Question

Prove that for all numbers a and b, \[|a+b| \leq |a| + |b|\]Prove that for all numbers a and b, \[|a+b| \leq |a| + |b|\]@Mathematics

Answer 1

dear math, please grow up and solve your own problems! :P

Answer 2

just consider each of these cases in turn to get your proof:
a<0, b<0
a<0, b=0
a<0, b>0
a=0, b<0
a=0, b=0
a=0, b>0
a>0, b<0
a>0, b=0
a>0, b>0

Answer 3

the book said you only need 4 cases, and not nine :-D

Answer 4

true, you could make use of the fact that a+b=b+a to eliminate some

Answer 5

by the way, I was just wondering if there some story behind why you chose the name @agdgdgdgwngo?

Answer 6

consider the case that a geq 0 and b geq 0,
it follows that |a + b| = a + b, which is equal to |a| + |b|

consider the case that a leq 0 and b leq 0.
it follows that |a + b| = -(a + b) = -a + (-b) = |a| + |b|

consider the case that a geq 0 and b leq 0,
hence, we must prove that |a + b| leq a - b
there are two subcases here: a - b geq 0 or a - b leq 0
for the first subcase we can show that a + b leq a - b by simply subtracting a, proving that -b geq b
for the second subcase we can show that -a - b leq a - b, and by simply adding b, proving that a geq -a

same thing with the case that a leq 0 and b geq 0, but with a and b switched.

tombstone

Answer 7

yup - so 4 cases is indeed necessary and sufficient

Answer 8

:( silly book. the next page says there is a much easier proof

Answer 9

I guess the corollary to "think ahead" is "read ahead" in this case! :-)

Answer 10

alright here is the books new and improved proof (I won't be surprised to see an even easier proof 400 pages later though):

using |a| = sqrt{a^2}, (|a + b|)^2 = (a+b)^2 = a^2 + 2ab + b^2
leq a^2 + 2|a|*|b| + b^2
= |a|^2 + 2|a|*|b| + |b|^2
= (|a| + |b|)^2

now root both sides to get that |a+b| leq |a| + |b| since if x^2 < y^2 then x < y provided that x geq 0 and y geq 0 (the book says I have to prove that soon) :(

Answer 11

wow - neat proof