- let a,b >=1 and n>=3 numbers from the set of natural numbers N ,
- how can prove it that always for any value of n exist numbers a and b such that this equation
a+b+1=n is true ?
- is this prove with reductio ad absurdum correct ?
in a reducto ad absurdum argument, we assume the opposite is true. in this case, that would be:
for some natural number n >=3 there are no natural numbers a >=1, b >=1 with a+b+1 = n.
we don't know what n may be (except that it is greater or equal 3).
since n >=3, n - 1 >=2.
since n - 1 >=2, (n - 1)/2 >=1.
(n - 1)/2 + (n - 1)/2 + 1 = (n - 1) + 1 = n.

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Let n be 4. Then n-1 is 3. (n-1)/2 is 3/2 which is not a natural number.

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