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Having the luxury to just wander about in mathematics, I asked if there is a number system with a base (radix ?) that is not an integer. A short search on the internet provides a yes answer. The one article I skimmed showed the golden mean can be used as a base. The golden mean is irrational, thus, a base can be irrational. My question then is are irrational numbers in base 10 irrational in other number systems with an irrational base?

I first thought of this question for more specific numbers such as Pi and e. So, the number represented by the symbols 10 is an integer in base e, for example. At least it looks like an integer. It also appears to be rational.

It may well be that many of the numbers, I will call them counting numbers to identify them, in a base e system are irrational. I realized that I think I can not write what is 1 in base 10 in base e. If I go outside and pick up a rock and put on the picnic table, then there is a symbol that represents the number of rocks on the table. If I put a second rock on the table, I can still see a symbol in my mind that represents those rocks. It is when I pick up another rock and put it on the table that things get complicated. What is that number? Is it rational?

So, in general, are numbers that are irrational in base 10 numbering, irrational in other number systems that have an irrational base?

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    $\begingroup$ Irrational numbers are irrational numbers. $\endgroup$
    – Angina Seng
    Apr 3, 2020 at 21:38
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    $\begingroup$ I feel like the core issue is "how do you define an integer?" (Or rational, irrational, etc.) Our positional base notation does not change the nature of what is an integer or not, just its representation, how it appears to us. However, there is a formal construction of the set of integers, so when $5$ is said to be an integer, this is independent of any actual representation of it. Perhaps what you instead might refer to as an integer might be "a number with no fractional part/decimal part"? $\endgroup$
    – PrincessEev
    Apr 3, 2020 at 21:40
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    $\begingroup$ Any transcendental number will be "irrational" in any algebraic base. For example, $e$ will appear to be irrational in base $\sqrt p$ for all primes $p$, or $\pi$ will be irrational in any base that is a real solution of any given polynomial with algebraic coefficients. What may be interesting is to discover whether there are rational appearances among a set of algebraic numbers in a given algebraic base, or similar appearances from a set of transcendentals given a specific transcendental base. $\endgroup$
    – abiessu
    Apr 3, 2020 at 21:40
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    $\begingroup$ What you appear to be really asking is, given two numbers $a,b$, does there exist a polynomial $p(x)$ having rational coefficients such that $p(a)=b$. $\endgroup$
    – abiessu
    Apr 3, 2020 at 21:48
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    $\begingroup$ Consider the polynomial $p(x)$. Written in coefficient/exponent form, it looks like $p(x)=c_nx^n+c_{n-1}x^{n-1}+\dots+c_2x^2+c_1x^1+c_0x^0$. If you consider the decimal number $N=d_nd_{n-1}d_{n-2}\dots d_2d_1d_0$ where each $d_i$ is a decimal digit, then this number may be considered as a polynomial in $x$ taken at the value $x=10$, because we have $d_n10^n+d_{n-1}10^{n-1}+\dots+d_210^2+d_110^1+d_010^0$. $\endgroup$
    – abiessu
    Apr 3, 2020 at 23:15

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Being irrational (or not) is an intrinsic property of a real number, which is to say a property that the number has completely independently of what symbols you choose to use to describe it. Either the number is a ratio of integers, or it isn't, and using a different number base (even an irrational one) isn't going to change that.

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