3
$\begingroup$

When proving $\mathrm{LHS}=\mathrm{RHS}$, the most common way of doing it is by manipulating it in such a way to show that $\mathrm{LHS}$ equals to some expression which equals to $\mathrm{RHS}$. But what about these methods:

Method 1: Showing that $\mathrm{LHS}-\mathrm{RHS}=0$

For instance, if we are required to prove

$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$

We instead show that $\mathrm{LHS}-\mathrm{RHS}=0$ as follows:

$\mathrm{LHS} - \mathrm{RHS} = x^2\cos^2(x)+\sin^2(x) - x^2 + x^2\sin^2(x) - 1 + \cos^2(x)$

$ = x^2(\cos^2(x) + \sin^2(x)) - x^2 + (\sin^2(x) + \cos^2(x)) - 1$

$= x^2 - x^2 + 1 - 1 = 0$

Method 2: Showing that $\mathrm{LHS}=\mathrm{RHS}$ is equivalent with another equation which is true (taking care that we can always reverse the steps and showing it by putting $\iff$):

$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$

$ \iff x^2\cos^2(x)+x^2\sin^2(x) - x^2 = 1 - \sin^2(x) - \cos^2(x)$

$ \iff x^2 - x^2 = 1 - 1$

$ \iff 0 = 0$

Are these two methods of proving valid? Are there any cases where we can make fallacious argument by using these methods? Are any one of them better than the other?

$\endgroup$
4
$\begingroup$

Yes, they are perfectly valid. I don't know if we can say that any of these methods is better than the other ones — it's a matter of convenience, efficiency, and sometimes even personal preferences. Some problems are better handled with one method, and some other problems with another.

And if used correctly, they wouldn't lead to fallacious proofs. The mistake in many fallacious "proofs" (of the kind that are popular out there) is precisely a violation of the logic of a valid method. For example, what makes the last method in your post correct is the fact that each step is an equivalence $\color{magenta}{\Longleftrightarrow}$. And some fallacious proofs make us believe that this is the case in that "proof" too, while in fact they sneak in a one-way implication $\color{red}{\Longrightarrow}$ somewhere in the middle.

$\endgroup$
1
$\begingroup$

One of the related questions here, seems to be more suitable for method $2$:

$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x} (*)$$

Let $a =\sin x$, $b=\cos x$, then:

$(*)\Leftrightarrow \frac{1+a}{b}=\frac{1+a+b}{1-a+b}$

$\Leftrightarrow (1+a)(1-a+b) = b(1+a+b)$

$\Leftrightarrow 1-a^2+b(a+1) = b^2 + b(a+1)$

$\Leftrightarrow a^2+b^2=1$, which is provable.

For method $1$, I think you can use it to prove this simple expression:

$$a^2-b^2=(a-b)(a+b)$$

For this problem, using method $1$ or $2$ is actually the same, no method is better than the other.

For some problems when either side is hard to be made "equal" to the other side, or if you are stuck with this method because of that annoying "$\Rightarrow$", using substraction is recommended.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.