16
$\begingroup$

Between any two real numbers, there is an algebraic number and also a transcendental number. I understand what algebraic numbers and transcendental numbers are, but how can I prove that they both exist between any two real numbers?

$\endgroup$
  • 9
    $\begingroup$ For the first part just use the fact that every rational number is algebraic. Now for the second, if there are $x,y\in\mathbb R$ with $x<y$ such that $(x,y)$ contains only algebraic numbers then $(x,y)$ is countable $\endgroup$ – CIJ Dec 5 '16 at 1:04
  • $\begingroup$ math.stackexchange/questions/1690121/… $\endgroup$ – Adam Hughes Dec 5 '16 at 16:55
  • 1
    $\begingroup$ @AdamHughes Your link is broken. You typed in math.stackechange/ instead of math.stackechange.com/. $\endgroup$ – Kevin Dec 5 '16 at 18:32
21
$\begingroup$

Hint: rational numbers are algebraic, and (non-zero) rational multiples of $\pi$ are trancendental.

$\endgroup$
21
$\begingroup$

Here's one way to do it without doing too much work, assuming you already know that between any two reals there is a rational:

  • Every rational is algebraic; so, between any to reals, there is an algebraic real.

  • Meanwhile, the set of algebraic numbers is countable, but every nonempty interval $(x, y)$ is uncountable. So "most" elements of $(x, y)$ are transcendental - in particular, there's at least one! So between any to reals, there's a transcendental, as well.

$\endgroup$
10
$\begingroup$

Here's a concrete example of how to construct a rational $q$ and a transcendental $t$ between two real numbers $r_1$ and $r_2$.

Take the decimal expansions of the two real numbers $r_1$ and $r_2$. At some point (say the $n$th digit) they must differ (otherwise they are the same number).

Now take the number composed of the first $n$ digits of $r_2$. As it has a finite decimal expansion it is rational, and clearly it falls between $r_1$ and $r_2$. Call it $q$.

To obtain a transcendental number in the range, look at the difference $r_2 - q$ and find a positive rational less than this difference (say by taking its decimal expansion as far as the first nonzero digit). Call it $d$.

Consider a value $x = \frac{\pi d}{4}$. This is transcendental and $0 < x < d$. Therefore $t = q + x$ is also transcendental and $r_1 < t < r_2$.

$\endgroup$
  • 1
    $\begingroup$ you mean "take a decimal expansion for each of the two real numbers..." $\endgroup$ – djechlin Dec 5 '16 at 16:40
  • 1
    $\begingroup$ What if the first digit of $r_2$ differing from $r_1$ is also the last non-zero digit of $r_2$? Then you would have $q = r_2$. Of course there will always be a number $r$ with finite decimal expansion such that $r_1<r<r_2$, and it should be easy enough to tweak this procedure to find such a number. $\endgroup$ – David K Dec 5 '16 at 17:55
4
$\begingroup$

Def'n: A set $T\subset \mathbb R$ is dense in $\mathbb R$ iff $T\cap (a,b)\ne \emptyset$ whenever $a<b.$

If $x$ is transcendental and $y$ is rational then $x+y$ is transcendental.

If $S$ is dense in $\mathbb R$ and $x\in \mathbb R$ then $x+S=\{x+y:y\in S\}$ is dense in $\mathbb R.$

So take transcendental $x$ and $S=\mathbb Q.$ Then $T=x+S$ is a dense set of trancendentals.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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