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jhonyy9

  • 5 years ago

- let a,b and n - numbers naturals from N, a>=1,b>=1 and n>=3, - how can prove it that for any values of n always will be one number a and one number b such that this equation n=a+b+1 is true ? dem: if n>=3 so (n-1)>=2 so (n-1)/2 >=1 - if (n-1)/2 >=1 so are a and b >=1 so n=a+b+1 will be n=(n-1)/2 +(n-1)/2 +1 - if n=3 3=(3-1)/2 +(3-1)/2 +1 3=1+1+1 3=3 - is this prove correct please ?

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  1. anonymous
    • 5 years ago
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    In your proof above you moved away from natural numbers. How about something like below ? For any natural number n>=3 , prove there exist natural numbers a>=1 and b>=1 such that: n=a+b+1 Let m=n-1 : For any natural number m>=2 , prove there exist natural numbers a>=1 and b>=1 such that: m=a+b Pick a=1 ; satisfies a>=1 Pick b=(m-1) ; satisfies b>=1 since m>=2 Then - m = a + b = 1 + (m-1) = m

  2. jhonyy9
    • 5 years ago
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    - can may be accepting correct this prove ?

  3. jhonyy9
    • 5 years ago
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    what i have wrote

  4. anonymous
    • 5 years ago
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    When you say "n=a+b+1" , and then say "n=(n-1)/2 +(n-1)/2 +1 " you are saying that "a=(n-1)/2" and "b=(n-1)/2" . When n is an odd number, it will work because "a=(n-1)/2" and "b=(n-1)/2" will be natural numbers. But when n is an even number, then your a and b will not be natural numbers.

  5. anonymous
    • 5 years ago
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    You could say that for odd values of n (n>=3) , what you originally did, i.e. pick: a=(n-1)/2 and b=(n-1)/2 And for even values of n (n>=4) , pick: a=n/2 and b=(n-2)/2 ; n=a+b+1 ; n=(n/2) + (n-2)/2 + 1 ; n= ( n + n - 2)/2 + 1; n=(n-1)+1 ; n=n

  6. jhonyy9
    • 5 years ago
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    - so i have asked this in this way because a and b are grater or equal 1 and from n>=3 have resulted that (n-1)/2 >=1 too - hence i have thought that this equation can be writing like n=(n-1)/2 +(n-1)/2 +1 is it this correct ?

  7. anonymous
    • 5 years ago
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    The point is that the problem asks you to prove that there are NATURAL numbers a , b, N such that a>=1 and b>=1 and n>=3 for which n=a+b+1 . So you need to show which values of a and b will satisfy ALL the requirements (including them being natural numbers). For odd values of n ( n = 3, 5, 7, 8, 9, ... ) your proof works, because you can pick: a=(n-1)/2 and b=(n-1)/2 But for even values of n ( n=4, 6, 8, 10, ... ) you need to pick different values for a and b so that they will be natural numbers: a=(n/2) and b=(n-2)/2

  8. anonymous
    • 5 years ago
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    The problem is that in your original proof, if n is an even number, then a and b won't be natural numbers. For example, when n=4 then (n-1)/2 = (4-1)/2 = 3/2 = 1.5 which is not a natural number (a counting number). See the definition of natural numbers here: http://en.wikipedia.org/wiki/Natural_numbers

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