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Is there a rational number between any two irrationals? Watch

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    Question in the title.

    I feel intuitively the answer should be yes, but can't really seem to prove this... Any help?
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    Yes, all you have to do is look at the digits of both irrational numbers until there is a difference between two digits, truncate the irrationals after this digit (so you have a rational number) add the two rational numbers together, divide by two and you've got a rational number between the two irrationals.

    So for example, suppose you've got the two following irrational numbers \alpha,\beta and wlog write them as

    \alpha = 0.a_{1}a_{2}a_{3}...

    \beta = 0.b_{1}b_{2}b_{3}...

    Where the a_i, b_i are the digits.

    The reason we can say they're between 0 and 1 is because it doesn't matter what the digits are to the left of the decimal point, it's just extra notation if we add them in.

    Let's say that a_i = b_i for 1 \leq i \leq n-1, so what we're saying is there is a difference between the two irrationals at the digit in place a_n

    So denote the rational numbers

    \alpha' = 0.a_{1}a_{2}a_{3}...a_{n}

    \beta' = 0.b_{1}b_{2}b_{3}...b_n

    Then the number \dfrac{\alpha' + \beta'}{2} is also rational, but note that

    \dfrac{\alpha' + \beta'}{2} = \dfrac{0.a_{1}a_{2}a_{3}...a_{n} + 0.b_{1}b_{2}b_{3}...b_n}{2}

    This number will have the first n-2 digits identical to the digits of \alpha',\beta' but the last two digits will be such that a < \dfrac{\alpha' + \beta'}{2} < b
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    Yes, there is.

    Hint for the proof: consider the decimal expansions of the numbers.
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    (Original post by Maths Failure)
    Question in the title.

    I feel intuitively the answer should be yes, but can't really seem to prove this... Any help?
    Yes

    Edit : as Noble said

    Edit : even more what Noble said
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    Oh ok thanks guys
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    (Original post by Maths Failure)
    Oh ok thanks guys
    I've just updated my reply with a more mathematical explanation which you might find more helpful in trying to understand.
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    (Original post by TenOfThem)
    Yes

    Edit : as Noble said

    Edit : even more what Noble said
    :eek: I think you accidentally negged me, with that god awful -8 reputation power.

    Either that or you're getting your own back for me finding your facebook picture the other day :lol:
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    (Original post by Noble.)
    Yes, all you have to do is look at the digits of both irrational numbers until there is a difference between two digits, truncate the irrationals after this digit (so you have a rational number) add the two rational numbers together, divide by two and you've got a rational number between the two irrationals.

    So for example, suppose you've got the two following irrational numbers \alpha,\beta and wlog write them as

    \alpha = 0.a_{1}a_{2}a_{3}...

    \beta = 0.b_{1}b_{2}b_{3}...

    Where the a_i, b_i are the digits.

    The reason we can say they're between 0 and 1 is because it doesn't matter what the digits are to the left of the decimal point, it's just extra notation if we add them in.

    Let's say that a_i = b_i for 1 \leq i \leq n-1, so what we're saying is there is a difference between the two irrationals at the digit in place a_n

    So denote the rational numbers

    \alpha' = 0.a_{1}a_{2}a_{3}...a_{n}

    \beta' = 0.b_{1}b_{2}b_{3}...b_n

    Then the number \dfrac{\alpha' + \beta'}{2} is also rational, but note that

    \dfrac{\alpha' + \beta'}{2} = \dfrac{0.a_{1}a_{2}a_{3}...a_{n} + 0.b_{1}b_{2}b_{3}...b_n}{2}

    This number will have the first n-2 digits identical to the digits of \alpha',\beta' but the last two digits will be such that a < \dfrac{\alpha' + \beta'}{2} < b
    Why does i have to be bounded? If i was infinite could an argument be made that \alpha and \beta are still possibly different numbers?
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    (Original post by Maths Failure)
    Why does i have to be bounded? If i was infinite could an argument be made that \alpha and \beta are still possibly different numbers?
    No, if the numbers do not differ for some finite i then they are the same number. You can argue it in a similar way to how 0.999... = 1 using infinite series and sequences because if the digits do not differ for any finite i then the infinite sum of both are exactly the same.
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    (Original post by Noble.)
    :eek: I think you accidentally negged me, with that god awful -8 reputation power.

    Either that or you're getting your own back for me finding your facebook picture the other day :lol:
    Balls

    Let me see if I can redress

    xx

    I will come back and personal rep tomorrow but I am out today

    Sowwy
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    (Original post by TenOfThem)
    Balls

    Let me see if I can redress

    xx
    Haha don't worry about it, I've got plenty of rep.
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    (Original post by Noble.)
    Haha don't worry about it, I've got plenty of rep.
    You call a +6 "plenty"

    Hah



















    [obviously I have more cos i am soooooooo much older]
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    (Original post by TenOfThem)
    You call a +6 "plenty"

    Hah



















    [obviously I have more cos i am soooooooo much older]
    Haha, I've just seen you joined this site in Sep 2011 and have 15.5k posts - I think we can safely conclude you like helping people too much, what with you being a maths teacher. You and Mr M should start some Avengers like team of maths teachers.
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    (Original post by Noble.)
    Haha don't worry about it, I've got plenty of rep.
    Attempted to redress balance
 
 
 
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