Just reading Langs calculus... (proof question, probably easy)

RHRN

Just started reading Lang's calculus, and I'm having real difficulty with some of the absolute value proofs (I'm terrible at proving things).

Im asked to prove:
|x+y|=>|x|-|y|

He gives a hint saying to let x=x+y-y and apply theorem 3 (stating that |x+y|<=|x|+|y|)

While the solution is probably obvious i havent a clue what I'm doing!

I can prove |x-y|<=|x|+|y| but thats it!
Also, if anyone can suggest a way to get better at proofs in some way, I'd love to hear it, they're definitely my weak point in maths!

Thanks

hivizman

You probably did |x-y|<=|x|+|y| by thinking of x-y as x+(-y). Then using your Theorem 3 you get |x+(-y)|<=|x|+|(-y)|. Then, knowing that |(-y)|=|y|, your result follows.

Try splitting the expression x+y-y into two terms and using brackets in an analogous way and you should get the required result.

With regard to mathematical proofs, I could never go beyond memorising a bundle of "tricks" so, when I went to university to study mathematics, it was a bit of a struggle to keep up with students who could "see" proofs directly (or perhaps just had a bigger set of tricks).

sponsoredwalk

What a beautiful book

I checked my copy of the book and the question is prove

|x + y| ≥ |x| - |y|

|x| - |y| ≤ |x + y| is the same thing

Well I'll give you a hint and write the rest in spoiler

Work with JUST this side

|x| - |y|

and follow Lang's advice!

|x + y - y | - |y|

Don't read what's in the spoiler!!!

EDIT: I cleaned it up, it was messy and incoherent a minute ago, this is a lot better and a lot more intuitive

Start with just x

01) |x|
02) |x + y - y|
03) using theorem III |a +b| ≤ |a| + |b| set a = (x + y) and b = (-y)
04) |(x + y) + (- y)| ≤ |x + y| + |-y|
05) subtract |y| from both sides
06) |(x + y) + (- y)| - |y| ≤ |x + y| + |-y| - |y|
07) Follow Lang's advice and recognise that |-y| = |y| as the definition stated
08) |(x + y) + (- y)| - |y| ≤ |x + y| + |y| - |y|
09) |(x + (y - y))| - |y| ≤ |x + y| + (|y| - |y|)
10) |x + 0| - |y| ≤ |x + y| + 0
11) |x| - |y| ≤ |x + y|

Q.E.D.

If you get absolutely crazy sick of it I wrote the solution in one line
at a time so that you can get a hint if you're pulling your hair out and
then after reading one line stop and try the rest yourself

Great book, seriously, the lesson I learned from this book is to religiously
stick to & remember definitions as they come up!

hivizman

Another thing to do is to turn the inequality round so that it's a less than or equal expression rather than a greater than or equal to expression:

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

Then add |y| to both sides - as an absolute value must be zero or positive this does not affect the direction of the inequality:

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

Does this way of expressing what you have to prove ring any bells?

By the way, this is another "trick" that can be used a lot in mathematical proofs - rather than working towards the required expression, see what happens if you rearrange the required expression. You can often restate the required expression in a way that makes things a little less obscure, and this can give you a clue as to the shape of your proof.

sponsoredwalk

Did you get it? You mentioned you don't understand what you're doing and
I can see why that may be the case, you have to understand what this is
really saying and it'll all fall into place. Like so many things in mathematics it
comes straight from the Pythagorean theorem & if you re-read the proof of
|x + y| ≤ |x| + |y| knowing this information it'll all make sense.

|x| just means a length, i.e. it can't be a negative number.

Lets just imagine we're on a single number line. If we want to add
|x| to |y| we will get |x + y| = |x| + |y|.



But now if we think of a triangle:



We see that the |x + y| ≤ |x| + |y|

The equation just says that no matter what way you shape the
triangle, be it as I have it in the picture, skewed to some weird angle
or just as a straight line, the |x + y| length will never be bigger than
|x| + |y| although it certainly can be shorter. At most it can be equal.

Think about the Pythagorean theorem, in my picture above set
x = 3
y = 4
Usually the hypoteneuse is called z or something:

x² + y² = z²
3² + 4² = z²
9 + 16 = z²
25 = z²
|z| = 5

We can think of z = |x + y| so z² = |x + y|²
|x + y|² = 25
Taking a square root of both sides:
|x + y|= 5

Think of z ≤ |x| + |y| [or, better yet: |x + y| ≤ |x| + |y| ] and plug in the numbers:

5 ≤ 3 + 4
5 ≤ 7

is true. This is a lot better understoof with vectors, to tell you the honest
truth I didn't understand any of the above until I read the first chapter of
Lang's Intro to Linear Algebra so I can see how this might confuse.

sponsoredwalk

Did you get it? You mentioned you don't understand what you're doing and
I can see why that may be the case, you have to understand what this is
really saying and it'll all fall into place. Like so many things in mathematics it
comes straight from the Pythagorean theorem & if you re-read the proof of
|x + y| ≤ |x| + |y| knowing this information it'll all make sense.

|x| just means a length, i.e. it can't be a negative number.

Lets just imagine we're on a single number line. If we want to add
|x| to |y| we will get |x + y| = |x| + |y|.



But now if we think of a triangle:



We see that the |x + y| ≤ |x| + |y|

The equation just says that no matter what way you shape the
triangle, be it as I have it in the picture, skewed to some weird angle
or just as a straight line, the |x + y| length will never be bigger than
|x| + |y| although it certainly can be shorter. At most it can be equal.

Think about the Pythagorean theorem, in my picture above set
x = 3
y = 4
Usually the hypoteneuse is called z or something:

x² + y² = z²
3² + 4² = z²
9 + 16 = z²
25 = z²
|z| = 5

We can think of z = |x + y| so z² = |x + y|²
|x + y|² = 25
Taking a square root of both sides:
|x + y|= 5

Think of z ≤ |x| + |y| [or, better yet: |x + y| ≤ |x| + |y| ] and plug in the numbers:

5 ≤ 3 + 4
5 ≤ 7

is true. This is a lot better understood with vectors, to tell you the honest
truth I didn't understand the above until I read the first chapter of
Lang's Intro to Linear Algebra so I can see how this might confuse.

RHRN

Thanks to both of you, I managed to prove it without highlighting any of the spoiler

Once I realised that I should have split x+y-y into (x+y) and (-y) it all clicked into place!

Thanks a lot, I'm loving this book!