Hello math aces, I have a question that bugs me. I guess I possibly misunderstood something, maybe you can explain it to me.
As I understand it, in irrational numbers like pi, no sequence of numbers repeats. But if one looks at the combination of numbers from 00 to 99, for example, there are a hundred possible combinations, so that in my eyes think a combination of numbers must repeat at some point, after all possibilities of combination have been "exhausted". But as I said, I am no expert and I am grateful for clarifications.
The first thing to note here is that a repeating decimal doesn't have to be made up of only two digits, and there are many more than a hundred possible arrangements for a 10-digit number made up of the digits from 0 to 9, for example. A number can repeat twice or more in a decimal sequence of 10, so we may have 0
113485926. But just to give you an idea of how many possible arrangements there are, let's say that we can't reuse a single number in a given sequence of 10.
The above means our possible permutations are 10P10 = 10! = 3,628,800 possible arrangements. These are permutations, not combinations: order matters here, because 0113 is different from 1103 even though they both use the same set of numbers.
Now, the second part is how to prove that an irrational number is non-repeating and non-terminating. One way to go about this is to use proof by contradiction: if we can show that an assumption of decimal repetition and termination in an irrational number leads to a contradiction, then we can say that
by definition, an irrational number is non-repeating and non-terminating.
Let
x be an irrational number. We will assume the negation of our initial assumption is true and then show it to be contradictory, thereby disproving it and showing that any number meeting these conditions must be rational. Our two assumptions that we want to examine here are these:
1) the irrational number will terminate at some point, or
2) the irrational number will be non-terminating but will repeat.
If 1 is true, then it is necessary that a positive integer
n exists such that the decimal expansion stops after
n digits. Consider this example:
0.25. This is the decimal representation of 1/4 (a rational number). Notice that this terminates two digits after the decimal point, so n = 2. Therefore, we can express this as an integer =
x × 10^n, which, in this case, is 0.25 × 10² = 0.25 × 100 = 25. As you can see, this is an integer. Let's call it m.
From the above, we can see that m =
x × 10^n, so
x = m/(10^n). But now we can see one contradiction against our initial assumption: since we can see that
x can be expressed as the ratio of the two integers m and 10^n, it is therefore a rational number. So we have just disproven the first assumption and shown that an irrational number will
not terminate at some point.
The second part has to do with disproving that the irrational number will have a repeating decimal expansion. Again, we assume that an irrational number will have a repeating decimal expansion and then show that this leads to a contradiction, proving the negation of our assumption via contradiction. Consider this repeating decimal:
x = 35.
34823232323 ad infinitum.
For a repeating decimal, there are an
m number of positive integers, in this case 3, followed by a repeating number of positive integers
n, in this case 2 (since 23 is made up of two digits). So, representing this using a base of 10, we get this:
x × 10^m × (10^n - 1). Note that this is the product of a few integers, which means it is also an integer itself. Let's call it
I, which gives us this equation:
I =
x × 10^m × (10^n - 1). This means the number we have assumed to be irrational,
x, can be expressed as
x = I/(10^m × (10^n - 1). But now we see the contradiction in our initial assumption:
x can be expressed as the ratio between two integers, which means it is rational.
Therefore, based on (1) and (2), any number
x whose decimal expansion is terminating or infinitely repeating must, by definition, be a rational number. Otherwise it is irrational, as in the case of pi.
