Is it possible to not have irrational numbers? (Math noob question): Is there a base that can be used like binary that produces no irrational numbers or numbers with an infinite amount of one number after the decimal (don't know the name)? I feel like this is a problem with using base 10. Any insight would be greatly appreciated. Thanks!
Please use layman's terms!
 A: No. The numbers are independent of the base you write them in. You may consider the fact that the decimal representation is not periodic as definition of an irrational number; however that's not the actual definition. Indeed, the irrational numbers were discovered already by the ancient Greek, who didn't know the decimal (or any other positional) number system.
The actual definition of an irrational number is a number that cannot be written as the quotient of two integers. The concept of an integer, as well as the concept of a quotient, are independent of how you write a number.
For example, in a square the quotient of the lengths of the diagonal and the side is an irrational number, namely $\sqrt 2$. It should be obvious that the concepts of square and diagonal don't depend on how you write down the numbers.
Also, the standard proof that $\sqrt 2$ is irrational does not make any use of the representation of any number.
A: We can define the number system independently of the base in which the numbers are written. Euclid uses the length of a line to represent a number, based on the multiples of a reference length (which would be called $1$ today). From this, whole numbers, including negative numbers, can be worked out.
The development of the number system since has been based on thinking about (i) what other numbers can we construct and (ii) what other numbers are useful. The pursuit of these questions is the story of a whole era of mathematics. One of the constructions investigated is through decimal expansions.
It is possible to confine ourselves to working with rational numbers - the field $\mathbb Q$ as it is now called. But this does not include all the numbers Euclid could construct, nor does it solve all the equations we want solved, nor does it deal well with taking limits or modelling continuity.
Choosing a base does not solve these problems, though it can help us to explore and understand why they exist.
A: The question has already been quite satisfactorily answered by celtschk and Mark Bennet. But let me try to answer a modified question "Q2" that may be what the OP was thinking.
In a positional number system where the base is an integer – for example, our familiar decimal, octal and binary systems – it is well known that all rational numbers have terminating or repeating representations, and all irrational numbers have nonterminating, nonrepeating representations. So it is natural to ask (and this may be what the OP intended to ask):
Q2: Is it possible to have a positional number system with (noninteger) base $b$, and a finite collection of digits, such that all real numbers have a terminating or repeating representation?
Note that this question does not ask about "not having irrational numbers". It asks about "not having numbers whose representation in base $b$ resembles what the irrationals look like in the decimal system", a completely different question.
Certainly it is possible to choose $b$ such that some irrational numbers have such representations – like, choose $b=\pi$, then for example $\pi = 10_b$ and $1/\pi = 0.1_b$.
However, the answer to Q2 is no. For any $b$ you choose, the number of terminating or repeating representations is countably infinite. That is because any such representation can be specified by a finite sequence of digits before comma, a finite sequence of digits after the comma before the repeating part, and a finite sequence of digits representing the repeating part. For example, the usual decimal expansion
$$
3/28 = 0.10\overline{714285}
$$
could be specified by the triple of finite digit sequences $(0, 10, 714285)$. The collection of such triples is countably finite.
But the set of real numbers is uncountably infinite, so it is impossible for every real number to have such a representation. Whatever $b$ you choose, some (in fact most) real numbers will have infinite, nonrepeating representations in your positional system. Let's conclude with a concrete example: if you choose $b=\pi$ (and allowed digits $0,1,2,3$), then already the simple rational number $1/2$ will have a nonterminating, nonrepeating representation  $0.1121120210201223000101001\ldots$
