Can two equations when multiplied sometimes not give graph of both the equations?

Question

I wanted graph of this equation: $$\left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{5}\right)^2+\left(\frac{y}{4}\right)^2-1\right)\left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{3.4}\right)^2+\left(\frac{y}{2.72}\right)^2-1\right)\ \ \ \ \ \left(x-y\right)=0$$

I want a x=y line as well. So I kept (x-y) on the function side since (x-y)=0 should ultimately give x=y as graph along with graph of other functions. But I don't get the x=y line as a whole. It's not complete and I don't understand why.

I think there's something to do with which other equations I have used. But that doesn't seem to click. Isn't it mathematically correct to multiply two equations with 0 on right sides and get both equations' graph?

Answer 1

You're completely right about "the other equations" being the source of the problem.

Let me start with something simple. What is the value of $$ x \cdot 0 ? $$

It's very tempting to say, "It's $0$, of course, because anything times zero is zero!" But what if $x$ is, for instance, a cheese omelet? Then...well...then multiplication by zero isn't even defined, and you'd have to say "$x \cdot 0$ is undefined" rather than "it's zero." You might say that "cheese omelet" isn't really math, but what if we replace $x$ like this: $$ (1/0) \cdot 0 $$ Then our first factor is undefined, and multiplication by undefined things is undefined, so once again the product is "undefined" rather than zero.

Now let's look at a more complicated example. I'm going to stick with the real numbers -- no complex numbers allowed! -- and I want to look at $$ \sqrt{x} \cdot 0 $$ You might say "Well, that's definitely zero!", but what if $x = -3$? Then the square root is undefined, and so the product is undefined as well.

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Here's a theorem that your question hints at:

Theorem: Suppose $f, g : \Bbb R^2 \to \Bbb R$, and $G$ is the solution set of the equation $g(x, y) = 0$, and $F$ is the solution set of the equation $f(x, y) = 0$. Then the solution set of the equation $$ f(x,y) g(x, y) = 0 $$ is exactly $F \cup G$.

In your case, you've got $g(x, y) = x - y$, and the solutions for $x-y = 0$ form a line in the plane; that's the set $G$ in the theorem. You've multiplied it by some complicated mess that I'll temporarily call $f(x, y)$, and looked for solutions to $$ \tag{1} f(x, y) \cdot g(x, y) = 0 $$ (perhaps using Desmos or some other graphing tool), and you're wondering why the solution set of this doesn't include all of $G$.

The answer is "because $f$ is not defined on all of $\Bbb R^2$ --- for certain $(x,y)$ pairs, $f(x, y)$ is undefined, and those pairs cannot be part of the solution to equation (1)."

For instance, let's look at the point $(a, b) = (-1, -1)$. Clearly this point satisfies $g(a, b) = a - b = (-1) - (-1) = 0$. But what is $f(a, b)$? Well, your formula for $f$ is $$ f(x, y) = \left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{5}\right)^2+\left(\frac{y}{4}\right)^2-1\right)\left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{3.4}\right)^2+\left(\frac{y}{2.72}\right)^2-1\right) $$ and plugging in $-1$ for each of $x$ and $y$ gives, deep in the middle, this: $$ \ldots \left(\frac{\left((-1)\sqrt{\frac{\left|\left|(-1)\right|-5\right|}{-(-1)-5}}+5\right)}{5}\right)^2 \ldots $$ and the middle square root in that simplifies to \begin{align} &(-1)\sqrt{\frac{\left|\left|(-1)\right|-5\right|}{-(-1)-5}}\\ &= -\sqrt{\frac{\left|1-5\right|}{1-5}}\\ &= -\sqrt{\frac{\left|-4\right|}{-4}}\\ &= -\sqrt{\frac{4}{-4}}\\ &= -\sqrt{-1} \end{align} which is undefined. So because $f(a, b)$ is undefined, so is $f(a, b) \cdot g(a, b)$, and hence this product cannot be zero...so the point $(a, b) = (-1, -1)$ is not part of the plot of the solutions to equation (1).

The modified theorem that gets to what's going on here is this:

Theorem: Suppose $D, E \subset \Bbb R^2$, $f:D \to \Bbb R$, $g:E \to \Bbb R$, $G$ is the solution set of the equation $g(x, y) = 0$, and $F$ is the solution set of the equation $f(x, y) = 0$. Then the solution set of the equation $$ f(x,y) g(x, y) = 0 $$ is exactly $(F\cap E) \cup (G \cap D)$.

which says that only those solutions of $g = 0$ (i.e., $G$) that lie in the domain of $f$ (i.e., $D$) get included in the result, and only solutions of $f = 0$ that lie in the domain of $E$ get included in the result.

Answer 2

Notice that if $f(x) = \left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{5}\right)^2+\left(\frac{y}{4}\right)^2-1\right)\left(\left(\frac{\left(x\sqrt{\frac{\left|\left|x\right|-5\right|}{-x-5}}+5\right)}{3.4}\right)^2+\left(\frac{y}{2.72}\right)^2-1\right)$

then $f(x)$ is not defined for $x=-5$ because then $\sqrt {\frac {|x|-5|}{-x-5}}$ is dividing by $0$. And if $x > -5$ then $-x-5 < 0$ so $\sqrt {\frac {|x|-5|}{-x-5}}$ is the square root of a negative number.

So $f(x)$ simply doesn't work and does not exist if $x \ge -5$.

And if $f(x)$ does not exist for $x \ge -5$ then $f(x)\cdot (x-y)$ can't exist for $x \ge -5$ either.

It doesn't matter that multiplying by $f(x)(x-y) = 0$ and $0$ exists always. If $f(x)$ doesn't exist then neither does $f(x)\times something else$. So we have $f(x)(x-y)= 0$ AND $x < -5$. We can't magic things into existence just because they "cancel out".