jhonyy9
  • jhonyy9
- 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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
jhonyy9
  • jhonyy9
- can may be accepting correct this prove ?
jhonyy9
  • jhonyy9
what i have wrote

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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.
anonymous
  • anonymous
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
jhonyy9
  • jhonyy9
- 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 ?
anonymous
  • anonymous
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
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.