A 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 104 we get:
52345.2345
(3) So, subtracting (2) by (1) gives us:
104 - 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:
d1...dm.a1...an
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 10n*the number by the original number which after canceling for the repeating decimal gives us:
10n * (d1...dm.a1...an...) - (d1...dm.a1...an...) =
(d1..dma1..an) - (d1..dm).
(3) Now, our rational number is equal to the value in step #2 divided by 10n - 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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