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Homework Help: Proving discontinuity for rational numbers (reduced form)

  1. Dec 19, 2016 #1
    Hello! This is my first post on these forums.
    So I was stuck with this question in my Mathematical Analysis exam, and it is as follows:
    ƒ(x) = 0 if x ∉ ℚ and (p + π) / (q + π) - (p / q) if x = (p / q) ∈ ℚ (reduced form).

    1- Prove ƒ is discontinuous at all rational numbers except 1:
    This is easy. Suppose a ∈ ℚ not equal to 1 and for all sequences Xn ∉ ℚ, Xn → a, with ƒ(Xn) → ƒ(a) (that it's continuous at a). Now ƒ(Xn) → 0, because Xn is an irrational sequence. Since a is rational, ƒ(a) = (p + π) / (q + π) - (p / q), after unifying, we get ƒ(a) = (π(q - p)) / (q2 + π × q). since a is not 1, p can never be equal to q, and thus, ƒ(a) is not equal to 0, a contradiction.

    2- ƒ is continuous at every irrational number.
    Here comes your part.
    Using the same method (Sequence Characterization method), we can prove for Xn ∉ ℚ. As for Xn = Pn / Qn ∈ ℚ, suppose Xn → a ∉ ℚ, we want to show ((Pn + π) / (Qn + π)) - (Pn / Qn) → 0, same as ƒ(a), to prove the continuity in all cases. Using algebra, I couldn't find anything to wrap things around. Any ideas?
    Last edited: Dec 19, 2016
  2. jcsd
  3. Dec 19, 2016 #2

    Ray Vickson

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    Is PIE supposed to be ##\pi##? If so, just write it at pi (definitely not pie or PIE), and you can also go to the grey ribbon at the top of the input panel and click on the menu labeled "Σ", then click on the pi symbol, to get π.
  4. Dec 19, 2016 #3
    Yeah, it is pi (all the time I refer to it as pie :P). I can't find the symbol though.
    EDIT: Nevermind.
  5. Dec 21, 2016 #4


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    Some algebra will give you [tex]
    \frac{p + \pi}{q + \pi} - \frac pq = \frac{\pi}{q} \left(1 - \frac{p}{q}\right) \left(1 + \frac{\pi}{q}\right)^{-1}.[/tex] Now you just need the result that if [itex]r_n = p_n/q_n[/itex] is a rational sequence (in lowest terms with [itex]q_n > 0[/itex]) such that [itex]r_n \to a \in \mathbb{R} \setminus \mathbb{Q}[/itex] then [itex]q_n \to \infty[/itex].

    Now assuming that [itex]q_n[/itex] is bounded above leads to a contradiction (there would exist an [itex]\epsilon > 0[/itex] such that for sufficiently large [itex]n[/itex] there is no integer [itex]P[/itex] such that [itex]|a - \frac{P}{q_n}| < \epsilon[/itex], which is a condition that [itex]p_n \in \mathbb{Z}[/itex] must satisfy), and if a sequence of integers has no upper bound then ... we don't quite have what we need.
    Last edited: Dec 21, 2016
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