Triangle Inequality Proof

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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!
 

Answers and Replies

  • #2
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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.
 
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  • #3
LCKurtz
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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
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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
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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
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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|##.

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
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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.
 

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