Just a definition by a mathematician? But there should be some good reasons for adopting the definition.
Topics
About: junior high school mathematics
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Why Are Mirror Images Regarded to Be Congruent in the Plane Geometry?
- 2: It Seems More Natural to Not Call Mirror Images 'Congruent', Supposing That 'Congruent' Means Being Same-Shaped and Same-Sized
- 3: Some Reasons Why We Should Think of Map Instead of "Move"
- 4: 2 Figures Are Congruent iff 1 of Them Is the Image of the Other Under an Isometry
- 5: Isometry Is Really Any Combination of Translation, Rotation, and Mirroring
- 6: But Anyway, Why Is Isometry Used in the Definition of 'Congruence'?
- 7: Mirroring on a Plane or in a 3-Dimensional Space as a Rotation in the Ambient Space
Starting Context
- The reader knows the background of this site.
Target Context
- The reader will know some reasons why mirror images are regarded to be congruent in the plane geometry.
Orientation
There is an article on becoming a benefactor of humanity by being a conduit of truths
Main Body
1: Why Are Mirror Images Regarded to Be Congruent in the Plane Geometry?
Special-Student-7-Rebutter
In the (2-dimensional) plane geometry, 2 figures one of which is a mirror image of the other seems to be regarded to be congruent, at least in the Japanese junior high school mathematics.
Special-Student-7-Hypothesizer
The Japanese junior high school mathematics seems to be saying that 2 figures one of which can be "moved" to coincide with the other is regarded to be congruent.
Special-Student-7-Rebutter
What does "move" exactly mean?
Special-Student-7-Hypothesizer
It is the problem that "move" is not defined explicitly in the Japanese junior high school mathematics, although I understand that junior high school students in general are not expected to understand rigorous definitions.
Special-Student-7-Rebutter
But saying "move" is meaningless if the term is not understood accurately.
Special-Student-7-Hypothesizer
Intuitively speaking, a triangle is imagined like a physical object, which is moved in the space: as normal junior high school students cannot teleport objects, they will imagine continuously moving the object in the space, without deforming the object.
Special-Student-7-Rebutter
I see 2 problems there.
1st, how can you be sure that the object is not being inadvertently deformed while you are moving the object?
Special-Student-7-Hypothesizer
Well, the physical object is supposed to be very hard and be never deformed unintentionally by my just moving it.
Special-Student-7-Rebutter
Is "very hard" a legitimate mathematical concept?
Special-Student-7-Hypothesizer
"very hard" is a physical assumption that if the object is wooden or something, the object will not be deformed, at least much, by being moved without being intentionally tortured. ... I know that it is not a refined mathematical concept, but it seems to have been a historical motivation for thinking of 'congruence'.
Special-Student-7-Rebutter
2nd, what does "the space" mean when you say "is moved in the space"? As we are talking about a plane, "the space" should be naturally construed to be the plane, I think. Then, a figure cannot be "moved" to coincide with its mirror image by moving the figure continuously in "the space".
Special-Student-7-Hypothesizer
Apparently, the Japanese junior high school mathematics "moves" the figure in the ambient 3-dimensional space.
Special-Student-7-Rebutter
I do not say that they cannot introduce the ambient 3-dimensional space, but it seems unwise: while we should have been able to concentrate on the 2-dimensional plane, why do they have to complicate the situation by introducing the extra dimension?.
Special-Student-7-Hypothesizer
They seem to have cursorily introduced the ambient 3-dimensional space, because they are familiar with the 3-dimensional space. However, they will have to introduce the ambient 4-dimensional space for the 3-dimensional space geometry, if they are consistent with the strategy.
Special-Student-7-Rebutter
Certainly, 3-dimensional figures can be "moved" in the ambient 4-dimensional space, but it seems to be against the purpose of being intuitive: why do they have to make students think of the hard-to-imagine 4-dimensional space while we are concerned only with the 3-dimensional space?
Special-Student-7-Hypothesizer
Introducing the ambient space seems to be a strategy that is natural only for the plane geometry, and employing a strategy that lacks generality seems unwise.
2: It Seems More Natural to Not Call Mirror Images 'Congruent', Supposing That 'Congruent' Means Being Same-Shaped and Same-Sized
Special-Student-7-Hypothesizer
In fact, supposing that 'congruent' means being same-shaped-and-same-sized, calling mirror figures 'same-shaped' seems unnatural, in the first place.
Special-Student-7-Rebutter
Are there some grounds for that supposition?
Special-Student-7-Hypothesizer
At least most general (not mathematical) dictionaries say that 'congruent' means being same-shaped-and-same-sized.
Special-Student-7-Rebutter
Mirror images are not same-shaped in my vocabulary.
Special-Student-7-Hypothesizer
If someone feels mirror images being same-shaped, the reason seems to be that he or she has imagined moving figures in the ambient higher-dimensional space.
Special-Student-7-Rebutter
But most people do not naturally imagine moving 3-dimensional figures in the ambient 4-dimensional space, I guess.
3: Some Reasons Why We Should Think of Map Instead of "Move"
Special-Student-7-Hypothesizer
In fact, introducing "move" seems not wise, while "move" is understood as 'continuously move', as is naturally so.
A reason is that a figure has to be taken out of the plane into the ambient 3-dimensional space, if mirroring wants to be regarded to be a kind of "move".
Special-Student-7-Rebutter
As has been discussed above.
Special-Student-7-Hypothesizer
Another reason is that we have to think of the entire continuous movement of the figure in order for us to talk in terms of "move", while we are interested only in the original figure and the terminal figure.
So, why do we not skip the unnecessary middle?
Special-Student-7-Rebutter
If there is a reason, it seems to be that they are not freed from the intuitive mental picture of moving a physical object.
Special-Student-7-Hypothesizer
But in order for us to guarantee that the figure is not deformed in being moved, we have to guarantee that the figure is not deformed at each position, and then, why do we not guarantee that the figure is not deformed just at the terminal position?
Special-Student-7-Rebutter
I do not guess that there is a legitimate reason.
Special-Student-7-Hypothesizer
So, let us think of 'map' instead of "move", where 'map' means mapping each point of the original figure to a point of the terminal figure, without bothering to move the point continuously in the space.
The merits are the reverse of the unwiseness of introducing "move": we do not need the ambient space and we do not need to think of the middle.
Any map is prevalently denoted like \(f: S_1 \rightarrow S_2\), where \(S_1\) and \(S_2\) and some sets, and \(S_1\) is called 'the domain of the map' and \(S_2\) is called 'the codomain of the map'. The map does not need to really map the domain to the whole codomain, and the really mapped part of the codomain is called 'the range of the map'; if the range equals the codomain, the map is called to be a surjection. If any different 2 elements of \(S_1\) are mapped to different elements, the map is called to be an injection. Any map that is a surjection and an injection is called a bijection.
4: 2 Figures Are Congruent iff 1 of Them Is the Image of the Other Under an Isometry
Special-Student-7-Hypothesizer
In fact, mathematically speaking, 2 figures are congruent iff 1 of them is the image of the other under an isometry, where any isometry is any map that preserves the distance between any 2 points.
In the plane geometry, with the sets of 2 figures denoted as \(S'_1 \subseteq \mathbb{R}^2\) and \(S'_2 \subseteq \mathbb{R}^2\), \(S'_1\) and \(S'_2\) are congruent if and only if there is an isometry, \(f: \mathbb{R}^2 \rightarrow \mathbb{R}^2\), such that \(f (S'_1) = S'_2\).
Special-Student-7-Rebutter
Are mirror images congruent by the definition?
Special-Student-7-Hypothesizer
Yes: the lengths of any corresponding sides of any 2 mirror triangles are the same.
Note that "preserves the distance between any 2 points" means that with the distance between 2 points, \(p_1, p_2\), denoted as \(dist (p_1, p_2)\) and the image of \(p_i\) denoted as \(f (p_i)\), \(dist (f (p_1), f (p_2)) = dist (p_1, p_2)\).
Special-Student-7-Rebutter
Is any angle automatically preserved?
Special-Student-7-Hypothesizer
Yes. Although the definition of isometry talks only about lengths, any triangle, \(ABC\), is mapped to \(A'B'C'\), and the angle, \(\angle ABC\), equals the angle, \(\angle A'B'C'\), because the side length, \(CA\), equals the side length, \(C'A'\), in addition that \(AB = A'B'\) and \(BC = B'C'\).
5: Isometry Is Really Any Combination of Translation, Rotation, and Mirroring
Special-Student-7-Hypothesizer
Any isometry is really any combination of translation, rotation, and mirroring.
Special-Student-7-Rebutter
Well, is that obvious?
Special-Student-7-Hypothesizer
Although we do not show any rigorous proof, it is intuitively obvious.
Any translation preserves distances.
Any rotation preserves distances.
Any mirroring preserves distances.
And any isometry has to be a combination of translation, rotation, and mirroring, because when a triangle, \(ABC\), is mapped to \(A'B'C'\) by the isometry, \(A'B'C'\) is same-shaped-and-same-sized with or a mirror image of \(ABC\) (what else can it be indeed?), and \(A'B'C'\) has to have been mapped from \(ABC\) by a combination of translation, rotation, and mirroring.
Special-Student-7-Rebutter
We are supposing that if \(A'B'C'\) is same-shaped-and-same-sized, it has to be a combination of translation and rotation, and if \(A'B'C'\) is a mirror image, it has to be a combination of translation, rotation, and mirroring.
Special-Student-7-Hypothesizer
Although we have not proved the supposition rigorously, it seems intuitively obvious.
6: But Anyway, Why Is Isometry Used in the Definition of 'Congruence'?
Special-Student-7-Rebutter
I understand the definition of 'congruence' with isometry, but why "isometry"?
Special-Student-7-Hypothesizer
That is the point on which we have begun this article. Is a definition, "2 figures are congruent iff 1 of them is the image of the other under a combination of translation and rotation.", not more natural?
Someone may say that it is just that a mathematician (in fact, Euclid, I guess) just adopted the definition with isometry, but I think that that mentality is not good: mathematics is not any discipline in which students have to blindly accept an unreasonable definition just because an authority made the definition; we can and should ask whether there are good reasons to adopt the definition.
Special-Student-7-Rebutter
What are the good reasons for the definition of 'congruence' with isometry?
Special-Student-7-Hypothesizer
Some typical concerns in the plane geometry is that the length of a line segment equals the length of another line segment and an angle equals another angle, and isometry is sufficient for addressing such concerns.
We are typically concerned with whether an angle \(\angle ABC\) equals another angle \(\angle A'B'C'\), and it does not really matter whether the triangle, \(A'B'C'\), is same-shaped-and-same-sized with or a mirror image of the triangle, \(ABC\).
Special-Student-7-Rebutter
Although there may be some cases in which it matters, but it is more economical to distinguish between being same-shaped-and-same-sized and being a mirror image only in such rare cases.
Special-Student-7-Hypothesizer
That is the good reason why we have the concept that is defined with isometry.
Special-Student-7-Rebutter
But there is the issue that whether the word, "congruent", is appropriate to represent the concept.
Special-Student-7-Hypothesizer
As is discussed above, supposing that the general meaning of 'congruent' is being same-shaped-and-same-sized, the mathematical "congruent" is against the general meaning or along the general meaning only by introducing the ambient space. Either way, I feel that that use of "congruent" somehow misleading.
Special-Student-7-Rebutter
To summarize, 2 figures are said to be congruent iff 1 of them is the image of the other under an isometry, and the reason why isometry is used there is that isometry is along our typical concerns of whether 2 line segment lengths are equal and whether 2 angles are equal. Whether "congruent" is a desirable word is an issue, but an mirror image can be regarded to be same-shaped-and-same-sized by introducing the ambient space, although whether introducing such an extra space is wise is another issue.
7: Mirroring on a Plane or in a 3-Dimensional Space as a Rotation in the Ambient Space
Special-Student-7-Hypothesizer
Although we do not really need the ambient space in order to define 'congruence', "One of the 2 figures can be moved in the ambient space to coincide with the other figure." seems to be a view of congruence, and let us see how any mirroring is a rotation in the ambient space.
A mirroring on a plane will be easy to imagine as a rotation in the ambient 3-dimensional space.
Let us suppose that there is a triangle, \(ABC\), on the \(0 \lt x\) half of the \(x-y\) plane, and \(ABC\) is mirrored with respect to the \(y\) axis, with the mirror image, \(A'B'C'\) on the \(x \lt 0\) half of the plane.
As one will be able to imagine, when \(ABC\) is rotated \(\pi\) around the \(y\)-axis in the ambient 3-dimensional \(x-y-z\) space, \(ABC\) coincides with \(A'B'C'\).
More specifically, when any point, \((x, y, z)\), is rotated \(\theta\) around the \(y\) axis, the image, \(x', y', z'\), is \(\begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} cos \theta & 0 & - sin \theta \\ 0 & 1 & 0 \\ sin \theta & 0 & cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}\), by a matrix notation. If someone is not familiar with matrix notations, what that means is nothing but \(x' = cos \theta x - sin \theta z, y' = y, z' = sin \theta x + cos \theta z\).
Special-Student-7-Rebutter
\(y\) is not changed as we are rotating the point around the \(y\) axis.
Special-Student-7-Hypothesizer
When \(z = 0\), which means that the point is on the \(x-y\) plane, and \(\theta = \pi\), \(x' = - x, y' = y, z' = 0\), which is indeed the mirroring.
Let us see a mirroring in a 3-dimensional space as a rotation in the ambient 4-dimensional space.
Let us suppose that there is a tetrahedron, \(ABCD\), in the \(0 \lt x\) half of the \(x-y-z\) space, and \(ABCD\) is mirrored with respect to the \(y-z\) plane, with the mirror image, \(A'B'C'D'\) in the \(x \lt 0\) half of the space.
Special-Student-7-Rebutter
While the mirroring on the plane was done with respect to a line, the mirroring in the 3-dimensional space is done with respect to a plane.
Special-Student-7-Hypothesizer
Yes. In fact, a mirror you use to look at your face is a plane, not a line, right?
Special-Student-7-Rebutter
I have not used any line mirror before.
Special-Student-7-Hypothesizer
More generally, a mirroring in a \(\mathbb{R}^n\) space is done with respect to a \(n - 1\)-dimensional hyperplane. In fact, the line on the plane is a \(1\)-dimensional hyperplane.
When \(ABCD\) is rotated \(\pi\) around the \(y-z\) plane in the ambient 4-dimensional \(x-y-z-w\) space, \(ABCD\) coincides with \(A'B'C'D'\).
Special-Student-7-Rebutter
How can I imagine "rotated \(\pi\) around the \(y-z\) plane"?
Special-Student-7-Hypothesizer
Let us think in this way: rotations around the \(y\)-axis in the \(x-y-z-w\) space are 3-dimensional rotations in the \(x-z-w\) space.
Special-Student-7-Rebutter
Ah, as only the \(y\)-axis is fixed, it is free in the \(x-z-w\) space.
Special-Student-7-Hypothesizer
Let us restrict the 3-dimensional rotations in the \(x-z-w\) space to only the rotations around the \(z\)-axis, which is what we are talking about.
Special-Student-7-Rebutter
So, they are the rotations with the \(y\)-axis and the \(z\)-axis fixed.
Special-Student-7-Hypothesizer
And such any rotation can be parameterized by a single angle, \(\theta\), which is the angle of the 3-dimensional rotation in the \(x-z-w\) space around the \(z\)-axis.
When any point, \((x, y, z, w)\), is rotated \(\theta\) around that way, the image, \(x', y', z', w'\), is \(\begin{pmatrix} x' \\ y' \\ z' \\ w' \end{pmatrix} = \begin{pmatrix} cos \theta & 0 & 0 & - sin \theta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ sin \theta & 0 & 0 & cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\), by a matrix notation. If someone is not familiar with matrix notations, what that means is nothing but \(x' = cos \theta x - sin \theta w, y' = y, z' = z, w' = sin \theta x + cos \theta w\).
Special-Student-7-Rebutter
It has to be that because it is the 3-dimensional rotation in the \(x-z-w\) space around the \(z\)-axis while the \(y\)-axis is already fixed.
\(y\) and \(z\) are not changed of course.
Special-Student-7-Hypothesizer
When \(w = 0\), which means that the point is in the \(x-y-z\) space, and \(\theta = \pi\), \(x' = - x, y' = y, z' = z, w' = 0\), which is indeed the mirroring.
Special-Student-7-Rebutter
So, while most people probably cannot easily imagine rotations in the ambient 4-dimensional space, any mirroring in any 3-dimensional space is indeed a rotation in the ambient 4-dimensional space.
Special-Student-7-Hypothesizer
A way to imagine the rotation in the 4-dimensional space is to think of the projections of the rotation into the \(x-y-z, x-y-w, x-z-w, y-z-w\) spaces.
For example, the projection of \((x, y, z, w)\) into the \(x-y-w\) space is \((x, y, w)\), and \(B: (1, 0, 0, 0)\) and \(D: (1, 0, 1, 0)\) map to the same \((1, 0, 0)\); the projection of \((x, y, z, w)\) into the \(y-z-w\) space is \((y, z, w)\), and \(B: (1, 0, 0, 0), C: (4, 0, 0, 0), B': (-1, 0, 0, 0), C: (-4, 0, 0, 0)\) map to the same \((0, 0, 0)\).
Special-Student-7-Rebutter
We have seen the projections, so, what?
Special-Student-7-Hypothesizer
Well, an important thing is that when you look at a 3-dimensional space, you are aware that the space is not the whole space but a projection, which means that each point is not really a point but a set of points with 1 parameter.
Special-Student-7-Rebutter
So, when I extend my finger to touch a point in the projection, I have to be aware that I may not really touch the point in the 4-dimensional space, because my finger is touching the \(w = 0\) one in the set of points while the point is at \(w = 2\) for example.
Special-Student-7-Hypothesizer
"Should be touching the point but not really touching the point" may seem supernatural but is not: the point is just at another \(w\).
Special-Student-7-Rebutter
I see, ..., but so what?
Special-Student-7-Hypothesizer
Well, beyond that, it is a matter of your imagination; in an analogue, you cannot really see any 3-dimensional object, either: you see 2-dimensional projections and your brain imagines the shape of the 3-dimensional object.
Special-Student-7-Rebutter
So, it should be the same with a 4-dimensional object, you mean.
Special-Student-7-Hypothesizer
I am guessing that it is a matter of practices that you begin to imagine the 4-dimensional object.
