Is this backwards reasoning?

Question

Yesterday I was answering a question on induction: Certain step in the induction proof $\sum\limits_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ unclear

Basically, I was proving a certain formula using induction.

$$\sum\limits_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$

The base case it's okay. Now let's assume the formula is valid for $N$, we want to demonstrate the following, that is

$$\sum\limits_{i=0}^{N+1} i^2 = \frac{(N+1)(N+2)(2(N+1)+1)}{6} \ \ (1) $$

that is to say

$$\sum\limits_{i=0}^{N} i^2 + (N + 1)^2 = \frac{(N+1)(N+2)(2(N+1)+1)}{6}\ \ (2) $$

$\Rightarrow$ (thanks to induction hypothesis)

$$\frac{N(N+1)(2N+1)}{6} + (N+1)^2 = \frac{(N+1)(N+2)(2(N+1)+1)}{6}\ \ (3) $$

Then I concluded that if one show that (3) is true (by simplifying, and getting $0 = 0$) then the proof is valid and complete.

Some argue this is backwards reasoning; but I can't understand why.

The equalities that I use to go from (1) to (2) to (3) can be used for going from (3) to (2) to (1)

My argument is, if (3) simplifies to $0=0$, so it is equivalent with that and therefore True, also (2) is True, and also (1) is true, which was what I wanted to prove.

Is this backwards reasoning, and if so someone please explain me why

Answer 1

EDIT: The question has been updated, so the answer is mostly irrelevant (but it still shows how one word here or there can change quite a lot :-) )

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I don't think your reasoning was backwards, but its presentation and wording might have been understood so. Specifically, stating "we have" implies it's something we either assume or have already proven rather than something we're trying to prove. Presenting the same reasoning slightly differently could avoid the ambiguity:

Let's evaluate the sum for $N+1$: $$\sum_{i=1}^{N+1} i^2 = \sum_{i=1}^{N} i^2 + (N+1)^2$$ The induction hypothesis tells us that $$\sum_{i=1}^{N} i^2 = \frac{1}{6}N(N+1)(2N+1)$$ so $$\sum_{i=1}^{N} i^2 + (N+1)^2 = \frac{1}{6}N(N+1)(2N+1) + (N+1)^2$$ Simplifying the right-hand side yields $$\frac{1}{6}N(N+1)(2N+1) + (N+1)^2 = \frac{1}{6}(N+1)\left((N+1)+1\right)\left(2(N+1)+1\right)$$

Finally, combining the equalities shows that the statement holds for $N+1$ as well, thus completing the inductive step.

Answer 2

It is a common misconception for students that you can prove $A=B$ by showing that it simplifies to $0=0$. Often times this is a bad habit picked up from observing others who find it convenient to work this way (and who may or may not understand when it is logically sound). But this is generally a bad practice because it really is backwards reasoning; like Peter Košinár, I won't go so far as to accuse you of this, but the order that your proof is written can easily be interpreted this way.

When doing things in this order (starting from $A=B$ and simplifying down to $0=0$), once has to take extra care that every operation is perfectly reversible. Adding the same thing to both sides is reversible by subtraction. Multiplying both sides by the same quantity is reversible by division...unless it's possible that the multiplier is $0$! Squaring both sides is potentially reversible, provided they were known to be non-negative to begin with.

So when you write out a proof in this order, what you are doing is placing a burden on yourself to consider the reversibility of every single step (or if you don't mention this, you place the burden on the reader). And if you're going to have to explicitly reverse each step anyway, it's almost always the case that it's simpler to just write it down in logical order in the first place (starting from a known truth and ending up at $A=B$).

One of the conventions of mathematical writing is that we strive to present arguments in a logical order of deduction, starting from true facts and building more and more true facts until we reach the desired conclusion. This convention is not necessarily obvious or natural: it might feel better to explain your thinking process in chronological order ("first I did this, then I calculated that, and that's why I did this to both sides").

Mathematical analysis is full of proofs that are written in exactly the opposite order as the discovery process — this might seem strange to you, but it is the standard convention and it is what your readers will expect. (That's not to say every proof needs to be written mechanically from basic premises to final conclusion: many times an author can improve the readability of a long proof by using abridgments like "we'll use this fact, whose proof is deferred to the appendix", or annotations like "we chose this because such-and-such will be needed later in Section 3").

Answer 3

Equation (1) uses the formula to sum the terms from $0$ to $N+1$, by replacing $N$ with $N+1$ everywhere on the right side. Check.

The left side of equation (2) sums the terms from $0$ to $N$, and adds the $(N+1)$th term explicitly. Check.

The left side of equation (3) makes use of the assumption that the equation holds for terms summed from $0$ to $N$. Check.

Explicit calculation of the left side of (3) results in the same expression as you found on the right side of (1). Check.

So you've got all of the pieces there; you just need to arrange them a bit differently and change the phrasing in between the steps.

Answer 4

If you want to present a proof in the way you discovered it you could indeed do it this way:

To prove $A$, it suffices to prove $B$, which is equivalent to $C$, ...

Or in symbolic form:

We have:

  $A$

  $\ \Leftarrow B$

  $\ \Leftrightarrow C$.

  $\ \cdots$

In both cases, the logical reasoning is made clear without going in the 'opposite direction' to the way you found the proof.

Just to make sure people don't think that simplification to $0=0$ means that an equation is true, find the logical error in the following fake proof:

We want to prove $\sqrt{x^2+1} = x \sqrt{1+\frac1{x^2}}$ for every $x \ne 0$, and we proceed by squaring and simplifying:

  $x^2+1 = x^2 ( 1 + \frac1{x^2} )$.

  $x^2+1 = x^2+1$.

Therefore we (WRONGLY) conclude that the claim is proven.

Many students would see nothing wrong, so dear reader, please make sure you know precisely what is wrong.

Answer 5

How about this.

To show that $$L(n)=\sum_{i=1}^{n}a_i=R(n)$$ holds for all $n$, you use induction. Step 1: Show that $$L(1)-R(1)=0.$$ Step 2: Show that $$L(n)=R(n)\quad\Rightarrow\quad L(n+1)=R(n+1).$$ Since $L(n+1)=L(n)+a_{n+1}$, we have $$L(n)=R(n)$$ $$L(n)+a_{n+1}=R(n)+a_{n+1}$$ $$L(n+1)=R(n)+a_{n+1}$$ so we need to show that $$R(n)+a_{n+1}=R(n+1)\qquad(1)$$ The standard method (let us call it method A) to show $(1)$ is to try to construct the form of $R(n+1)$ from $R(n)+a_{n+1}$. An alternative method (method B) is simply to check if $$R(n)+a_{n+1}-R(n+1)=0.$$ It is usually much easier to do method B, as you don't need to come up with any tricks from algebra. Just expand and simplify. If you get $0$, you're home free. In fact we can do better than this. It is sufficient to show that $$k\cdot[R(n)+a_{n+1}-R(n+1)]=0$$ for some nonzero $k$. This means that you can get rid of troublesome fractions on your way by multiplying by them. If you do that, you should write $\propto$ instead of $=$.