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For example, let's say that we want to prove that $A=B$ and also know that $X=Y$. If $A=B \Rightarrow X=Y$, does this not imply that $A=B$?

I understand that there are some potential issues with this sort of reasoning. If both sides of any equation are multiplied by $0$ then the conclusion $0=0$ is reached. This obviously can't prove anything. From this (and logically) we can infer that all steps taken must be reversible. It seems to me that this is valid for any equation (see below for specific example), but I am not sure. So I have two questions:

If $X=Y$ and the equation $A=B$ will lead to $X=Y$ when reversible steps are taken, can it be said that $A$ is in fact equal to $B$? And if so, does this also work for any two statements? That is, if $p \rightarrow q$ and $[m \rightarrow n] \Rightarrow [p \rightarrow q]$ with reversible steps, does $m\rightarrow n$?

Here's a specific example: Prove that $\frac{d}{dx}(fg)=f'g +g'f$ given the definition of a derivative.

$\frac{d}{dx}f(x)=\lim_{h\to0} \frac{f(x+h)-f(x)}{h} \therefore \frac{d}{dx}(f(x)g(x))=\lim_{h\to0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}$ and $f'(x)g(x)+g'(x)f(x)=\lim_{h\to0}[\frac{f(x+h)-f(x)}{h}*g(x)+\frac{g(x+h)-g(x)}{h}*f(x)]$. For the sake of convenience, I will no longer type out "$\lim_{h\to0}$." This equation can be re-written as follows:

$\frac{f(x+h)g(x+h)-f(x)g(x)}{h}=\frac{f(x+h)-f(x)}{h}*g(x)+\frac{g(x+h)-g(x)}{h}*f(x)$

${f(x+h)g(x+h)-f(x)g(x)}=({f(x+h)-f(x)})*g(x)+({g(x+h)-g(x)})*f(x)$

${f(x+h)g(x+h)-f(x)g(x)}=f(x+h)g(x)-f(x)g(x)+f(x)g(x+h)-f(x)g(x)$

$f(x+h)g(x+h) = f(x+h)g(x) + f(x)g(x+h) - f(x)g(x)$

Dividing by $f(x)g(x)$ gives $\frac{f(x+h)g(x+h)}{f(x)g(x)} = \frac{f(x+h)}{f(x)}+\frac{g(x+h)}{g(x)}-1$.

And finally, evaluating the limit gives $\frac{f(x)g(x)}{f(x)g(x)} = \frac{f(x)}{f(x)}+\frac{g(x)}{g(x)}-1 \Rightarrow 1=2-1 \Rightarrow 1=1$.

Does this prove the above?

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  • $\begingroup$ The answer to the question in the title: no. Consider $0=1$ as a statement. Then $0=1-1=1-0=0-0=0$, so we can conclude $0=0$. The latter is true while the former is clearly absurd. $\endgroup$ – Clayton Mar 27 '18 at 2:36
  • $\begingroup$ Dividing by $f(x)g(x)$ is not a "reversible" step, i.e. $xy=z$ is not logically equivalent to $y=\frac{z}{x}$. $\endgroup$ – Derek Elkins Mar 27 '18 at 2:50
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No. The statement $0=1$ can be used to prove any other statement at all. This is known as the principle of explosion, and quite definitively shows that simply leading to truth is not enough to promise truth.

If as you say in your post, "reversible" steps are taken, you are not working with the implication $A=B \Rightarrow X=Y$, but rather the biconditional $A=B \iff X=Y$, in which case yes, the truth of one implies the other. But what is important in this second case is that until you independently verify the truth of one of the sides, you don't actually know if either are true or false.

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  • $\begingroup$ See this Wikipedia article for more: en.wikipedia.org/wiki/Principle_of_explosion $\endgroup$ – Valborg Mar 27 '18 at 2:34
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    $\begingroup$ Also, you have committed a grievous sin in your example of proving the product rule for derivatives by dropping those limits. You absolutely cannot multiply through by $h$ in this case, because it is locked behind those limit signs! This is why professors and TAs will harp about proper notation to everyone taking calculus. This account doesn't have enough reputation to comment on other posts, so someone please let the poster know that he has violated several rules here. $\endgroup$ – Valborg Mar 27 '18 at 2:39
  • $\begingroup$ How exactly is $h$ locked behind the limit signs? (Just a bit confused by the wording) $\endgroup$ – Shawhin Layeghi Mar 27 '18 at 3:14
  • $\begingroup$ I think that one of the main rules of limits is that $\lim_{h\to a} [f(h)] = f(\lim_{h\to a}[h])$. $\endgroup$ – Shawhin Layeghi Mar 27 '18 at 3:16
  • $\begingroup$ @ShawhinLayeghi $h$ is the variable you are taking the limit with respect to, so it is not any fixed number you can multiply through the equation. It is in fact tending towards $0$, so you are in effect multiplying both sides by $0$ when you attempt this, which while allowed, is a tautology, and thus nothing can be recovered. Consider if I had attempted the same operation with $\lim_{h\rightarrow 0}\frac{h}{h}=1$. $\endgroup$ – Valborg Mar 27 '18 at 3:17
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Let's say that we want to prove that $A=B$ and also know that $X=Y$. If $A=B \Rightarrow X=Y$, does this not imply that $A=B$?

No. In classical (ordinary, everyday) logic, as can be verified with a truth table, we have the tautology:

$$P\implies (Q\implies P)$$

If $P$ is true, then the implication $Q\implies P$ will always be true regardless of whether $Q$ is true or false.

By the way, this tautology is also the basis for a useful method of proof: To prove $Q\implies P$, we need only prove that $P$ is true.

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No.

Showing that a statement leads to a true statement does not prove that former statement.

Simple counterexample:

The statement $0=1$ leads to $1=1$ since if $0=1$, then $0+1=1+0$

And if you don't like the fact that this relies on arithmetical assumptions, here is a purely logical proof:

If $0=1$, then we can replace the $0$ with a $1$, and therefore $1=1$

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