# Does getting $\pi$ as the result of an expression mean the expression must involve an irrational (transcendental?) number?

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?

Thanks.

• Possible duplicate of Can we ever get an irrational number by dividing two rational numbers? – arseniiv Dec 16 '17 at 19:26
• 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. – Dave L. Renfro Dec 16 '17 at 19:26
• @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? – RickSanch3z Dec 16 '17 at 20:05
• 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. – Dave L. Renfro Dec 16 '17 at 21:35
• 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. – Dave L. Renfro Dec 16 '17 at 21:42

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.