Proving discontinuity for rational numbers (reduced form)

In summary: Any ideas?In summary, ƒ is discontinuous at all rational numbers except 1. Proving this continuity is easy using sequence characterization.
  • #1
cppabstract
2
0
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?
 
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  • #2
cppabstract said:
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 + PIE) / (q + PIE) - (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 + PIE) / (q + PIE) - (p / q), after unifying, we get ƒ(a) = (PIE (q - p)) / (q2 + PIE × 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 + PIE) / (Qn + PIE)) - (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?

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.
 
  • #3
Ray Vickson said:
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.

Yeah, it is pi (all the time I refer to it as pie :P). I can't find the symbol though.
EDIT: Nevermind.
 
  • #4
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.
 
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Related to Proving discontinuity for rational numbers (reduced form)

1. What is discontinuity in the context of rational numbers?

Discontinuity refers to a break or interruption in the pattern of a rational number. It occurs when there is a gap between two values of a rational number, indicating that the number is not continuous.

2. How can we prove that a rational number is discontinuous?

To prove discontinuity for a rational number in reduced form, we need to show that there is a gap between two values of the number. This can be done by finding two numbers in its reduced form that have different values but are very close to each other, indicating a break in the pattern.

3. What is the significance of proving discontinuity for rational numbers?

Proving discontinuity is important in mathematics as it helps us understand the behavior and properties of rational numbers. It allows us to identify and classify rational numbers based on their continuity, which is crucial in various mathematical applications.

4. Can a rational number be both continuous and discontinuous?

No, a rational number can only be either continuous or discontinuous. A number cannot have both properties at the same time. If a rational number is continuous, it means that there are no gaps or breaks in its values, while a discontinuous number will have at least one break in its pattern.

5. Are there any real-life examples of discontinuous rational numbers?

Yes, there are many real-life examples of discontinuous rational numbers. For instance, the time taken for a car to travel a certain distance can be represented by a rational number, but if the car suddenly stops in the middle of the journey, the time taken will not be continuous. This shows that the rational number representing the time is discontinuous.

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