math issue

[Brad Jennings]

Damn dude, it makes so much sense but I don't remember how to prove it. Brad Jennings "You're mom is nice. Mind if I go out with her?" - Jörgen Sigvardsson

[Stuart van Weele]

Google for "triangle inequality".

[Chris Maunder]

|a+b|2 = (a+b)2
= a2 + 2ab + b2
<= |a|2 + 2|a||b| + |b|2
= (|a| + |b|)2
=> |a+b| <= |a| + |b|

or

For all a,b in R:

We have -|a| <= a <= |a| (defn of absolute value)
Sim'ly -|b| <= b <= |b|

So -|a|-|b| <= a+b <= |a|+|b|
+> |a+b| <= |a|+|b|

cheers, Chris Maunder

[Wesner Moise]

Proof: |x| = x , if x >= 0 |x| = -x , otherwise Case 1: Both a and b >= 0. Then, |a| = a, by definition. Then, |b| = b, by definition Adding both inequalities yields a+b >= 0, so |a+b| = a+b So, |a| + |b| = a + b = | a + b | Thus, the two sides are equal Case 2: A >= 0 , B < 0 |A| = A |B| = -B |A| + |B| = A - B Since B < 0, -B > 0 by multiplication of -1 Thus A - B > 0 and |A| + |B| > 0 if A+B < 0, Then |A+B| < 0 < |A| + |B| if A+B > 0, then |A+B| = A + B, by definition | A + B | - ( | A | + | B | ) = A + B - ( A - B ) = 2B < 0, since B < 0 Adding |A| + |B| to both sides, we have | A + B | < | A | + | B | Case 3: A is negative, B is nonnegative By the commutative property of addition, |A| + |B| = |B| + |A| and |A + B| = |B + A| Reverse the roles of A and B, and proof follows from Case 2 Case 4: Both A<0 and B<0 |A| = -A |B| = -B The addition of two negative numbers is also negative. Since A < 0 and B < 0, adding both inequalities results yields A + B < 0 Thus, by definition, |A+B| = -(A+B) Thus, |A| + |B| = (-A) + (-B), by substitution, = -(A+B), by distributive property of multiplication over addition = | A + B |, by equality Both sides of the equation are equal All cases are proved. QED

[mystro_AKA_kokie]

base assumption: a>b(basic principle would work otherwise) first list all possible configurations of a+b(in terms of the signs they may have.then evaluate).

left side of inequality

(both a&b positive)a+b = c0,|c0| == c0 == (|a|+|b|(right side)); (both a&b negative)-a+(-b) = -c0,|-c0| = -(-c0) = c0; ('a' negative)-a+b = -c1,|-c1|<|c0|; ('b' negative)a+(-b)= c2,(|-c1|==|c2|)<|c0|; proff: all results on left side satisfy inequality. :) It's a sh*tty world. Take advantage of whomever,whenever,whereever. And oh.. becarefull what you say to me,am too sensitive.Or i might just show up at your house.

[nay]

AAAAAAAHHHHHHHHHHHRRRRRRRRRRGGGGGGGGGHHHHHHHHHHHHHHH!!!!!!!!!!!. I just took a final in Non-Euclidean Geometry today! Luckily I didn't have to prove this (we proved it on an earlier test). I'm glad that class is over. nay

[ColinDavies]

For it to be equal the difference between the numbers must be 1 and when multiplied the product must be 1. So a constant of 1.6180339887498948482045868343656.... solves it for equality, or (Sqrt 5 + 1)/2 For anything less than this it will fail. Of course things are different with irrational numbers. :-) Quite shocking that other CPIans have given you all these bogus algebraic answers, when you need your homework done. :-) Regardz Colin J Davies

Sonork ID 100.9197:Colin

Warning Link to the minion's animation, do not use. It's a real shame that people as stupid as you can work out how to use a computer. said by Christian Graus in the Soapbox

[Xiangyang Liu]

Proof by contradiction: the opposite can't be true. If      |a+b| > |a|+|b| is true, then either (a)     a+b > |a|+|b| or (b)     -(a+b) > |a|+|b| From (a) we get      a > |a|+|b|-b = |a|+(|b|-b) >= |a|, so a > |a| From (b) we get      -a > |a|+|b|+b = |a|+(|b|+b) >= |a|, so -a > |a| Both conclusions are false. ;) Click here to see my articles and software tools

[brianwelsch]

A formal proof is not needed because it falls under the realm of Euclid's Rule of Common Sense. Euclid said: "Any Equations which upon first glance seem natural and true, need not be proven, for your time would be wasted, and I abhor wasting time." I swear it's true, he did say that. Really. BW "I always wanted to be somebody, but now I realize I should have been more specific." - Lily Tomlin

[Chris Maunder]

Nice. cheers, Chris Maunder