Triangle Inequality Proof

I was trying to get things to cancel in the original inequality, which wasn't working. I knew I needed something like this but I just couldn't see it. Thanks again.In summary, we can use the reverse triangle inequality to prove that if the absolute value of the difference between two numbers is less than 1, then the absolute value of the first number must also be less than the absolute value of the second number plus 1.
  • #1
Bennigan88
38
0

Homework Statement


Use the triangle inequality to prove that [itex] \left| s_n - s \right| < 1 \implies \left| s_n \right| < \left| s \right| +1[/itex]

Homework Equations


The triangle inequality states that [itex]\left| a-b \right| \leq \left| a-c \right| + \left| c-b \right|[/itex]


The Attempt at a Solution


So far I have [itex] \left| s_n - s \right| < 1 \implies -1 < s_n - s < 1 \implies s_n < s + 1[/itex]. But now I don't see how I can use the triangle inequality to move forward. Hints greatly appreciated, no spoilers please!
 
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  • #2
Bennigan88 said:

Homework Statement


Use the triangle inequality to prove that [itex] \left| s_n - s \right| < 1 \implies \left| s_n \right| < \left| s \right| +1[/itex]

Homework Equations


The triangle inequality states that [itex]\left| a-b \right| \leq \left| a-c \right| + \left| c-b \right|[/itex]

The Attempt at a Solution


So far I have [itex] \left| s_n - s \right| < 1 \implies -1 < s_n - s < 1 \implies s_n < s + 1[/itex]. But now I don't see how I can use the triangle inequality to move forward. Hints greatly appreciated, no spoilers please!
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:
  • #3
Bennigan88 said:

Homework Statement


Use the triangle inequality to prove that [itex] \left| s_n - s \right| < 1 \implies \left| s_n \right| < \left| s \right| +1[/itex]

Homework Equations


The triangle inequality states that [itex]\left| a-b \right| \leq \left| a-c \right| + \left| c-b \right|[/itex]


The Attempt at a Solution


So far I have [itex] \left| s_n - s \right| < 1 \implies -1 < s_n - s < 1 \implies s_n < s + 1[/itex]. But now I don't see how I can use the triangle inequality to move forward. Hints greatly appreciated, no spoilers please!

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
I found that in general, [itex] a < b + 1 \nRightarrow \left| a \right| < \left| b \right| + 1[/itex]. Setting [itex] c = 0 [/itex] would give me [itex] \left|{s_n-s}\right| < \left|{s_n}\right| + \left|{s}\right|[/itex]. I feel like what I really need to show is [itex] \left|{s}\right| - \left|{s_n}\right| \leq \left|{s_n-s}\right|[/itex].
 
  • #5
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
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
LCKurtz said:
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|##.

Bennigan88 said:
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?

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
LCKurtz said:
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.

Thank you, that's much nicer.
 

1. What is the Triangle Inequality Theorem?

The Triangle Inequality Theorem states that the sum of any two sides of a triangle is always greater than the third side.

2. How is the Triangle Inequality Theorem used in geometry?

The Triangle Inequality Theorem is used to determine whether three given side lengths can create a valid triangle. It is also used in proofs and constructions in geometry.

3. How is the Triangle Inequality Theorem proven?

The Triangle Inequality Theorem can be proven using the Law of Cosines or the Pythagorean Theorem. It can also be proven using algebraic manipulation and the fact that the shortest distance between two points is a straight line.

4. Can the Triangle Inequality Theorem be applied to all triangles?

Yes, the Triangle Inequality Theorem can be applied to all triangles, regardless of their shape or size.

5. How does the Triangle Inequality Theorem relate to other theorems in geometry?

The Triangle Inequality Theorem is closely related to the Triangle Sum Theorem and the Exterior Angle Theorem. It is also used in conjunction with other theorems to prove various geometric properties and relationships.

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