Prove (4) by breaking the proof into cases akin to the proof of Theorem 1.1. x·y...
Prove (4) by breaking the proof into cases akin to the proof of Theorem 1.1.
x·y ≤ |x·y| = |x|·|y| for all x,y∈R. (4)
(for reference)
Theorem 1.1 (Triangle inequality).
|x+y|≤|x|+|y| forallx,y∈R. (3)
Proof. To prove (3), we consider each possible case so to be
able to exploit the definition (1).Case 1: x ≥ 0, y ≥ 0. We then
have by (1) that |x| = x, |y| = y, and |x + y| = x + y, and so (3)
holds as an equality.
Case 2: x < 0, y < 0. We then have by (1) that |x| = −x, |y|
= −y, and |x + y| = −x − y, and so again (3) holds as an
equality.
Case3: x<0,y≥0,x+y≤0.
Wethenhaveby(1)that|x|=−x,|y|=y,and|x+y|=−x−y. We see now
|x + y| = −x − y ≤ −x + y = |x| + |y|since −y ≤ 0 ≤ y and so (3) holds.
Case 4: x < 0, y ≥ 0, x + y ≥ 0. We now have |x| = −x, |y| = y, and |x + y| = x + y, and deduce|x + y| = x + y ≤ −x + y = |x| + |y|
2
since x < 0 < −x.
Case 5: The only other cases are where the roles of x and y in
cases 3 and 4 are switched, and so (3) holds then as well.
Homework Answers
x·y |x·y| = |x|·|y| for
all x,y
R. (4)
To prove (4), we consider each possible case.
Case 1: x 0, y
0 , then |x| =
x, |y| = y => |x| . |y| = x . y
Also, x 0, y
0 => x . y
0. Thus, |x·y| = x . y
So, |x·y| = x . y = |x| . |y| , and so (4) holds as an equality.
Case 2: x < 0, y < 0. then |x| = -x, |y| = -y => |x| . |y|
= (-x) . (-y) = x . y
Also, x < 0, y < 0 => x . y 0. Thus, |x·y| = x
. y
So, |x·y| = x . y = |x| . |y| , and so (4) holds as an equality.
Case3: x<0, y 0, then |x| = -x,
|y| = y => |x| . |y| = (-x) . (y) = - x . y
Also, |x . y| = (-x) .y = -x . y
Thus, |x . y| = |x| . |y|
Also, x<0, y 0, => x . y <
0
As, |x| > 0, |y| > 0 => |x| . |y| > 0
=> x . y < 0 < |x| . |y|
Thus,
x . y < |x . y| = |x| . |y|
and so (4) holds.
Case 4: x 0, y < 0,
then |x| = x, |y| = -y => |x| . |y| = (x) . (-y) = - x . y
Also, |x . y| = x . (-y) = - x. y
Thus, |x . y| = |x| . |y|
Also, x 0, y < 0, =>
x . y < 0
As, |x| > 0, |y| > 0 => |x| . |y| > 0
=> x . y < 0 < |x| . |y|
Thus,
x . y < |x . y| = |x| . |y|
and so (4) holds.
Add Answer to:
Prove (4) by breaking the proof into cases akin to the proof of
Theorem 1.1.
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