"Let's look at the worlds which are created by the idea of dimensions. A mathematical world can exist on a single point, on a single line, on a plane, in space, in a hypercube (tesseract)."
This is the only page of the section "The Worlds of Dimensions." By the way, the name "tesseract" for the 4D hypercube appears in Madeleine L'Engle's Wrinkle in Time -- a movie which will be released in about three months. (L'Engle refers to the tesseract as 5D, because she's including time as Einstein's fourth dimension.)
Here are some excerpts from this Pappas page:
"Imagine your world and your life on a flat plane. Three dimensional creatures can invade your world without you even knowing by simply entering your domain from above or below. Dimensions beyond the third have always been intriguing. The cube was one of the first 3D objects to be introduced into the fourth dimension by becoming a hypercube. Computer programs have even been devised to derive glimpses of the fourth dimension by picturing 3D perspectives of the various facets of the hypercube."
There is only one picture on this page. It shows all the figures, or worlds, mentioned in the second sentence -- from the point to the hypercube. Its caption simply reads: "The four dimensions."
Of course, we read lots about dimensions in Szpiro's book about Poincare and Perelman. Notice that there wasn't much talk about hypercubes or tesseracts in that book. That's because a cube, in any dimension, is topologically equivalent to the sphere of the same dimension -- and it's the sphere that appears in the Poincare Conjecture. This sphere is the boundary of a 4D hypersphere, and so it's considered to be a 3D manifold.
Lesson 7-8 of the U of Chicago text is called "The SAS Inequality." Two years ago, I kept moving lessons around and in the end, I never covered Lesson 7-8. But don't worry -- it doesn't appear in the modern Third Edition of the U of Chicago text at all!
And so this is what I wrote three years ago about this lesson:
But what is in Section 7-8? This section covers the SAS Inequality -- often known as the Hinge Theorem, according to the Exploration Question at the end of the section. Dr. Franklin Mason also calls this theorem the Hinge Theorem.
I'm of two minds as to whether to include Lesson 7-8 on this blog. Notice that Dr. M includes the Hinge Theorem, Triangle Inequality, and all other inequalities related to triangles in his Chapter 5. But we've been waiting because many of these theorems depend on indirect proof, so we were waiting for Chapter 13. Yet on this blog, I posted the Circumcenter Concurrency Theorem even though it also requires an indirect proof (but, as Dr. Wu pointed out, the part that requires indirect proof can be handwaved over).
[2017 update: I wrote more about the significance of the SAS Inequality in my September 12th post. Also, three years ago I included a worksheet with an SSS Inequality -- that's what the "indirect proof" was for. In the end, indirect proof and the SSS Inequality have been dropped. Oh, what the heck -- let me reblog what I wrote about SSS Inequality in this post below, even though it's no longer on the worksheet.]
As I mentioned before, this section gives a proof of the SAS Inequality using the Triangle Inequality. As usual, I have decided to convert the proof to two-column format. The figure accompanying the proof gives two triangles, ABC and XYZ.
SAS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the measure of the included angle of the first triangle is less that the measure of the included angle of the second, then the third side of the first triangle is shorter than the third side of the second.
Given: AB = XY, BC = YZ, angle B < angle Y
Prove: AC < XZ.
Proof:
Statements Reasons
1. AB = XY, etc. 1. Given
2. exists isometry T s.t. A'B' is 2. Definition of congruent
XY, C' same side ofXY as Z
3. C'Y = ZY 3. Isometries preserve distance
4. Let m symmetry line C'YZ 4. Isosceles Triangle Symmetry Theorem
(m intersectsXZ at Q)
5. m perp. bis.C'Z 5. In isosceles triangle, angle bis. = perp. bis.
6. QC' = QZ 6. Perpendicular Bisector Theorem
7. A'C' < A'Q + QC' 7. Triangle Inequality
8. AC < XQ + QZ 8. Substitution
9. XQ + QZ = XZ 9. Betweenness Theorem (Segment Addition)
10. AC < XZ 10. Substitution
We see that the proof is similar to that of SAS Congruence Theorem, except that this isometry puts C' on the same side ofXY as Z, rather than the opposite side. Dr. M gives two proofs of the SAS Inequality (which he calls "the Hinge Theorem," a name mentioned in Exploration Question 19 in our text) -- his second proof is nearly identical to that given in the U of Chicago. In his first proof, Dr. M uses TASI (Unequal Angles Theorem) directly without invoking the Triangle Inequality -- but we would still be dependent on a theorem not proved until Chapter 13 in the U of Chicago.
Both Dr. M and Glencoe state a converse to SAS Inequality -- Glencoe calls it SSS Inequality. Dr. M hints at this proof -- we can use the same strategy that we used to derive Unequal Angles Theorem from its converse, Unequal Sides Theorem. We prove it indirectly:
SSS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the third side of the first triangle is shorter than the third side of the second, then the measure of the included angle of the first triangle is less that the measure of the included angle of the second.
Given: AB = XY, BC = YZ, AC < XZ.
Prove: angle B < angle Y
Indirect Proof:
Assume not. Then angle B is either less than or equal to angle Y.
Case 1: angle B = angle Y. Then triangles ABC and XYZ are congruent by SAS Congruence, and so AC = XZ, a contradiction.
Case 2: angle B > angle Y. Then AC > XZ by SAS Inequality, a contradiction.
In either case we have a contradiction of AC < XZ. Therefore angle B < angle Y. QED
For Euclid, the SAS Inequality is his Proposition 24. Dr. M's first proof is based on Euclid:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI24.html
and the converse, the SSS Inequality, is Euclid's Proposition 25:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI25.html
As I mentioned before, this section gives a proof of the SAS Inequality using the Triangle Inequality. As usual, I have decided to convert the proof to two-column format. The figure accompanying the proof gives two triangles, ABC and XYZ.
SAS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the measure of the included angle of the first triangle is less that the measure of the included angle of the second, then the third side of the first triangle is shorter than the third side of the second.
Given: AB = XY, BC = YZ, angle B < angle Y
Prove: AC < XZ.
Proof:
Statements Reasons
1. AB = XY, etc. 1. Given
2. exists isometry T s.t. A'B' is 2. Definition of congruent
XY, C' same side of
3. C'Y = ZY 3. Isometries preserve distance
4. Let m symmetry line C'YZ 4. Isosceles Triangle Symmetry Theorem
(m intersects
5. m perp. bis.
6. QC' = QZ 6. Perpendicular Bisector Theorem
7. A'C' < A'Q + QC' 7. Triangle Inequality
8. AC < XQ + QZ 8. Substitution
9. XQ + QZ = XZ 9. Betweenness Theorem (Segment Addition)
10. AC < XZ 10. Substitution
We see that the proof is similar to that of SAS Congruence Theorem, except that this isometry puts C' on the same side of
Both Dr. M and Glencoe state a converse to SAS Inequality -- Glencoe calls it SSS Inequality. Dr. M hints at this proof -- we can use the same strategy that we used to derive Unequal Angles Theorem from its converse, Unequal Sides Theorem. We prove it indirectly:
SSS Inequality Theorem:
If two sides of a triangle are congruent to two sides of a second triangle, and the third side of the first triangle is shorter than the third side of the second, then the measure of the included angle of the first triangle is less that the measure of the included angle of the second.
Given: AB = XY, BC = YZ, AC < XZ.
Prove: angle B < angle Y
Indirect Proof:
Assume not. Then angle B is either less than or equal to angle Y.
Case 1: angle B = angle Y. Then triangles ABC and XYZ are congruent by SAS Congruence, and so AC = XZ, a contradiction.
Case 2: angle B > angle Y. Then AC > XZ by SAS Inequality, a contradiction.
In either case we have a contradiction of AC < XZ. Therefore angle B < angle Y. QED
For Euclid, the SAS Inequality is his Proposition 24. Dr. M's first proof is based on Euclid:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI24.html
and the converse, the SSS Inequality, is Euclid's Proposition 25:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI25.html
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