Question

Don't understand what the instructor is doing here, and how she went from one step to the next:

(See attached img)

Answer 1

you have to break up your absolutes into individual results .... but its hard to see the picture

Answer 2

Ok I'll try writing out what she did.

Answer 3

|x| = |x - y + y| ≤ |x - y| + |y|

She proceeded from that step.

Answer 4

She is trying to prove part (iii), that says :

||x| - |y|| ≤ |x - y|

Answer 5

||x| - |y|| ≤ |x - y|
^ that is the original

Answer 6

|x| = |x - y + y| ≤ |x - y| + |y|

does this step make sense to you?

Answer 7

No

Answer 8

would you say that x = x+y-y ?

Answer 9

x = x + 0

x = x+(y-y)

x = x+y-y

x = x-y+y

x = (x-y) + (y)

Answer 10

Yes this makes sense.

Answer 11

x+y-y = (x-y) + (y)

then

|x+y-y| <= |x-y| + |y|

Answer 12

What you said after the then is what I do not understand.

Answer 13

in order change this into an absolute value; it no longer remains "equal" at all times

Answer 14

Why

Answer 15

simply because an absolute value is NOT the same as a negative integer and so we have to adjust for it, thru some thrm most likely

Answer 16

Could you show me how this "adjustment" works?

Answer 17

I want to know exactly as to how one comes from: x + y - y = (x - y) + y
to |x + y - y| <= |x-y| + |y|

Answer 18

|x+y-y| <= |x-y| + |y|

|10+5-5| <= |10-5| + |5|

|10| <= |5| + |5|

10 <= 5 + 5 is true


|10+(-5) -(-5)| <= |10-(-5)| + |-5|

|10-5 +5| <= |10+5| + |-5|

|10| <= |15| + 5

10 <= 20

Answer 19

\[(x+y-y)^2=(x-y+y)^2=(x-y)^2+2(x-y)y+y^2 \le (x-y)^2+y^2\]

what about this?

example of this being true:
what if x=1 and y=2

\[(1-2)^2+2(1-2)2+(2)^2=1+(-4)+4=1 \le 5=1+4=(1-2)^2+(2)^2\]

Answer 20

.... another great proof destroyed by a short screen :)

Answer 21

poor amistre

Answer 22

recall
\[|x|=\sqrt{x^2}\]
if
\[(x+y-y)^2 \le (x-y)^2+y^2\]
then
\[\sqrt{(x+y-y)^2} \le \sqrt{(x-y)^2+y^2} \le \sqrt{(x-y)^2}+\sqrt{y^2}\]

Answer 23

So for:
|x + y - y | = |x - y| + y, the x and y values have to be positive numbers?

Answer 24

myin I do not understand how you go from:
|x| = \sqrt{(x)^2}
to the next steps.