**fraction**is any ratio between whole numbers. In mathematical terms, it is called a

**rational number**. An

**irrational number**is any number such as

**π**which cannot be represented as the ratio of two whole numbers.

In today's blog, I will go over a single lemma regarding fractions:

**Lemma: for any given rational number let's say a/b, there exists an integer let's say**.

**c**such that: absolute(a/b - c) ≤ (1/2)(1) To prove this, we need only consider the case where abs(a) is greater abs(b). [If abs(a) ≤ abs(b), then the conclusion follows from Corollary 2.1, here]

(2) From the division algorithm (see Theorem 1, here), we know that a = bq + r where r ≥ 0 and less than abs(b).

[For example, if a=-3, b=-2, then q=2, r=1 where r is greater than b but less than abs(b).]

(3) Let a'=abs(a), b'=abs(b)

(4) We know that a' - b'q is less than b' (since a' - b'q = r and r is less than b')

(5) So, it follows that: a'/b' - q is less than 1.

(6) Now, if both a,b are positive or a,b are negative, it follows that a/b = a'/b' and abs(a/b - q) is less than 1.

(7) If a,b are of different sign, than -a/b = a'/b' and -a/b - q = -(a/b + q) so that abs(a/b + q) is less than 1.

(8) The conclusion follows from Lemma 2, here.

QED

Corollary: if a/b is a rational number, then there exists an integer c such that: absolute(a/b - c/2) ≤ (1/4).

(1) From the lemma above, we know that for 2*(a/b), there exists a number c such that:

abs(2*(a/b) - c) ≤ (1/2).

(2) Dividing both sides by 2, gives us:

abs(a/b - c/2) ≤ (1/4)

QED

Now, it turns out that any number with a repeating decimal can be represented as a rational number.

Let me start with an example

(1) Let's assume that we have a decimal such as 5.234523452345... We can represent this decimal as a repeating decimal such as 5.2345.

(2) Now, we know if we multiply the number by 10

^{4}we get:

52345.2345

(3) So, subtracting (2) by (1) gives us:

10

^{4}- 1 = 52345 - 5 = 52340.

(4) So, the rational form of this repeating decimal is:

52340/9999.

Now, let's look at the proof that demonstrates this:

Lemma: Any number with a repeating decimal is rational.

(1) Any number with a repeating decimal can be represented with the following form:

d

_{1}...d

_{m}.a

_{1}...a

_{n}

NOTE: If a number has a nonrepeating portion, then we multiply this number by the number of nonrepeating digits, to get a number of the above form. Later, we divide our result by this same number.

(2) We can get an integer result by subtracting 10

^{n}*the number by the original number which after canceling for the repeating decimal gives us:

10

^{n}* (d

_{1}...d

_{m}.a

_{1}...a

_{n}...) - (d

_{1}...d

_{m}.a

_{1}...a

_{n}...) =

(d

_{1}..d

_{m}a

_{1}..a

_{n}) - (d

_{1}..d

_{m}).

(3) Now, our rational number is equal to the value in step #2 divided by 10

^{n}- 1.

QED

## 2 comments :

Hi Larry, I might be wrong but I think you have two typos - one in point 4 of the lemma, where I think you meant 'Assume (a/b - q) is greater then (1/2)', not (a/b), and the other one in the last line of the proof about the repeating decimals. I think that should be 'divide (d1...dma1...am) - ( d1...dm) by [(10^n) - 1] not by 10^(n-1). Thanks for the posts. They are really helpful

Hi Kazbich,

Thanks very much for your comments. I noticed that the previous proof for (a/b -q) had flaws (it only worked for positive a,b) so I updated it.

You are right about your second comment. I have update the proof.

Thanks very much for noticing the typos.

-Larry

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