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Let $\mathbb{R}$ be the set of all real numbers under the usual metric $d$ defined as follows: $$d(x,y) \colon= |x-y|$$ for all $x$, $y$ in $\mathbb{R}$, and let $\mathbb{Q}$ be the set of all rational numbers.

Then how to rigorously prove that, given a positive integer $n$ which is a non-perfect square, the number $\sqrt{n}$ is in the closure of $\mathbb{Q}$?

For example, how to show that $\sqrt{3}$ is in the closure of $\mathbb{Q}$?

Generalising the above, how to show that, for a given positive integer $n$ which is not the $k$th power of any positive integer, the number $\sqrt[k]{n}$ is in the closure of $\mathbb{Q}$?

How to explicitly find a sequence of rational numbers that converges to our number? And how to rigorously demonstrate this convergence?

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  • $\begingroup$ Continued fractions might be a good approach in general. $\endgroup$
    – Hayden
    Oct 31, 2014 at 10:47
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    $\begingroup$ There is more than one method of constructing $\mathbb R$ from $\mathbb Q$ - they are equivalent, but there are proofs which follow each. One, for example, would be to find a Cauchy Sequence in the rationals which would converge to the appropriate limit. Methods of approximation to roots of $x^n-k=0$ can be used for this. Another would be to find a set of rationals whose least upper bound is the square root you are looking for (prove that the square can't be greater or less than $n$). This is easier than finding an explicit sequence. $\endgroup$ Oct 31, 2014 at 10:51

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$\mathbb{Q}$ is dense in $\mathbb{R}$. That is for any $x \in \mathbb{R}$ you can find a sequence of rationals converging to $x$.

To see this, let $\lfloor x \rfloor$ denote the floor function. Note that $$\lfloor x \rfloor \leq x < \lfloor x \rfloor + 1.$$

Show that $$ l_n := \frac{\lfloor 2^n x \rfloor}{2^n} \to x. $$

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Newton's method starting at $x_0=N$ finds a sequence of rational numbers that converges to $\sqrt[k] N$ : $$ x_{n+1} = f(x_n), \qquad f(x)=x-\frac{x^k-N}{kx^{k-1}} = \frac1k\big((k-1)x+\frac{N}{x^{k-1}}\big) $$

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