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Translating Statements

  1. Jul 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Express each of these mathematical statements using predicates, quantifiers, logical connectives, and mathematical operators.

    c) Every positive real number has exactly two square roots.

    d) A negative real number does not have a square root that is a real number.

    2. Relevant equations



    3. The attempt at a solution

    For part c), the answer in the textbook is, [itex]\forall x \exists a \exists b(a ≠ b \wedge \exists c(c² = x ↔ (c = a \vee c = b)))[/itex] my answer is, [itex]\forall x \exists y \exists z [((x > 0) \wedge (a \ne b)) \implies ((\sqrt{x} = y) \wedge (\sqrt{x} = z))][/itex]

    For part d), the answer in the textbook is, [itex]\forall x ((x < 0) \implies \neg \exists y (x = y^2))[/itex]; my answer is, [itex]\forall x[(x < 0) \implies (\sqrt{x} \notin \mathbb{R})[/itex]

    I was wondering, are my answers as valid as the ones provided in the textbook?
     
    Last edited: Jul 5, 2013
  2. jcsd
  3. Jul 5, 2013 #2

    mfb

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    Your answer to (c) introduces y and z, but uses a and b afterwards?
    You cannot use the square root symbol here - it is a function with only one value (the positive root). It cannot be a AND b at the same time, as you want a!=b.
    You probably don't want the effect of "=>" in your formula. It is satisfied even if both sides are wrong (with a=b, for example), so it does not make any statement about the existence of roots at all.

    For (d): You cannot write ##\sqrt{x}## if you don't know if that is defined at all. With complex numbers, that is possible, but even then you need a definition of the square root function first.

    The short answer: No.
     
  4. Jul 5, 2013 #3
    Okay, well I understand the textbook's answer for part d); however, I am having a little more difficulty with part c). My question is, where does it answer specify that x is positive? And to be clear, a and b are the two roots that are mentioned in the original statement?
     
  5. Jul 5, 2013 #4

    mfb

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    That is the magic of "<=>". For negative x, there is no such c, both sides are false and the equality holds. I would write ##\forall x \in \mathbb{R}^+##, however.

    ##\in \mathbb{R}## for all symbols is missing.
     
  6. Jul 5, 2013 #5

    verty

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    I think it has been assumed that all variables range over the real numbers. For this reason, (d) is incorrect as it assumes the square root of x is real for any candidate x.

    I didn't notice that the book's answer for (c) is wrong. I don't particularly like the book's answer anyhow. I think I would use this logic: x has two distinct roots and any third root is non-distinct.

    Bashyboy: see if you can translate my statement.
     
  7. Jul 5, 2013 #6

    haruspex

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    I can't make much sense of that. Are you sure you've copied this correctly?
     
  8. Jul 8, 2013 #7
    Haruspex, that is actually copied-and-pasted from my electronic textbook, so there can't possible exist any error in transcribing.
     
  9. Jul 8, 2013 #8

    haruspex

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    Well, it's wrong. It should be [itex]\forall x \exists a \exists b(a ≠ b \wedge \forall c(c² = x ↔ (c = a \vee c = b)))[/itex]. Note the ##\forall c##.
     
  10. Jul 8, 2013 #9
    Ah, yes, that is in fact more comprehensible.
     
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