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Help me please with a demonstration 'reductio ad absurdum'!
1. - substitution
- be a and b,two numbers greater or equal than 1 from the set of real numbers R,
- be n a number greater or equal than 3 from the set of real numbers R,
2. - conclusion
- there is always a and b such that equation is true: n=a+b+1
3. - demonstration
- ???
Thank you very much your all helps !
This very confusing. First you say "let a and b be two numbers greater or equal than 1 from the set of real numbers R"
But then you say "there is always a and b such that equation is true: n=a+b+1". You cannot select a and b first (which is what your first sentence means) and then say "there exist a and b such that".
I think what you mean is "If n is any real number larger than 3 then there exist real numbers a and b, larger than 1, such that n= a+ b+ 1".
That's obviously true but I can see no reason to use "reductio ad absurdum" (proof by contradiction). If n> 3, then n- 1> 2. Let a= b= (n-1)/2> 1. Then n-1= a+ b so n= a+ b+ 1.
(Of course, there exist an infnite number of choices for a and b. To show "there exists" you only need to demonstrate one.)
can you help me,please with one acceptable demonstration that in this condition for every n there always numbers a and b for the ecuation n=a+b+1 is true
thank you very much
in a reducto ad absurdum argument, you assume the opposite is true. in this case, that would be:
for some real number n > 3 there are no real numbers a >1, b >1 with a+b+1 = n.
we don't know what n may be (except that it is greater than 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.
but if n is a real number, so is n - 1, and so is (n - 1)/2.
therefore we have proved that for this n > 3, the real numbers (n - 1)/2 don't exist. this is absurd.
(the logical conclusion therefore being that no such real number > 3 actually exists, so our assumption was wrong,
so EVERY real number > 3 has such an a and b > 1 with a+b+1 = n).
HI Deveno ! can you tel me please if those numbers are not real numbers ,they are natural numbers - so in this case this prove is correct too ?
- Thank you very much for your reply !