Fake induction proofs

Claim: $\frac{d}{dx}x^n=0$ for all $n\ge0$.

Base case: ($n=0$): $\frac{d}{dx}x^0=\frac{d}{dx}1=0$

Inductive step: Assume that $\frac{d}{dx}x^k=0$ for all $k\le n$. Then by the product rule,

$$\frac{d}{dx}x^{n+1}=\frac{d}{dx}(x^n\cdot x^1)=x^n\frac{d}{dx}x^1+\left(\frac{d}{dx}x^n\right)x^1=x^n\cdot0+0\cdot x^1=0.$$

Flaw:

In order for this to be a valid proof, the inductive step must be valid for all $n\ge0$. However, when $n=0$, one can’t use the inductive hypothesis to rewrite $\frac{d}{dx}x^1$ as $0$.

This “spoof” appears in Martin V. Day’s “An Introduction to Proofs and the Mathematical Vernacular.” Day gives its source as Edward J. Barbeau’s “Mathematical Fallacies, Flaws and Flimflam.”

Here is a "proof" of a famous identity by Ramanujan: $$\sqrt{1+\sqrt{1+2\sqrt{1+3{\sqrt{1+4\sqrt{\dots}}}}}}=2.$$

Claim: Let us prove this more general result for all $n\geq 0$: $$\sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{\ldots}}}}}=n+1.$$

Base case: When $n=0$, we have $\sqrt{1+0\sqrt{\dots}}=0+1$, and this is true.

Inductive step: Assume that the identity holds for some $n$ and let us prove it holds for $n+1$. By squaring both sides we get $$1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{\ldots}}}}=n^2+2n+1.$$ Subtracting $1$ and dividing by $n$, we get $$\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{\ldots}}}}=n+2,$$ which is what we wanted to show. $\blacksquare$

Flaw:

We cannot divide by $n$ because at the very beginning of induction $n=0$. Moreover, one should justify the existence of the infinite nested radical.

Comment: This "proof" is funny because, in fact, it gives a correct result!

Claim: $a^n=1$ for all nonnegative integers $n$, whenever $a$ is a nonzero real number.

Base case: $a^0=1$ is true by the definition of $a^0$.

Inductive step: Assume that $a^m=1$ for all nonnegative integers $m$ with $m\leq k$. Then notice that $$ a^{k+1}=\frac{a^k\cdot a^k}{a^{k-1}}=\frac{1\cdot 1}{1}=1. $$

Flaw:

The flaw occurs in the inductive step where we implicitly assume that $k\geq 1$ in order for us to talk about $a^{k-1}$ in the denominator; otherwise, the exponent is not a nonnegative integer, meaning we cannot apply the inductive hypothesis. We checked the base case only for $n=0$; thus, we are not justified in assuming that $k\geq 1$ when we try to prove the statement for $k+1$ in the inductive step. It is exactly at $n=1$ that the proposition breaks down.

Here is one published by Knuth.

Claim: $$\underbrace{\frac1{1\cdot2}+\frac1{2\cdot3}+\ldots}_{n\text{ terms}}=\frac32-\frac1n$$

Base case: For $n=1$, we have $\frac32-\frac11=\frac1{1\cdot2}$

Inductive step: $$\left(\frac1{1\cdot2}+\ldots+\frac1{(n-1)\cdot n}\right)+\frac1{n\cdot(n+1)}=\frac32-\frac1n+\frac1{n\cdot(n+1)}$$ $$=\frac32-\frac1n+\frac1n-\frac1{n+1}=\frac32-\frac1{n+1}$$

Flaw: