I'm proving algorithm to justify using inductive hypothesis.
$T(k)$ is the minimum round value the number of handshaking within $N=k$ people.
If $T(k-2)=T(k-1)=k-2$, can I prove about $T(k)$?
I used proof by inductive hypothesis.
1) Base case, $N=3, N=4$
T(3) = minimum number of handshaking within 3 people is 3
T(4) = minimum number of handshaking within 4 people is 3
2) Inductive Hypothesis ($k$ is odd number)
Until $N=k-1$, it is true that $T(k-2)=T(k-1)=k-2$.
3) Prove when $N=k$
How can I prove?
Letting n = k - 1, then T(n-1) = T(n) = n - 1.
If this is going to be true for any n, there is a contradiction in this statement. For example, take n = 5 just to show this. You said:
$$ T(5 - 1) = T(4) = 5 - 1 = 4 \\ T(5) = 5 - 1 \\ \therefore T(4) = t(5) = 4 $$
Then if you do this for n = 6, then:
$$ T(6-1) = T(5) = 6 - 1 = 5 \\ T(6) = 6 - 1 = 5 \\ \therefore T(6) = T(5) = 5 $$
In the top example, T(5) = 4 and in the bottom T(5) = 5. Therefore, your original statement can't be possible; its impossible for both $T(n-1)$ and $T(n)$ to be equal to $n - 1 $