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Math Help - Irrational numbers.Why ?

  1. #1
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    Irrational numbers.Why ?

    Hi all,

    I would like to know why rational numbers can encompass numbers that are infinite but periodical ( like 1/3) and cannot comprise numbers that are infinite but non-periodical (like Pi).

    An explanation made in simple terms would be welcome.

    Ps : I do know that infinite-non-periodical numbers are called irrationals because they cannot be written as a ratio such as a/b, my question is : why can't they ?

    Many thanks
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  2. #2
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    It depends on the numbers in question.

    Square roots of prime numbers are the easiest to prove irrationality for example...
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  3. #3
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    I'm not sure what you mean by "infinite but periodical." Are you speaking of a decimal representation?

    The rational numbers are defined as the set a/b where a and b are integers and b is non-zero. Any real numbers that are not rational are classified as "irrational." It's a definition.

    -Dan
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  4. #4
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    The fact that the LeTeX complier is not working make it difficult to do this in proper notation.

    If we add up 10^{-n}+10^{-2n}+…we get 1/(10^{n}-1).
    If we have the decimal .231231231… we can write that as (231) [10^{-3}+10^{-6}+… ] we get (231)/(10^{3}-1). This is rational.
    So what you are calling periodic is what makes it possible to reduce the decimal to a rational number.

    Given a number like \pi in which the decimal part is not eventually periodic, it is impossible to establish that sum of negative powers tens which gives the rational part.
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  5. #5
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    A couple of years ago, a person asked, on a different forum, how to prove that any rational number could be written as a fraction. I was puzzled by such a question because the definition I had learned, of rational numbers, was "numbers that could be written as a fraction". Then I realized that he must have learned to define "rational number" as a number whose decimal expansion (and so expansion in any base) is "ultimately repeating". I believe that is what werecat means by "infinite but periodical".

    Just as I eventually learned the proof that any fraction was equivalent to a repeating decimal, so he was asking about the proof that any repeating decimal is equivalent to a fraction.

    Werecat is asking about "the proof that any fraction was equivalent to a repeating decimal". Werecat, any rational number can, by definition, be written as a fraction m/n where m and n are integers. Now, we change to decimal form by actually dividing m by n. Imagine doing that- we determine how many (integer times) m will divide into n (k such that kn<m< (k+1)n) and subtract. Each time we do a "trial division", we get a remainder which, of course, cannot be greater than n-1. That is, there are only "n" possible remainders, 0 to n-1.

    Eventually, at most after n divisions, a remainder will repeat. Now, we bring down a "0" and divide again. But we will now be doing exactly the same division as the first time we got that remainder and so will get both the same quotient and the same remainder so the next division will be exactly the same again. Do you see how that guarentees that we will continue repeating the same division? Do you see how that guarentees that the same quotient will repeat and so we will have a repeating decimal?
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  6. #6
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    if the terminal part of a decimal expansion for a number x is periodic, say with period of length m, than (10^m)x - x has a decimal expansion that terminates (which is obviously rational).

    but if (10^m)x - x = a/b, then (10^m - 1)x = a/b, and x = a/(b(10^m - 1)), which is also clearly rational. here is an example of this in practice:

    suppose x = 0.45123123123123.... which terminates in a cycle of length 3 (123 repeating). so 10^3x = 1000x is:

    451.23123123123123.... and when we subtract, we get 999x = 450.78 = 45078/100, so x = 45078/99900.

    on the other hand, if we have a fraction a/b, we can write a = (q1)b + r1, where r1 < b. this gives us the "integer part", q1.

    to find the "next decimal", we write 10r1 = (q2)b + r2. this gives us a = (q1)b + r1 = q1(b) + (q2)b/10 + r2/10 = (q1 + q2/10)b + r2/10.

    moral of the story: decimal expansions of fractions have to terminate or eventually repeat, so if a decimal expansion does neither, it cannot be a fraction.

    (if 10r1 < b, q2 will be 0, and we go the "the next decimal place", with 100r1 = (q3)b + r3).

    so we wind up with a sequence q1.q2q3q4..... where each qj for j > 1 is 0-9.

    now for some m, b < 10^m (we can always take b > 0, by considering -a/b if b < 0). and so there are at most 10^m

    "sequences" of remainders q(j1).....q(jm) possible before we get to a sequence that repeats indefinitely

    (which will happen as soon as we get a remainder we've gotten before. in fact 10^m is a rather generous upper bound,

    since each rj < b, there can be at most b of them).

    with our example, 45078/99900:

    the integer part is 0, since 45078 < 99900. so now we consider 450780 = 4*(99900) + 51180. so the first decimal is 0.4.......

    next we consider 511800 = 5*(99900) + 12300, so the second decimal is 5, and we have so far 0.45......

    then 123000 = 1*(99900) + 23100, so the third decimal is 1, and we have so far 0.451....

    continuing, 231000 = 2*(99900) + 31200, so the 4th decimal is 2, and we have 0.4512....

    and 312000 = 3*(99900) + 12300, so the 5th decimal is 3, and we have 0.45123....

    but wait! we've gotten this remainder before, after step 2. so we know what will happen next, we'll get a decimal of 1, with a remainder of 23100.

    and step 7 will be just like step 4, and step 8 just like step 5, we have a repeating cycle of 3: 45078/99900 = 0.45123123123123.....

    moral of the story: decimal expansions of fractions have to terminate or repeat. if a decimal expansion does neither, it cannot be a fraction.
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  7. #7
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    Hey there. Well according to the definition of a rational number "The rational numbers are defined as the set x/y where x and y are integers and y is non-zero. Any real numbers that are not rational are classified as irrational."
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