that inequality could confuse her Fool

Answer 16

There's something called the Triangle Inequality which states\[|a+b|\leq|a|+|b|.\]If we modify it a litte:\[|a-b|=|a+(-b)|\leq|a|+|(-b)|=|a|+|b|,\]it still holds.

Answer 17

@TuringTest: I always like reading your answer such a simple way of explaining things as someone rightly asserted "simple is beautiful" :)

Answer 18

If you're still skeptical, do you want me to prove the Triangle Inequality?

Answer 19

Mistake made above:
think of it this way:
the MOST |a-b| can be is |a|+|b|
if
|a-b|=|a|+|b|
then either
a<0 and b>0
or
a>0 and b<0
or
a=0
and/or
b=0
choosing any two positive a and b will show you that
|a-b|<|a|+|b|
does this help?

Answer 20

thanks FFM :)

Answer 21

any time the signs of a and b are the same
|a-b|<|a|+|b|

Answer 22

unless
a=0
and/or
b=0

Answer 23

okeee so the rule is

|a-b| ≤ |a|+|b|

but what about

|a+b| is that equal to |a|+|b|?

Answer 24

no, across gave you that as the "triangle inequality"
which she will now explain...

Answer 25

@Jessica: As across gave you what you are asking, read him/her answer.

Answer 26

@jessica_nl: that only holds when both a and b have the same sign.

Answer 27

or if
a=0
and/or
b=0

Answer 28

I guess 0 has the same sign as 0 though :/

Answer 29

you typed:

There's something called the Triangle Inequality which states
|a+b|≤|a|+|b|.
If we modify it a litte:
|a−b|=|a+(−b)|≤|a|+|(−b)|=|a|+|b|,
it still holds.

BUT!

If i dont know the sign of a and b,

will can i always use

|a+b|≤|a|+|b|?

Answer 30

it is clear that
\[\leq\] means "less than OR equal to" right?

Answer 31

so it means just that.
\[|a+b|\] is always less than OR equal to
\[|a|+|b|\]

Answer 32

yes you can always use
|a+b|≤|a|+|b|
if you don't know the sign of a and b

Answer 33

meaning basically it cannot be bigger than
\[|a|+|b|\]

Answer 34

only if you know they have the SAME SIGN then
|a+b|=|a|+|b|

Answer 35

ahh oke :)

clear ;)

i didnt know that -.-' so i lost some points at the exam because i uses the = sign in stead of the ≤ sign :S

Answer 36

but thank you guys for your help :)

Answer 37

Oke i can always use

|a+b|≤|a|+|b|

And

|a-b|≤|a|+|b|

Is the same right? to be extra sure for the next time ? :)

Answer 38

yes! that is correct!

Answer 39

Yes, but don't please memorize it as the word of god.

Answer 40

oke,

I had this simplified but in the same form question:

\[\int\limits\limits_{0}^{\infty}|2x ^{2}-3x| dx \]

How can i calculate this integral? Now i know that it is not equal but i have to use this sign ≤

Answer 41

Are you sure that's the integral? Because it doesn't converge (FTC).

Answer 42

owhh lol...

oke.. i will type the real one ;)

Answer 43

\[\int\limits_{0}^{\infty}|9e ^{-(1/4)t}-6e ^{-(1/2)t}|dt\]