Through any $n$ distinct points on a plane we can draw a straight line.

Question

I can't understand what is wrong with this paradox. How we should strictly mathematically explain it?

Mathematical induction:

1. The basis:

$n=1,n=2$. Through any two (one) points on a plane we can draw a straight line.

2. The inductive step:

$n=k$. Through any $k$ distinct points on a plane we can draw a straight line.

3. Fake-Paradox:

We have an arbitrary $k+1$ points on the plane: $P_1, P_2, ..., P_{k+1}$. From $2)$ (inductive step) we can draw a straight line $L_1$ through $k$ points $P_1, P_2, ..., P_{k}$ and line $L_2$ through $k$ points $P_2, ..., P_k, P_{k+1}$. Lines $L_1$ and $L_2$ have at least two common points $P_2$ and $P_k$. But any two distinct points of a straight line completely determine that line $\Rightarrow L_1=L_2$ and $P_{k+1} \in L_1$. And we prove that

Through any $n$ distinct points on a plane we can draw a straight line.

Answer 1

Since you've tagged this as homework I won't just give you the answer, but think about this: your stage 3. relies on $P_2 \ne P_k$. When might this not happen?

Edit (in response to comment): Let $\phi$ be a statement about natural numbers; we say $\phi(n)$ if $\phi$ holds for some particular $n$. Let $n_0 \in \mathbb{N}$. The principle of mathematical induction states that $\phi(n)$ holds for all $n \ge n_0$ if and only if both of the following conditions are satisfied:

• $\phi(n_0)$ holds; and
• For each $n \ge n_0$, if $\phi(k)$ holds then $\phi(k+1)$ holds.

Here we can set $n_0=2$. But the second condition is not satisfied, since if it were to be satisfied we'd need $\phi(2) \Rightarrow \phi(3)$, but as shown this is not the case since the argument that would make it so would require $P_2 \ne P_2$.

Answer 2

The argument in the induction step doesn’t work in going from $n=2$ to $n=3$: the two lines aren’t forced to be the same line. Try drawing a picture of such an example.

Answer 3

Hint: We know that through any 2 points in the plane we can draw a straight line, but that is not true for any 3 points. Thus, you should focus your attention on the argument that the claim is true for $n=2$ implies that it is true for $n=3$. Does the argument in 3. (Fake Pair Of Ducks) work for $k=2$ (3 points)?