Vtyjkyuk

I was triying to do this but I couldn't then, I tried to see the answer, and, the answer not satisfies me. The book, and my search on the internet use this.

$$|x + y| le |x| + |y|$$
Which is true, then they use someone called "Triangular Inequality"

$$|(x - y) + y| le |x - y| + |y|$$

$$|x| le |x - y| + |y|$$

$$|x| - |y| le |x - y|$$

And that's the proof, but, it doesn't seem so good for me.

If I have the first inequality again (Which is true)

$$|x + y| le |x| + |y|$$ Then, let $x$ be $(n - k)$ and $n,k in Re$

$$|(n - k) + y| le |n - k| + |y|$$

$$|(n - k) + y| - |y| le |n - k|$$

Then, my proof is based in the fact that if k = y, then

$$|n| - |y| le |n - k|$$ Which is I want to prove. That's why I don't get the first proof, because to me seems so arbitrary

You've proven $|n-k+y|-|y|le|n-k|$ for any $n,,k,,y$. The special case $k=y$ of this theorem is therefore also valid, giving $|n|-|y|le|n-y|$. Although any other choice of $k$ wouldn't obtain this second result, as long as some value of $k$ lets you obtain it it's proven.

One way into this is just to go through the cases so there are basically eight options:

1) $x$ positive $y$ positive and $x$ greater than $y$.

2) $x$ positive $y$ positive and $x$ less than $y$.

3) $x$ positive $y$ negative and $x$ greater than $y$.

4) $x$ positive $y$ negative and $x$ less than $y$.

5) $x$ negative $y$ positive and $x$ greater than $y$.

6) $x$ negative $y$ positive and $x$ less than $y$.

7) $x$ negative $y$ negative and $x$ greater than $y$.

8) $x$ negative $y$ negative and $x$ less than $y$.

We need to show that $|x|-|y| leq |x-y|$.

In case 1), all the modulus signs fall away (because $x$, $y$ and $x-y$ are all positive). So we need to prove $x-y leq x-y$. This is easy $x-y = x-y$ :)

In case 2) the left hand side is negative the right hand side cannot be and a negative number is smaller than a non negative number.

In the third case the right hand side will be $|x| + |y|$ which is always going to be bigger than $|x|-|y|$.

The same applies in case 4).

Actually the same applies in 5) and 6) because here, on the right hand side, we will be taking a negative number from a negative number and then turning it positive which is just the equivalent of adding the modulus of both numbers.

In case 7) $x$ is greater than $y$ and they are both negative numbers so if we think of their absolute values then $x$ will be less than $y$. This lets us know that the left hand side is negative (we are taking a larger number from a smaller number) where the right hand side cannot be (because it is an absolute value).

This just leaves us with case 8). In this case, we know that all of the modulus operators are trimming off minus signs (as $x$, $y$ and $x-y$ will all be negative). So here we just need to prove that $(-x)-(-y) leq -(x-y)$ or $y-x leq y-x$ which is true as $y-x=y-x$.