What is wrong with this proof based on induction?

Question

Premise:

1. Let P(n) return the real part of the nth nontrivial root of the zeta function
2. The first several roots of the zeta function are already known

Proof:

1. Pick any integer n. For example, 1.
2. Solve for P(n)
3. Solve for P(n+1)
4. Both (2) and (3) are 1/2 as established by Alan Turing and others
5. By induction all P(n) are 1/2

QED the real part of all nontrivial roots of the riemann zeta function is 1/2

Any problems?

Answer 1

The problem with this proof is that it isn't a proof. Step number 4 claims something that was not proven in previous steps. We don't know what $f$ is, that's the whole point.

Also, even if this "proof" was correct, this is not how a proof by induction. A proof of induction would look like:

1. Prove that the real part of $f(1)$ is $\frac12$.
2. Assume that the real part of $f(n)$ is $\frac12$.
3. Prove, from assumption $(2)$, that the real part of $f(n+1)$ is $\frac12$

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An example of your proof, showing something very wrong:

1. Define $P(n) = n$
2. The first several values of $P(n)$ are already known

Claim: for all values $n$, the value $P(n)$ is smaller than $1000000$.

Proof:

1. Pick any integer $n$, for example $1$.
2. Solve for $P(n)$
3. Solve for $P(n+1)$
4. Both (2) and (3) are smaller than $1000000$ as $1<1000000$ and $2<1000000$.
5. By induction all $P(n)$ are smaller than $1000000$

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An example of a good proof by induction:

Claim: For all values $n\in\mathbb N$, the sum $\sum_{i=1}^n i$ is equal to $\frac{n(n+1)}{2}$

Proof:

First of all, the claim is true for $n=1$ since $$\sum_{i=1}^1 i = 1 = \frac{1\cdot(1+1)}{2}$$

Now, assume that the claim is true for $n$.

1. By assumption, $$\sum_{i=1}^n i = \frac{n(n+1)}{2}.$$
2. By definition, $$\sum_{i=1}^{n+1}i = \sum_{i=1}^n i + (n+1)$$
3. From (1) and (2), it follows that $$\sum_{i=1}^{n+1}i = \sum_{i=1}^n i + (n+1) = \frac{n(n+1)}{2} + n+1$$
4. Simplifying the expression, we get $$\frac{n(n+1)}{2} + n+1 = \frac{n^2+n}{2} + \frac{2n+2}{2} = \frac{n^2+3n+2}{2} = \frac{(n+1)(n+2)}{2}$$
5. From (4), it follows that $$\sum_{i=1}^{n+1}i = \frac{(n+1)(n+2)}{2}$$ in other words, the claim is true for $n+1$.

So, assuming that the claim is true for $n$, we have proven (in point (5)), that the claim is true for $n+1$. We have also proven that the claim is true for $1$. This allows us to conclude that the claim is true for all values of $n$.

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Unlike your proof, this valid induction proof cannot be refuted by the same arguments. It is valid for all values of $n$.

Answer 2

There are a considerable number of problems. First, you've omitted the base case; induction can only work if the claim holds of the starting point (usually $0$). Second, steps (2) and (3) have no content; $f(n)$ and $f(n+1)$ simply have values, no solving is required. Third, step (4) draws a conclusion that isn't there; why do you think that $f(n)$ is $\frac{1}{2}$? Or that $f(n+1)$ is?