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anonymous

  • 5 years ago

why x raised to power 0 equals one?

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  1. anonymous
    • 5 years ago
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    this is another definition

  2. anonymous
    • 5 years ago
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    however, be careful 0^0 is undefined i believe

  3. anonymous
    • 5 years ago
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    \[x^n \div x^n = x^{n-n} = x^0 = 1 \]

  4. anonymous
    • 5 years ago
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    we want that x^m / x^n = x^(m-n) , so what if m = n , then you have x^(0) , which should equal 1

  5. anonymous
    • 5 years ago
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    Of course this doesn't hold if x = 0, as it is division by zero. But that is whole different topic.

  6. anonymous
    • 5 years ago
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    newton, but thats not a proof

  7. anonymous
    • 5 years ago
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    excellent newton

  8. anonymous
    • 5 years ago
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    It's a way to explain it - they asked 'why', not for a proof.

  9. anonymous
    • 5 years ago
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    ok

  10. anonymous
    • 5 years ago
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    by defining x^0 = 1, we are allowed to use the x^m/x^n = x^(m-n) rule without qualifications or caveats

  11. anonymous
    • 5 years ago
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    THATS the reason. i dont want to hear anything else

  12. anonymous
    • 5 years ago
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    If you write it the other way around it is a proof. View \[x ^{n} \div x ^{n}\] as a fraction. Any number divided by itself is 1.

  13. anonymous
    • 5 years ago
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    no it isnt

  14. anonymous
    • 5 years ago
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    its a definition, thats all. but there is a reason for the definition

  15. anonymous
    • 5 years ago
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    ok prove it, show me the deductive steps , axioms, etc

  16. anonymous
    • 5 years ago
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    sorry my style of learning is argumentative, dont take it personally

  17. anonymous
    • 5 years ago
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    but ive thought this through, so im fairly confident

  18. anonymous
    • 5 years ago
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    "In mathematics, a proof is a convincing demonstration (within the accepted standards of the field) that some mathematical statement is necessarily true." \[x ^{0} = x ^{n-n} = x ^{n} / x ^{n} = 1\] unless I have the definition of "definition" wrong ;)

  19. anonymous
    • 5 years ago
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    well why is x^n/x^n = x ^ (n-n) ?

  20. anonymous
    • 5 years ago
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    i would say thats false when n=0, and say x^0 is undefined

  21. anonymous
    • 5 years ago
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    Personally, I think it's a proof (more so than 0!). You use the definition \[x^n \div x^m = x^{n-m} \] Not every proof have to prove everything right down to the axioms, you know...

  22. anonymous
    • 5 years ago
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    so thats the weakest link here

  23. anonymous
    • 5 years ago
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    also you have to assume x is not zero

  24. anonymous
    • 5 years ago
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    Maybe it needs to have limits set on it? I think the basic question has been answered by INewton though, this is becoming a whole different topic.

  25. anonymous
    • 5 years ago
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    \[x^a \div x^b = x^{a-b}\ \forall \mathbb{R};\ x \not= 0\]

  26. anonymous
    • 5 years ago
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    Yes, assuming \[x \neq 0\] would solve that problem

  27. anonymous
    • 5 years ago
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    and 0^0 is indeterminate, well when we are dealing with functions

  28. anonymous
    • 5 years ago
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    You're looking for a problem which isn't there.

  29. anonymous
    • 5 years ago
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    hmmm, that does look like a proof,

  30. anonymous
    • 5 years ago
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    ie, as a counsequence of a previous definition

  31. anonymous
    • 5 years ago
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    i think thats what my point was. its a proof if you first accept the definition x^m / x^n = x^(n-n)

  32. anonymous
    • 5 years ago
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    err, you know what i mean

  33. anonymous
    • 5 years ago
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    It's a consequence of the rule \[x^a \div x^b = x^{a-b} ... \]

  34. anonymous
    • 5 years ago
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    exactly

  35. anonymous
    • 5 years ago
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    now ...

  36. anonymous
    • 5 years ago
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    Do you prove every rule you use in your proofs?

  37. anonymous
    • 5 years ago
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    you can use definitions, thats fine

  38. anonymous
    • 5 years ago
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    youre free to define as long as the definition is consistent

  39. anonymous
    • 5 years ago
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    yes i try to

  40. anonymous
    • 5 years ago
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    Then your proofs must be very boring.

  41. anonymous
    • 5 years ago
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    well i dont go back to ZFC , if thats what youre asking

  42. anonymous
    • 5 years ago
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    zermelo frankel cohen set theory (where we assume the continuum hypothesis is true)

  43. anonymous
    • 5 years ago
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    well one man's boredom is another man's delight. one man's garbaege is another man's delight

  44. anonymous
    • 5 years ago
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    one man's cow of a wife, is another man's wet dream

  45. anonymous
    • 5 years ago
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    so on and so forth

  46. anonymous
    • 5 years ago
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    shall i continue

  47. anonymous
    • 5 years ago
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    Topic has been exhausted. I'm out! Hope some help can be found in here lol. :)

  48. anonymous
    • 5 years ago
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    No, no - I'm pretty sure anyone would find a proof that went back to the axioms for evey last thing boring.

  49. anonymous
    • 5 years ago
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    yes i know i am pedantic. thats the way my brain works

  50. anonymous
    • 5 years ago
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    but he asked an axiomatic question

  51. anonymous
    • 5 years ago
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    anyways, you answered fine. i was just saying, merely calling it a definition is a bit unneceesary. it has a more fundamental reason

  52. anonymous
    • 5 years ago
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    if say , because i defined it that way. ummm, thats not necessary. but its true. you can get away with many theorems by just calling it a definition (in some cases)

  53. anonymous
    • 5 years ago
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    for example, you can define the real numbers as a complete ordered field. but thats inelegant

  54. anonymous
    • 5 years ago
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    Wait, now that I'm looking at it - why would it be indeterminate when n = 0? That doesn't put a 0 in the denominator...?

  55. anonymous
    • 5 years ago
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    id rather say, the real numbers is an example of a complete ordered field, and complete ordered fields are unique up to isomorphism

  56. anonymous
    • 5 years ago
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    hmmm

  57. anonymous
    • 5 years ago
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    you mean if x = 0 ?

  58. anonymous
    • 5 years ago
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    This is what you said: i would say thats false when n=0, and say x^0 is undefined

  59. anonymous
    • 5 years ago
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    well say x = 0 , and say n = 3. so 0^3 / 0^3 is undefined

  60. anonymous
    • 5 years ago
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    OK so it was a type-o

  61. anonymous
    • 5 years ago
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    no its fine if n=0, but x != 0

  62. anonymous
    • 5 years ago
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    Just wondering :) See ya!

  63. anonymous
    • 5 years ago
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    i have a contradiction

  64. anonymous
    • 5 years ago
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    0^2 = 0^4 / 0^2 which is undefined, but 0^2 = 0*0

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