No problem, I'm here to help! Is there anything else you need clarification on?

[mathdad]

Test for symmetry about the x-axis, y-axis and the origin.

x + |y| = 2

About y-axis:

-x + |y| = 2

Not symmetric about y-axis.

About x-axis:

x + |-y| = 2

x + y = 2

I say not symmetric about the x-axis.

About the origin:

-x + |-y| = 2

Not symmetric about the origin.

Correct? If not, why?

 

[Evgeny.Makarov]

Test for symmetry about the x-axis, y-axis and the origin.

x + |y| = 2

You should give the full problem statement. Currently it is not completely clear what has to be tested.

 

[MarkFL]

I would say you are correct regarding symmetry about the $y$-axis and the origin, however regarding the $x$-axis, consider:

$$\sqrt{a^2}=\sqrt{(-a)^2}$$

Now, since:

$$|a|\equiv\sqrt{a^2}$$

We then conclude:

$$|a|=|-a|$$

And so the given equation is symmetric about the $x$-axis. :D

 

[mathdad]

You should give the full problem statement. Currently it is not completely clear what has to be tested.

What are you talking about? Test the given absolute value equation for symmetry in terms of the x-axis, y-axis and origin.

 

[Evgeny.Makarov]

Test the given absolute value equation for symmetry in terms of the x-axis, y-axis and origin.

I am probably being too picky here. It is more common to apply the concept of symmetry to geometric figures. Thus, saying that a figure $S$ (a set of points on a plane) is symmetric with respect to a transformation $f$ means that whenever a point $P$ lies inside $S$, so does $f(P)$. In your case it would be clearer to ask whether the set of solutions to the equation $x + |y| = 2$ is symmetric about the $y$-axis. This would mean that the set $\{(x,y)\mid x + |y| = 2\}$ is symmetric w.r.t. the transformation $f(x,y)=(-x,y)$. This in turn means that
\[
x + |y| = 2\text{ implies }-x + |y| = 2\qquad(*)
\]
for all $x$ and $y$. My point is that talking about symmetries of figures w.r.t. transformations is more standard than talking about symmetries of equations, but if the latter concept was properly defined in a course, it is fine as well.

Next, checking whether one equation implies another is best done the way any universal statement is checked: either by constructing a general proof or by finding a single counterexample. If you need to refute the hypothesis that $x\le y$ implies $x^2\le y^2$ for all real numbers $x$ and $y$, it is best to note, for example, that $-2\le 1$, but $(-2)^2=4>1=1^2$. Similarly, to refute (*) it is sufficient to note that $x=y=1$ satisfies the premise $x + |y| = 2$, but violates the conclusion $-x + |y| = 2$. Therefore, (*) does not hold for all $x$ and $y$. The simple fact that the equation in the conclusion of (*) looks different than the equation in the premise is not the best explanation. For example, $x=y$ and $-x=-y$ are two different equations, but they are equivalent. But here I am talking about the final proof that can be written as an answer to the problem. Replacing $x$ with $-x$ and checking whether the equation becomes (essentially) different is a fine first step.

Concerning the symmetry about the $x$-axis, we have $x + |y| = 2$ implies $x + \lvert-y\rvert = 2$ for all $x$ and $y$ because $\lvert-y\rvert=y$, so the symmetry holds. The counterexample for the symmetry about the $y$-axis also works for the symmetry about the origin.

 

[mathdad]

I am probably being too picky here. It is more common to apply the concept of symmetry to geometric figures. Thus, saying that a figure $S$ (a set of points on a plane) is symmetric with respect to a transformation $f$ means that whenever a point $P$ lies inside $S$, so does $f(P)$. In your case it would be clearer to ask whether the set of solutions to the equation $x + |y| = 2$ is symmetric about the $y$-axis. This would mean that the set $\{(x,y)\mid x + |y| = 2\}$ is symmetric w.r.t. the transformation $f(x,y)=(-x,y)$. This in turn means that
\[
x + |y| = 2\text{ implies }-x + |y| = 2\qquad(*)
\]
for all $x$ and $y$. My point is that talking about symmetries of figures w.r.t. transformations is more standard than talking about symmetries of equations, but if the latter concept was properly defined in a course, it is fine as well.

Next, checking whether one equation implies another is best done the way any universal statement is checked: either by constructing a general proof or by finding a single counterexample. If you need to refute the hypothesis that $x\le y$ implies $x^2\le y^2$ for all real numbers $x$ and $y$, it is best to note, for example, that $-2\le 1$, but $(-2)^2=4>1=1^2$. Similarly, to refute (*) it is sufficient to note that $x=y=1$ satisfies the premise $x + |y| = 2$, but violates the conclusion $-x + |y| = 2$. Therefore, (*) does not hold for all $x$ and $y$. The simple fact that the equation in the conclusion of (*) looks different than the equation in the premise is not the best explanation. For example, $x=y$ and $-x=-y$ are two different equations, but they are equivalent. But here I am talking about the final proof that can be written as an answer to the problem. Replacing $x$ with $-x$ and checking whether the equation becomes (essentially) different is a fine first step.

Concerning the symmetry about the $x$-axis, we have $x + |y| = 2$ implies $x + \lvert-y\rvert = 2$ for all $x$ and $y$ because $\lvert-y\rvert=y$, so the symmetry holds. The counterexample for the symmetry about the $y$-axis also works for the symmetry about the origin.

You went out of your way to explain this in detail. Thanks.

 

[mathdad]

I am probably being too picky here. It is more common to apply the concept of symmetry to geometric figures. Thus, saying that a figure $S$ (a set of points on a plane) is symmetric with respect to a transformation $f$ means that whenever a point $P$ lies inside $S$, so does $f(P)$. In your case it would be clearer to ask whether the set of solutions to the equation $x + |y| = 2$ is symmetric about the $y$-axis. This would mean that the set $\{(x,y)\mid x + |y| = 2\}$ is symmetric w.r.t. the transformation $f(x,y)=(-x,y)$. This in turn means that
\[
x + |y| = 2\text{ implies }-x + |y| = 2\qquad(*)
\]
for all $x$ and $y$. My point is that talking about symmetries of figures w.r.t. transformations is more standard than talking about symmetries of equations, but if the latter concept was properly defined in a course, it is fine as well.

Next, checking whether one equation implies another is best done the way any universal statement is checked: either by constructing a general proof or by finding a single counterexample. If you need to refute the hypothesis that $x\le y$ implies $x^2\le y^2$ for all real numbers $x$ and $y$, it is best to note, for example, that $-2\le 1$, but $(-2)^2=4>1=1^2$. Similarly, to refute (*) it is sufficient to note that $x=y=1$ satisfies the premise $x + |y| = 2$, but violates the conclusion $-x + |y| = 2$. Therefore, (*) does not hold for all $x$ and $y$. The simple fact that the equation in the conclusion of (*) looks different than the equation in the premise is not the best explanation. For example, $x=y$ and $-x=-y$ are two different equations, but they are equivalent. But here I am talking about the final proof that can be written as an answer to the problem. Replacing $x$ with $-x$ and checking whether the equation becomes (essentially) different is a fine first step.

Concerning the symmetry about the $x$-axis, we have $x + |y| = 2$ implies $x + \lvert-y\rvert = 2$ for all $x$ and $y$ because $\lvert-y\rvert=y$, so the symmetry holds. The counterexample for the symmetry about the $y$-axis also works for the symmetry about the origin.

You went out of your way to explain this in detail. Thanks.