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jhonyy9
 5 years ago
 is this prove correct with math induction ?
let a,b and n natural numbers from N,
a>=1,b>=1,n>=3
 n=a+b+1  how may be prove it that always for any value of n will be one a and one b that this equation is true ?
 Assume there exists a natural number k such that k >= 3 and there exists a pair of natural numbers, a_k and b_k, such that (a_k + b_k + 1) = k.
Let a_(k+1) = a_k.
So, a_(k+1) is a natural number.
Let b_(k+1) = (b_k) + 1.
So, b_(k+1) is a natural number.
(a_(k+1) + b_(k+1) + 1) = [a_k + ((b_k) + 1) + 1] = [(a_k + b_k + 1) + 1] = (k + 1).
And (k + 1) is a natural number >= 3.
So,
jhonyy9
 5 years ago
 is this prove correct with math induction ? let a,b and n natural numbers from N, a>=1,b>=1,n>=3  n=a+b+1  how may be prove it that always for any value of n will be one a and one b that this equation is true ?  Assume there exists a natural number k such that k >= 3 and there exists a pair of natural numbers, a_k and b_k, such that (a_k + b_k + 1) = k. Let a_(k+1) = a_k. So, a_(k+1) is a natural number. Let b_(k+1) = (b_k) + 1. So, b_(k+1) is a natural number. (a_(k+1) + b_(k+1) + 1) = [a_k + ((b_k) + 1) + 1] = [(a_k + b_k + 1) + 1] = (k + 1). And (k + 1) is a natural number >= 3. So,

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0No need induction. Just choose \(a=n2\) and \(b=1\)
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