MATH 128AHomework 1 for MATH 128A.
MATH 128A: Homework 1 Due date: Thursday, April 9th
Instructions. You will submit your solutions on Gradescope, which you can access through our Canvas course page by clicking on the assignment.
You are welcome to submit your homework either handwritten or in LaTeX. If you choose to submit handwritten work, please use pen rather than pencil, and make sure that both your handwriting and your scans are as readable as possible in order to facilitate grading.
Please also feel free to attend office hours if you would like hints on the homework or if you would like to discuss more advanced material.
You will see that our Ximera course page proposes you to see the answers, but this will only become available once I publish the solution.
Answer. The correct answer is Corollary. The six parts discussed in class were:
Answer. The correct answer is that Euclid does not justify why the two circles intersect. His construction of an equilateral triangle on a given segment \(AB\) requires the two circles centered at \(A\) and \(B\) with radius \(AB\) to meet, but this is not guaranteed by the postulates stated so far.
Answer. The correct answer is Between. In class, primitive notions included objects such as points and lines, together with primitive relations such as betweenness.
Using GeoGebra, GeoGebra Classic, or pen and paper, perform the construction of Proposition I.3.
If you use pen and paper, include a drawing of your construction with your homework.
If you submit your homework in LaTeX, you may either include an image of your
construction (for example, as a .png file), or share your construction using the
sharing option provided by GeoGebra. In that case, your answer may simply be a link
to your construction.
Answer. https://www.geogebra.org/classic/mdyp9mnc
Remark. Observe how, when you move the initial points \(A\), \(B\), \(C\), and \(D\), the output (named \(f\) on my solution) moves with it. In general, when using GeoGebra, your inputs have the color blue. Everything else is something constructed from this initial data, and takes the color grey. If you construction is correct, your final construction point should move when you move your initial points. If it does not, please reach out to me as there is likely a gap in your understanding of GeoGebra.
Answer. We choose Proposition I.5.
Enunciation. In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.
Exposition. Let \(ABC\) be an isosceles triangle having the side \(AB\) equal to the side \(AC\), and let the straight lines \(BD\) and \(CE\) be produced further in a straight line with \(AB\) and \(AC\).
Specification. I say that the angle \(ABC\) equals the angle \(ACB\), and the angle \(CBD\) equals the angle \(BCE\).
Construction. Take an arbitrary point \(F\) on \(BD\). Cut off \(AG\) from \(AE\), the greater, equal to \(AF\), the less, by Proposition I.3, and join the straight lines \(FC\) and \(GB\).
Proof. Since \(AF\) equals \(AG\), and \(AB\) equals \(AC\), therefore the two sides \(FA\) and \(AC\) equal the two sides \(GA\) and \(AB\), respectively, and they contain a common angle, the angle \(FAG\).
Therefore the base \(FC\) equals the base \(GB\), the triangle \(AFC\) equals the triangle \(AGB\), and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle \(ACF\) equals the angle \(ABG\), and the angle \(AFC\) equals the angle \(AGB\), by Proposition I.4.
Since the whole \(AF\) equals the whole \(AG\), and in these \(AB\) equals \(AC\), therefore the remainder \(BF\) equals the remainder \(CG\), by Common Notion 3.
But \(FC\) was also proved equal to \(GB\), therefore the two sides \(BF\) and \(FC\) equal the two sides \(CG\) and \(GB\) respectively, and the angle \(BFC\) equals the angle \(CGB\), while the base \(BC\) is common to them. Therefore the triangle \(BFC\) also equals the triangle \(CGB\), and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Therefore the angle \(FBC\) equals the angle \(GCB\), and the angle \(BCF\) equals the angle \(CBG\), by Proposition I.4.
Accordingly, since the whole angle \(ABG\) was proved equal to the angle \(ACF\), and in these the angle \(CBG\) equals the angle \(BCF\), the remaining angle \(ABC\) equals the remaining angle \(ACB\), by Common Notion 3, and they are at the base of the triangle \(ABC\). But the angle \(FBC\) was also proved equal to the angle \(GCB\), and they are under the base.
Conclusion. Therefore in isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.
Comment on rigor. I do not see an actual logical gap in Euclid’s proof of Proposition I.5. Rather, the interesting point is that a modern reader may be tempted to derive the equality of the exterior angles from the equality of the interior base angles by subtracting from straight angles. However, Euclid cannot argue this way here, because the relevant result about adjacent angles on a straight line only appears later in Proposition I.13, and using it here would be circular. So the proof is not missing a step; instead, it is carefully written to avoid relying on facts that have not yet been established.
Answer. An axiomatic theory is a collection of primitive terms together with axioms describing how those terms are supposed to behave. A model for an axiomatic theory is a concrete interpretation of those primitive terms in which all the axioms are true.
For example, a theory might use the primitive words point and line and say that through any two distinct points there passes a line. A model would then specify what the points and lines actually are, and verify that this statement holds in that setting.
If so, describe your model clearly by explaining:
Answer. Yes.
Let the points be three distinct objects:
Let the lines be the sets
In other words, each line consists of exactly two of the three points.
Then Postulate 1 is satisfied, because given any two distinct points, there is a line containing them:
So this gives a model with three points and three lines satisfying Postulate 1.