# Flaw in proof that in a classroom with $n$ students, if there is a girl, then all students are girls

Problem: Prove that: In a classroom with n student, if there is a girl student, all students of this class are girl.

Solving: Let f(n) is the clause: In the class, if there is 1 girl student, all of n students are girl.

First: with n=1 ---> the clause f(1) is right (of course)

Second: with n=k, assume that the clause f(k) is right

we will prove that the clause f(k+1) is right.

Consider a set of (k+1) students {a1, a2, a3,....a(k),a(k+1)} (with a1 is the girl student)

• with first k student A= {a1, a2, a3,....ak}: using the inductive supposition above, we have all student in A are girl.

• Now we have B= {a1, a2, a3,....a(k-1),a(k+1)}: also using the inductive suppositon, we have have all student in B are girl.

So the f(k+1) clause is right.

Conclusion: In the class, if there is a girl student, all student are girl !!!!!!!!!!!

I can't find the wrong in this proof. Please help me. thanks

What happens to $A$ and $B$ for $k = 2$?
• exactly the proof breaks down at $k=2$, well-used induction example failure – Nikos M. Aug 16 '14 at 9:19
In a proof by induction, going from the $k$ step to the $k+1$ step must work for every specific value of $k$. Does it in this case?