# Triangle Inequality Proof

1. May 15, 2013

### Bennigan88

1. The problem statement, all variables and given/known data
Use the triangle inequality to prove that $\left| s_n - s \right| < 1 \implies \left| s_n \right| < \left| s \right| +1$

2. Relevant equations
The triangle inequality states that $\left| a-b \right| \leq \left| a-c \right| + \left| c-b \right|$

3. The attempt at a solution
So far I have $\left| s_n - s \right| < 1 \implies -1 < s_n - s < 1 \implies s_n < s + 1$. But now I don't see how I can use the triangle inequality to move forward. Hints greatly appreciated, no spoilers please!

2. May 15, 2013

### Mandelbroth

Well, if $s_n$ is less than $s+1$, what does that imply about $|s_n|$?

Edit: The utter wrongness of my previous post should be disregarded while I sit in the corner of shame. One can also consider the case where c = 0 in the given statement of the triangle inequality too get your answer.

Last edited: May 15, 2013
3. May 15, 2013

### LCKurtz

Do you have the equivalent inequality $|a+b|\le |a|+|b|$ to work with? That is what is usually called the triangle inequality. Anyway, here's my hint: $|a| = |(a-b)+b|$.

4. May 15, 2013

### Bennigan88

I found that in general, $a < b + 1 \nRightarrow \left| a \right| < \left| b \right| + 1$. Setting $c = 0$ would give me $\left|{s_n-s}\right| < \left|{s_n}\right| + \left|{s}\right|$. I feel like what I really need to show is $\left|{s}\right| - \left|{s_n}\right| \leq \left|{s_n-s}\right|$.

5. May 16, 2013

### davidchen9568

Write down the triangle inequality. You need a minus sign, which can be obtained by moving something to the other side of the inequality.

6. May 17, 2013

### Bennigan88

As far as I can figure, the way to prove it is this: $\left|{a+b}\right| \leq \left|{a}\right| + \left|{b}\right| \implies \left|{a}\right| \geq \left|{a+b}\right| - \left|{b}\right| \implies \left|{\left({a+b}\right)-b}\right| \geq \left|{a+b}\right| - \left|{b}\right| \implies \left|{x-y}\right| \geq \left|{x}\right| - \left|{y}\right|$ as long as $y = b$ and $x = b + a$ but since for any $b$ any other number can be expressed as $b+a$ for some $a$, this holds for all $x,y$.

Is there a flaw? Is there a more elegant way to show this?

7. May 17, 2013

### LCKurtz

Using my hint:$|a| = |(a-b)+b|\le |a-b| + |b|\implies |a|-|b|\le |a-b|$. That's it. You can get absolute values on the left if you switch the $a$ and $b$ so you have $|b|-|a|\le |a-b|$. Together those give $|\,|a|-|b|\,|\le |a-b|$. That is usually called the reverse triangle inequality.

8. May 17, 2013

### Bennigan88

Thank you, that's much nicer.