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Irrational numbers

  1. Jun 25, 2011 #1
    Prove that:
    1-If n^2 (n is a natural number) is even then n is even too .
    2-Product of infinit number of primes bigger than 2 is not even.


    Please do not "google it for me" :biggrin: .
     
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  3. Jun 25, 2011 #2

    tiny-tim

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    hi limitkiller! :wink:

    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Jun 25, 2011 #3
    I wanted too prove 1 (If n^2 (n is a natural number) is even then n is even too).

    If n^2 is even then it is divisible by 2.

    Then it has 2 as its divisors because it is an integer that divides n without leaving a reminder.

    Then I said :Who says?, 2 (Product of infinit number of primes bigger than 2 is not even) might be incorrect . :biggrin:
     
  5. Jun 25, 2011 #4

    tiny-tim

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    hi limitkiller! :smile:
    might be easier to prove the opposite …

    start by assuming n is odd :wink:
    what can you say about a prime bigger than 2?
     
  6. Jun 25, 2011 #5
    ummm I think I forgot to write one of the lines.Originally it should be like this:
    I wanted too prove 1 (If n^2 (n is a natural number) is even then n is even too).

    If n^2 is even then it is divisible by 2.

    Then it has 2 as its divisors because it is an integer that divides n without leaving a reminder.

    So n has 2 as its divisors too .

    Then I said :Who says?, 2 (Product of infinit number of primes bigger than 2 is not even) might be incorrect .





    About a prime bigger than 2 we can say its not divisible by 2, and product of a finit number of them is not divisible by 2...
    But I couldn't find anything about product of infinit number of them.
     
  7. Jun 25, 2011 #6

    tiny-tim

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    hi limitkiller! :smile:
    yes, but you have to give a reason for that

    (i still think my odd way is easier :wink:)
    ah i see … i had assumed that was a misprint

    there's no such thing as a product of an infinite number of numbers :redface:, i think it must mean "a product of any size, however large"
     
  8. Jun 25, 2011 #7

    HallsofIvy

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    This proves that if a number is even, then its square is even. What you want to prove is the "converse": if the square of a number is even, then the number is even.

    If n is odd then it is of the form 2k+ 1 for some integer k. What is the square of that?


    There is no such thing as a "product of an infinite number of primes". You can prove that the product of any number of primes, larger than 2, is odd.

    Why was this thread titled "irrational numbers"?
     
  9. Jun 25, 2011 #8
    Why?

    Because (1) was used in the proof of "[itex]\sqrt{2}[/itex] is not rational"
     
  10. Jun 25, 2011 #9

    HallsofIvy

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    Because of the way multiplication is defined!
    We define the product of two numbers, then define the product of three numbers as "xyz= (xy)z"- i.e. as the product of the two numbers xy and z (and use the associative property of multiplication to show that this is the same as x(yz)).

    We can then inductively define the product of n+ 1 numbers as [tex](x_1x_2x_3\cdot\cdot\cdot x_n)x_{n+1}[/tex]. But that only gives us the product of N numbers for N a positive integer.
     
  11. Jun 25, 2011 #10
    Then [itex]\prod_{k=1}^\infty k[/itex] is meaningless.But it is not,is it?
     
  12. Jun 25, 2011 #11

    micromass

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    If you would like to give it a meaning, then the product would still diverge. So you can't really talk about divisibility...
     
  13. Jun 25, 2011 #12
    I dont get you :uhh: . why does it mean that we can not talk about divisibility?
     
  14. Jun 25, 2011 #13

    micromass

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    Because the product

    [tex]\prod_{k=1}^{+\infty}{k}[/tex]

    diverges. It doesn't equal an integer, and you need integers to talk about divisibility.
     
  15. Jun 25, 2011 #14
    isn't it an integer itself?
    for example 2^+[itex]\infty[/itex] is divisible by 2.
     
  16. Jun 25, 2011 #15

    micromass

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    No: infinity is NOT an integer or a real number. You cannot calculate with infinity like you calculate with numbers.

    At the very best, you might say that infinity is an extended real number. But that still doesn't allow you to talk about divisibility...

    We have a good FAQ on the topic: https://www.physicsforums.com/showthread.php?t=507003 [Broken]
     
    Last edited by a moderator: May 5, 2017
  17. Jun 25, 2011 #16
    Thank you.
    But I don't think I used infinity illegaly...
    "Infinity is not a real number" does not mean that [tex]\sum_{k=1}^\infty \frac{1}{k}[/tex] is not a real number or even [tex]\prod_{k=1}^\infty k[/tex] is not a real number.
     
    Last edited by a moderator: May 5, 2017
  18. Jun 25, 2011 #17
    But wouldn't it be possible to talk of certain properties of an infinite series?

    I mean, for example, 1^(anything)=1. So, 1^(infinity) must also be 1, because regardless of the number of times we multiply, the answer will always be 1,right?
     
  19. Jun 25, 2011 #18

    micromass

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    Both of those or not real numbers, whether you like it or not.
     
  20. Jun 25, 2011 #19

    micromass

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    No, [itex]1^{+\infty}[/itex] is an undetermined form (even in the extended real numbers). The reason is the sequence

    [tex]\left(1+\frac{1}{n}\right)^n[/tex]

    Naively, you could say that [itex]1+\frac{1}{n}[/itex] converges to 1 and that n converges to infinity, and thus

    [tex]\left(1+\frac{1}{n}\right)^n\rightarrow 1^{+\infty}=1[/tex]

    But this is NOT true! Check it for yourself: take a calculator and compute the first several terms in the sequence, it will not converge to 1!

    The limit for the sequence is e and is defined as 2.718...
     
  21. Jun 25, 2011 #20
    then why is [tex]\sum_{k=1}^\infty \frac{1}{k!}[/tex] a real number?








    I got it...
     
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