Should we have an irrational (transcendental?) number in an expression to result in number $\pi$?

As an example: $$\pi = \frac{\text{circle circumference}}{\text{an integer diameter}}.$$

If the statement is true, then for any given integer diameter to result in $\pi$, we should have: $$\text{circle circumference} = \text{an irrational(transcendental?) number?}$$ In brief: Is it true that if we multiply an irrational (transcendental) number by an integer we always get an irrational (transcendental) number?


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    $\begingroup$ Possible duplicate of Can we ever get an irrational number by dividing two rational numbers? $\endgroup$
    – arseniiv
    Dec 16, 2017 at 19:26
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    $\begingroup$ Yes, this is correct. In fact, until relatively modern times (within the last few hundred years), the abstract idea of an irrational number was not really used or singled out for study, but instead the notion of two quantities being incommensurable was what people talked about. $\endgroup$ Dec 16, 2017 at 19:26
  • $\begingroup$ @arseniiv how could it be the same question? I asked: To get an irrational (transcendental in our case) number do we need to involve an irrational (transcendental) number in an expression? $\endgroup$ Dec 16, 2017 at 20:05
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    $\begingroup$ The product of $\pi$ and any nonzero integer is a transcendental number. This is also true if "nonzero integer" is replaced by "nonzero rational number", or even if "nonzero integer" is replaced by "nonzero algebraic number". More generally, the product of any transcendental number and a nonzero algebraic number will be a transcendental number. A proof of this is not all that immediate (unlike the case of a nonzero rational number multiplied by an irrational number being irrational), because I believe you pretty much have to show the algebraic numbers are closed under multiplication. $\endgroup$ Dec 16, 2017 at 21:35
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    $\begingroup$ Incidentally, the weaker result that any transcendental number times a nonzero rational number (note the restriction is to rational now) is transcendental is relatively easy to prove by contradiction --- Let $r$ be nonzero rational and $t$ be transcendental, and assume $rt$ is not transcendental (i.e. $rt$ is algebraic). Then $rt$ is a root of a polynomial with rational coefficients, and if you write this down symbolically and use things like $(rt)^3=r^3t^3,$ then you'll get an equation with rational coefficients that has $t$ as a root -- a contradiction. $\endgroup$ Dec 16, 2017 at 21:42

1 Answer 1


I'll answer a slight generalization of your brief question. Although much of this information is already in the comments, I feel that putting it all in one place would be helpful.

The statement I'll prove is the following: Suppose $\alpha$ is irrational (transcendental), then if $q\ne 0$ is rational, $q\alpha$ is also irrational (transcendental resp.)

I'll assume you're familiar with the facts that if $q,r$ are rational, then $qr$ is rational, and when $q\ne 0$, $1/q$ is also rational.

Then suppose $\alpha$ is irrational, then if $q\alpha$ is rational, say $q\alpha = r$ for a rational $r$, then $\alpha = (1/q)r$, which is the product of two rational numbers, and therefore rational. This contradicts the irrationality of $\alpha$. Hence $q\alpha$ must be irrational.

Now suppose $\alpha$ is transcendental. If $q\alpha$ were not transcendental, i.e. algebraic, then we have a nonzero polynomial with rational coefficients $p(x)$ such that $p(q\alpha)=0$. Let $$p(x)=a_nx^n + a_{n-1}x^{n-1}\cdots+a_1x +a_0.$$ Then $$p(q\alpha) =a_n(q\alpha)^n +a_{n-1}(q\alpha)^{n-1}+\cdots +a_0 =a_nq^n\alpha^n + a_{n-1}q^{n-1}\alpha^{n-1}+\cdots+a_0 =0, $$ so if we let $r(x)$ be the polynomial $$r(x)=a_nq^nx^n+a_{n-1}q^{n-1}x^{n-1}+\cdots +a_1qx+a_0$$, since each of the coefficients $a_iq^i$ is rational (because products of rational numbers are rational), this is a polynomial with rational coefficients. But by construction, we see that $r(\alpha)=0$, contradicting the fact that $\alpha$ is transcendental. Hence $q\alpha$ must in fact be transcendental.

  • $\begingroup$ +1 for "I feel that putting it all in one place would be helpful." I was kind of busy yesterday (among other things, right after those comments I had to leave to help my wife make some fudge for a small get-together where I live today), and thus I didn't give much thought to writing all this up, and anyway I figured several people would do this, but by the time I got back here nearly 12 hours later, it appears you're the only one who actually did it. $\endgroup$ Dec 17, 2017 at 9:32
  • $\begingroup$ @DaveL.Renfro I've understood your idea. But a proper explanation will serve other too. Thanks. BTW. And I still have notification that my answer is not different from another one. $\endgroup$ Dec 17, 2017 at 21:19

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