How can the proof by induction be reliable when it depends on the number of steps?

Question

Yesterday, I got a math problem as follows.

Determine with proof whether $\tan 1^\circ$ is an irrational or a rational number?

My solution (method A)

I solved it with the following ways.

I can prove that $\tan 3^\circ$ is an irrational, the proof will be given later because it takes much time to type.

Let's assume that $\tan 1^\circ$ is a rational number. As a result, $$ \tan 2^\circ = \frac{2 \tan 1^\circ }{1-\tan^2 1^\circ}$$ becomes a rational number. Here we don't know whether $\tan 2^\circ$ is an irrational or a rational. Let's consider each case separately as follows:

• If $\tan 2^\circ$ is actually an irrational then a contradiction appears in $\tan 2^\circ = \frac{2 \tan 1^\circ}{1 - \tan^2 1^\circ}$. Thus $\tan 1^\circ$ cannot be a rational.

• If $\tan 2^\circ$ is actually a rational then we can proceed to evaluate $$\tan 3^\circ = \frac{\tan 2^\circ + \tan 1^\circ}{1 - \tan 2^\circ \tan 1^\circ}$$ A contradiction again appears here because I know that $\tan 3^\circ$ is actually an irrational. Thus $\tan 1^\circ$ cannot be a rational.

For both cases whether $\tan 2^\circ$ is rational or not , it leads us to the conclusion that $\tan 1^\circ$ is an irrational. End.

My friend's solution (Method B)

Assume that $\tan 1^\circ$ is a rational number. If $\tan n^\circ$ ($1\leq n\leq 88$, $n$ is an integer) is a rational number, then $$ \tan (n+1)^\circ = \frac{\tan n^\circ + \tan 1^\circ}{1-\tan n^\circ \tan 1^\circ}$$ becomes a rational number.

Consequently, $\tan N^\circ$ ($1\leq N\leq 89$, $N$ is an integer) becomes a rational number. But that is a contradiction, for example, $\tan 60^\circ = \sqrt 3$ that is an irrational number.

Therefore, $\tan 1^\circ$ is an irrational number.

My friend's solution with shortened inteval (Method C)

Consider my friend's proof and assume that we only know that $\tan 45^\circ =1$ which is a rational number. Any $\tan n^\circ$ for ($1\leq n\leq 44$, $n$ is an integer) is unknown (by assumption).

Let's shorten his interval from ($1\leq n\leq 88$, $n$ is an integer) to ($1\leq n\leq 44$, $n$ is an integer).

Use his remaining proof as follows.

For ($1\leq n\leq 44$, $n$ is an integer), $$ \tan (n+1)^\circ = \frac{\tan n^\circ + \tan 1^\circ}{1-\tan n^\circ \tan 1^\circ}$$ becomes a rational number.

Consequently, $\tan N^\circ$ ($1\leq N\leq 45$, $N$ is an integer) becomes a rational number.

Based on the assumption that we don't know whether $\tan n^\circ$ for ($1\leq n\leq 44$, $n$ is an integer) is rational or not, we cannot show a contradiction up to $45^\circ$,

Questions

• Can we conclude that $\tan 1^\circ$ is a rational number in method C as there seems no contradiction?

• Is the proof by induction correctly used in method B and C?

• Is the proof by induction in method A the strongest?

Answer 1

The inductions in $B$ and $C$ are correct, and they simply continue what you did in $A$. Because you can express $tan(n+1)$ as a rational fraction in $tan(n)$ and $tan(1)$, you can have by induction $tan(n)$ as a fraction of $tan(1)$ only. So if $tan(1)$ is rational, $tan(n)$ too.

The three methods are the same idea, but the only problem with A is that, like you said, the fact that $tan(3)$ is irrational is not obvious, whereas the fact that $tan(30)$ or $tan(60)$ aren't is totally trivial..

In fact, there is also a problem in $C$, is that $tan(45)=1$ is clearly rational, so you cannot conclude anything on $tan(1)$^^

Answer 2

Method $A$ is overcomplicating things. You do not need to separate the cases based on $\tan(2^\circ)$. You already know that, if $\tan 1^\circ$ is rational, then $\tan 2^\circ$ is rational, so the first bullet of method $A$ is redundant.

Method $B$ is an expansion and improvement of method $A$. It is the only one of the three methods that is complete, because it is obviously true that $\tan 60^\circ$ is irrational, while the irrationality of $\tan 3^\circ$ is not obvious. In fact, the easiest way to prove that $\tan 3^\circ$ is irrational is to repeat the idea of method $B$.

Method $C$ does not prove anything else than the fact that method $C$ cannot prove that $\tan 1^\circ$ is irrational.

Answer 3

Can we conclude that $\tan1^\circ$ is a rational number in method C as there seems no contradiction?

No. B is proof by contradiction plus mathematical inducing.

1. Assume $\tan1^\circ$ is rational.
2. proof by inducing: for any $n>1,n\in\mathbb{N}$,$\tan(n^\circ)$ is rational.(note this uses 1 as basis)
3. find a contradiction: $\tan 60^\circ=\sqrt{3}$ is irrational. But the inducing in 2 is all right, thus disproving the basis 1, ie. $\tan1^\circ$ is rational.

C follows the same assumption 1) and inducing 2), but does not find contradiction. A single case that confirms the conjecture 2) cannot prove conjecture 2), so cannot prove the basis assumption 1).

Is the proof by induction correctly used in method B and C?

B is correct. C does not prove anything, as explained above.

Is the proof by induction in method A the strongest?

method A is not proof by induction. it is a direct proof by contradiction ($\tan1^\circ\to\tan2^\circ\to\tan3^\circ$), but over-complicated.

Answer 4

Let $S_n$ be the statement that $\tan n^\circ$ is rational. Then $$ \tan (n+1)^\circ = \frac{\tan n^\circ + \tan 1^\circ}{1-\tan n^\circ \tan 1^\circ} $$ shows that $S_1,S_n\implies S_{n+1}$. So by assuming $S_1$ we have by induction. $$ S_1\wedge S_1\implies S_1\wedge S_2\implies ...\implies S_1\wedge S_n\text{ etc.} $$ Note how $S_1$ is always a component in each step. Thus if we ever encounter any $S_n$ which is obviously false, we conclude by contraposition that all the statements $S_1\wedge S_k$ for $1\leq k\leq n$ must have been false. This does not leave out the possibility of some $S_k$'s being true. Only $S_1\wedge S_k$ is still not true since $S_1$ is false (because in particular $S_1\wedge S_1$ is false).

Methode A: Corresponds to taking $S_3$ and thus $S_1\wedge S_3$ as contradiction.

Method B: Corresponds to taking $S_{60}$ and thus $S_1\wedge S_{60}$ as contradiction.

Method C: Corresponds to only noticing that $S_{45}$ is true, but not considering that $S_1\wedge S_{45}$ is actually false because $S_1$ is false.

So to conlude, method A and B are very similar allthough $S_3$ and $S_{60}$ may not be equally simple to falsify. Method C tells us virtually nothing regarding $S_1$.

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Method B+C combined: We could, on the other hand, conclude from that fact that $S_{45}$ is true but $S_{60}$ is false, that since $$ S_1\wedge S_{45}\implies S_1\wedge S_{60} $$ where the latter is false, by contraposition $S_1\wedge S_{45}$ is false already implying $S_1$ to be false.