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Thread: Oh dear... I need homework help.

  1. #1 Oh dear... I need homework help. 
    Forum Professor Daecon's Avatar
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    I've received my first lot of homework assignments, and I'm already struggling.

    Specifically, the question that's got me confused is this:

    Find integers a, b and c so that a-(b-c)=(a-b)-c

    After playing around with positives and negatives, or a, b and c all having the same value, what I've come up with so far is that as long as c=0, then it doesn't matter what values a or b have.

    Which brings me to the problem, if I'm not overlooking something then how can I decide what integers to assign to a and b if it doesn't matter?

    Is there another solution I've not thought of, where c isn't 0 that I need to keep looking for?


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  3. #2  
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    Quote Originally Posted by Daecon View Post
    I've received my first lot of homework assignments, and I'm already struggling.

    Specifically, the question that's got me confused is this:

    Find integers a, b and c so that a-(b-c)=(a-b)-c

    After playing around with positives and negatives, or a, b and c all having the same value, what I've come up with so far is that as long as c=0, then it doesn't matter what values a or b have.

    Which brings me to the problem, if I'm not overlooking something then how can I decide what integers to assign to a and b if it doesn't matter?

    Is there another solution I've not thought of, where c isn't 0 that I need to keep looking for?
    You've figured out the right answer, although we can get there without experiment.

    First, note that a is common to both sides of the equation, so we can take it out right away (meaning that its value is irrelevant). Then we're left with -(b-c) = (-b)-c, allowing us to eliminate b from both sides (again meaning that its value is irrelevant). That ultimately leaves us to solve -(-c) = -c, or c = -c. That tells us that c = 0.


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  4. #3  
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    If you rearrange the equation you get a-b+c=a-b-c
    Then subtract a-b from both sides of the equation, to get c=-c
    Then c has to be 0. a and b could be anything.
    ETA: Oops, you beat me to it.
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    Forum Professor Daecon's Avatar
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    That's super, thanks guys.
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  6. #5  
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    Quote Originally Posted by Daecon View Post
    Find integers a, b and c so that a-(b-c)=(a-b)-c
    What's the subject? (Yes, I know it's mathematics) It seems like an unusual question to ask unless it's to demonstrate the non-associativity of the "" operation.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  7. #6  
    Forum Professor Daecon's Avatar
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    The course I'm doing is entitled "Introduction to Mathematical Thinking" and this is one of the questions from the first assignment.

    Most of the questions are about evaluating expressions and our understanding of the order of operations.

    A related question is finding a, b and c when a/(b/c)=(a/b)/c which I was planning on returning to after re-reading the lecture materials.

    I'm having difficulty with this one, but my first guess based on the previous question is that c=1, and that a and b can be any number. However this question is from a series that will be covered in a tutorial class on Wednesday, so I'm not too worried that I can't work it out at the moment.
    Last edited by Daecon; March 8th, 2014 at 05:08 AM. Reason: a and b
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  8. #7  
    KJW
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    Quote Originally Posted by Daecon View Post
    our understanding of the order of operations.
    I thought so, hence my mention of (non-)associativity. The division question confirms that. However, I did thought that it might've been about the axiomatic approach to algebra.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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