1. Joined
    15 Jun '06
    Moves
    16334
    14 Oct '11 21:37
    how do you prove that a decimal that neither repeats nor terminates is irrational?

    For example:

    0.1411411141114 . . . ,

    prove this number is irrational.
  2. Joined
    24 Apr '05
    Moves
    3061
    14 Oct '11 23:47
    Originally posted by tomtom232
    how do you prove that a decimal that neither repeats nor terminates is irrational?

    For example:

    0.1411411141114 . . . ,

    prove this number is irrational.
    One way is to first show the following: every rational number has a terminating or eventually repeating decimal expansion.** Then your claim in question (that a decimal expansion that neither repeats nor terminates is irrational) follows immediately.

    **This one should not be too hard to show. Here would be the basic idea and it could be made more rigorous:

    http://mathlesstraveled.com/2008/09/07/rational-numbers-and-decimal-expansions/
  3. Subscribertalzamir
    Art, not a Toil
    60.13N / 25.01E
    Joined
    19 Sep '11
    Moves
    45047
    15 Oct '11 07:331 edit
    If you divide m by n, the fraction left over is some value 0, 1, 2, ... , n-1. If at any point you get a zero, the division terminates there, e.g. dividing 1 by 4 gives a zero leaves 1; then 2 leaves 2, then 5 leaves zero, so 1 / 4 = 0.25 . If you don't get a zero, the division bounces around in values {1, .. , n-1}, until eventually it steps somewhere it has before. If the a'th fraction is the same as b'th, then a+1'th is the same as b+1'th etc, so dividing by n means that either the sequence terminates or then it repeats a cycle that is at most n-1 in length. Actually, if n is a compound number, the maximum length is probably shorter than n-1.. as a hunch, p-1 where p is the biggest prime by which n is divisible and for which lcd(10,p) = 1.

    The other half of the proof is to show that all fractions that do terminate or which do have a sequence are rational. If the fraction f terminates and there are d decimals then it can be expressed as f / 1 = (f * 10^d) / 10^d which is an integer over integer, that is, a rational number. If there is a sequence of length l, then

    f = f * (10^l - 1) / (10^l -1)

    expresses f in a way that terminates, and

    f = (f * 10^d * (10^l -1)) / (10^d * (10^l -1))

    expresses f as a fraction of two integers - that is, a rational number.

    Thus, any repeating or terminating sequence of decimals is a rational number, and every rational number leads to a terminating or repeating sequence of decimals, so non-repeating sequences have to be irrational.
  4. Joined
    15 Jun '06
    Moves
    16334
    15 Oct '11 19:51
    Good on ya mate.
  5. Joined
    09 Aug '06
    Moves
    5363
    30 Nov '11 00:04
    This is nice.
    Now prove this:

    e/Pi is not rational where e=2.7182818....
  6. Joined
    26 Apr '03
    Moves
    25837
    03 Dec '11 00:02
    Originally posted by smaia
    This is nice.
    Now prove this:

    e/Pi is not rational where e=2.7182818....
    That looks very hard! Are you sure it can be done?
  7. Joined
    18 Jan '07
    Moves
    6977
    03 Dec '11 23:28
    Originally posted by iamatiger
    That looks very hard! Are you sure it can be done?
    I believe it has been done, but it takes some serious mathematics. You don't just have to prove that e and pi are irrational - which are hard enough, but doable - but also that they're a different kind of irrational. Which is, well, obviously true, but non-obvious to prove.

    Richard
  8. Joined
    26 Apr '03
    Moves
    25837
    17 Jan '12 00:334 edits
    hmm, no ideas anyone?
  9. Standard memberSoothfast
    0,1,1,2,3,5,8,13,21,
    Planet Rain
    Joined
    04 Mar '04
    Moves
    2445
    11 Feb '12 07:40
    Originally posted by smaia
    This is nice.
    Now prove this:

    e/Pi is not rational where e=2.7182818....
    If e/pi were rational then sin(m*e)=0 for some nonzero integer m. Perhaps a contradiction could be derived from that.
Back to Top