Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be the function on $\mathbb{R}$ defined by $f(t) = \frac{p+\sqrt{2}}{q+\sqrt{2}}−\frac{p}{q}$ if $t = \frac{p}{q}$ with $p, q \in \mathbb{Z}$ and $p$ and $q$ coprime to each other, and $f(t) = 0$ if $t$ is irrational.

Answer the following:

  1. At which irrational numbers $t$ is $f$ is continuous?

  2. At which rational numbers $t$ is $f$ continuous?

I am totally stuck in this problem. How should I able to solve this problem?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

I'll start with 2, as it is easier.

Suppose $f$ is continuous on the rational number $t=\frac{p}{q}$. This means that if $x_n \to t$ then $f(x_n) \to f(t)=\frac{p+\sqrt{2}}{q+\sqrt{2}}-\frac{p}{q} = \frac{\sqrt{2}(q-p)}{q(q+\sqrt{2})}$. Since the irrationals are dense, there's a sequence of irrational numbers $y_n$ tending to $t$, in which case we must have $0=f(y_n) \to \frac{\sqrt{2}(q-p)}{q(q+\sqrt{2})}$, proving $\frac{\sqrt{2}(q-p)}{q(q+\sqrt{2})}=0$, i.e. $p=q$, or: $t=1$.

We'll prove that $f$ is indeed continuous on $1$. Let $x_n$ be a sequence of numbers tending to $1$. We can assume that it consists of rational numbers $\frac{a_n>0}{b_n>0}$ (since irrationals always give the value of the limit). We need to show $\frac{a_n+\sqrt{2}}{b_n+\sqrt{2}} - \frac{a_n}{b_n} \to 0$, or $\frac{a_n+\sqrt{2}}{b_n+\sqrt{2}} \to 1$, or $\frac{a_n-b_n}{b_n+\sqrt{2}} \to 0$ , or $\frac{\frac{a_n}{b_n}-1}{1+\frac{\sqrt{2}}{b_n}} \to 0$, which is true since the nominator goes to 0 and the denominator is bounded from below by 1.


Time for part 1. Say $f$ is continuous at the irrational $t$. Since $f(t)=0$, we need to determine when $\frac{a_n}{b_n} \to t$ implies $f(\frac{a_n}{b_n}) = \frac{\sqrt{2}(b_n-a_n)}{b_n(b_n+\sqrt{2})} = \sqrt{2} (1-\frac{a_n}{b_n})\frac{1}{b_n+\sqrt{2}}\to 0$. Since $\sqrt{2} (1-\frac{a_n}{b_n}) \to \sqrt{2}(1-t)\neq 0$, we need to understand when $\frac{a_n}{b_n} \to t$ implies $|b_n + \sqrt{2}| \to \infty$, i.e. $|b_n| \to \infty$.

This is always implied, since an irrational $t$ cannot be approximated by rational numbers with bounded denominator. Consider the set $A=\{ \frac{a}{b} | 1 \le |b| \le M \}$. We'll show that $c_{M}=\min_{x \in A} |x-t|$ exists and is positive. It exists since for any option of $b$ (there are finitely many), there's a best suitable $a$: this amounts to finding the closest integer to a number, $\min_{a \in \mathbb{Z}} |\frac{a}{b}-t| = \frac{1}{b}\min_{a \in \mathbb{Z}} |a-tb| $. It is positive since $t$ is irrational. This shows that for any $M$, there's $N$ such that $n \ge N \implies |b_n| > M$, just take $N$ such that $|\frac{a_n}{b_n}-t| < c_M$ for $n\ge N$.

In conclusion: $f$ is continuous on $1$ and on irrational numbers, i.e. whenever it attains 0.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.