Let's discuss Mathematics

[AdrienIer]

If at then end of the year half of my students were capable of doing half of what is in the official schoolbook I'd be super happy.

 

[Kyriakos]

If at then end of the year half of my students were capable of doing half of what is in the official schoolbook I'd be super happy.

I suspected that. Obviously it's like this everywhere.
Yes, I see your (and CKS's) point - given the entrance exam isn't for a math-focus anyway, I can accept that my characterization of the test was hyperbolic.
Still, would it kill them (those who wrote the test) to have even one "not as easy" problem? ^^ After all, by definition they are testing for the better students.

 

[Harv]

So if there is not broad awareness of these issues, then the person designing these tests might just believe the student is just careless. It is possible I have such an issue, but I did not ask my psychologist. Now I have a specific Greek word to ask about and a description of a specific issue. (Dysanagnosia)

I will put this into a spoiler, because it is somewhat off topic and could go in a separate thread.

Spoiler :

The issue would show up about as often as a 1 on a 20 sided die. So in your case, you correctly answered 24/25 questions and rolled that critical miss 1.

Where it shows up is mistakes tend to be of the "stupid" or "careless" kind. However, this is good enough to navigate the high school and university system. It shows up in work as trivial, but annoying mistakes. As your definition states, something trivial and menial compared to the work involved to get it.

I believe the workaround is the four-eyes principle. Just get somebody to check everything. It is more difficult when we have people in positions of power who believe these positions are not conditions or blind spots, but character deficits.

To counter that, we can go back to that game I was referencing, which was Axis and Allies Online. The user interface is designed to trap people with the issue you are describing. So the big task is to devise a strategy to destroy your position in my turn. The turn is split into phases as follows:
0. Plan the entire turn.
1. Buy my units.
2. Make my Combat Moves.
3. Resolve my Combat.
4. Make my Non-Combat Moves.
5. Place my Production Units.
6. Collect my Income.

My plan might to build an Aircraft Carrier, fly my planes out to combat, then return to the sea zone where we are building the Aircraft Carrier and land on it when complete. (This is really how the turn works.) However, in my Non-Combat Move, I forgot to land my planes in the sea zone and now I have an Aircraft Carrier with no Aircraft and the result is RIP Aircraft Carrier when it is your turn. This is entirely a UI issue, because you tell the system your Non-Combat Move is complete. It would cause zero disruption to the game flow if I discover my mistake, go back to non-combat move and fix it. They could combine phases 4 and 5 with no disruption to the game flow.

In a live game, the rule book says nothing about when phase 4 ends and phase 5 begins. It does not say when you touch that Aircraft Carrier, you have to place it. Most likely, I would have all my built units in my hand, look at the Aircraft Carrier, then decide phase 4 has not ended and just move the fighters back to where they came from, then move them to the intended sea zone, then place the Aircraft Carrier and land the planes.

To counter this, I have to write out exactly what I do, then go through the checklist to make sure it was done.

We might go so far as to change brains to check what I have done. In work, I have to come in a different day and check what I have prepared. On CFC, I have to go check what I wrote the day before - and discover I forgot to divide by 6 and as a result posted something ridiculous.

 

[Kyriakos]

I want to ask, without using the Pythagorean theorem (or anything more modern than it), is this a rigorous proof (or is it anywhere exposed in the second part?) that the perpendicular to a line from an external to it point A, is the shortest distance between that line and the point?
Also, can you think of a less verbose proof? (again without Pythagoras or more modern math; think of yourself as Thales )

eg @AdrienIer

(obv if one uses the P.T., it's just a matter of (say) AΓ^2-ΑΒ^2=ΒΓ^2 |AB unequal to AΓ, which leads to a negative equaling a positive)

 

[AdrienIer]

Lol literally proved it to my students last week (using Pythagoras of course).
I'm usually not great at doing maths with my hands tied behind my back (ie not using basic mathematical tools). Also some of what you posted is unclear : is epsilon the horizontal or vertical line ? Also I think some of the way you right stuff isn't using a convention I'm used to, are you using "< ABC" to mean the angle ABC ? If so I don't understand why the last >90 in the line before last isn't a <90. Unless you're going with a hidden cosine formula (which should be forbidden here) for a link between the lengths of AR and AD (D = delta here), and the angle ARD ?

 

[a pen-dragon]

This is more of a geometrical approach, and not rigorously written, but basically to determine the shortest path between a point and a line (i.e. their distance) draw circles with different radii around the point, increasing the radius if there is no intersection and decreasing it if there are two. When you obtain one intersection, you will have found the shortest distance (= the radius of that circle).

Since now we have a line that is tangential to a circle (ONE intersection), it will be perpendicular to the radial from the center of the circle to the intersection point.

 

[Kyriakos]

Lol literally proved it to my students last week (using Pythagoras of course).
I'm usually not great at doing maths with my hands tied behind my back (ie not using basic mathematical tools). Also some of what you posted is unclear : is epsilon the horizontal or vertical line ? Also I think some of the way you right stuff isn't using a convention I'm used to, are you using "< ABC" to mean the angle ABC ? If so I don't understand why the last >90 in the line before last isn't a <90. Unless you're going with a hidden cosine formula (which should be forbidden here) for a link between the lengths of AR and AD (D = delta here), and the angle ARD ?

I will type the answer and change Greek to Latin so as to be clear

This is more of a geometrical approach, and not rigorously written, but basically to determine the shortest path between a point and a line (i.e. their distance) draw circles with different radii around the point, increasing the radius if there is no intersection and decreasing it if there are two. When you obtain one intersection, you will have found the shortest distance (= the radius of that circle).

Since now we have a line that is tangential to a circle (ONE intersection), it will be perpendicular to the radial from the center of the circle to the intersection point.

A question on that: (sorry for a great multitude of edits)
Assume that what you typed can be formalized using pre-Pythagoras math. Then where is the part that shows that proving this you have more than the claim "the shortest distance from external to line point, to that line, can be constructed as a radius of circle with that point as center"? Which still is not a proof "that the shortest such distance is perpendicular to the line".
As in your proposed approach, we can't take as a given that the segment is perpendicular. On the other hand, while we can prove that such a radius is perpendicular to the tangent line (also a theorem of Thales), there we are based on an intermediate theorem that shows the perpendicular to be the shortest distance (ie what our current problem with pre-Pythagoras math is trying to reconstruct).

Also, is my own proof (which may be similar to what Thales did, I don't know) correct and rigorous in your view, and if not, can you help make it that?



PS: that Thales did prove it (the tangent one, and also the perpendicular being smaller), is well-known. I am just not sure what the progression of theorems there was and am trying to recreate it (but it has to be very known, just not by myself!)

There is, of course, also the approach of using Thales' theorem of analogy of sides in similar triangles, to conclude that the triangles AD(and point in z) and ABC are similar, therefore a segment of AC is in analogy with AD and thus the entire AC (which is in analogy to AB) can't equal AD. I will try to establish if this is a valid progression (no circular moments), but it seems that it is.

 

[a pen-dragon]

So what you find not proven is that AB from your picture is perpendicular to the line.

The easiest proof I can image is via symmetry. The angles on both sides of AB to the line add up to 180° by definition. Since both the line and the circle are symmetric to the line through A and B, the angles also have to be. Thus both are 180°/2=90°, meaning AB is perpendicular to our line.

 

[Kyriakos]

So what you find not proven is that AB from your picture is perpendicular to the line.

The easiest proof I can image is via symmetry. The angles on both sides of AB to the line add up to 180° by definition. Since both the line and the circle are symmetric to the line through A and B, the angles also have to be. Thus both are 180°/2=90°, meaning AB is perpendicular to our line.

In your view, is it correct to state that both the theorem of the singularity of perpendicular from external point to line, to that line, and the theorem of the perpendicular being the shorter distance from such a point to such a line, are not in different stages of a progression but both flow from an axiom having to do with defining angle sum from angle from the center of a tied circle?
But if that is so, wouldn't it imply my suggested proof at post 1684 (and at 1687 without symbols and with latin letters) has to somehow be circular? And I don't currently see how it is (nor where it could be flat-out wrong).

Thanks for the replies. I will look into it myself anyway, as I should

 

[a pen-dragon]

In your view, is it correct to state that both the theorem of the singularity of perpendicular from external point to line, to that line, and the theorem of the perpendicular being the shorter distance from such a point to such a line, are not in different stages of a progression but both flow from an axiom having to do with defining angle sum from angle from the center of a tied circle?

No.

The connecting line being perpendicular (and unique) are a consequence of symmetry.

This connecting line being the shortest from the line to the point is a distinct statement, "proven" by the construction via circles.

I guess one could also conclude that said connecting line is the shortest by only using symmetry arguments, but even then it is a distinct statement, even if the proof is almost identical.

This means that proving one using the other is not circular, as there are no other statements linking both prior to the proof at hand being completed.

 

[Kyriakos]

No.

The connecting line being perpendicular (and unique) are a consequence of symmetry.

This connecting line being the shortest from the line to the point is a distinct statement, "proven" by the construction via circles.

I guess one could also conclude that said connecting line is the shortest by only using symmetry arguments, but even then it is a distinct statement, even if the proof is almost identical.

This means that proving one using the other is not circular, as there are no other statements linking both prior to the proof at hand being completed.

Good to know, ty. No interest in commenting on whether my proof is correct or not? Implied that at least the distinctness of the statements doesn't make it have to be wrong.

 

[a pen-dragon]

Your proof is definitely correct, that is not in question.

Edit: maybe not in your condition of pre-Pythagorean maths, but you probably know about that better than me.

 

[Kyriakos]

I want to ask if the following approach is complete (doesn't need additional parts), as a proof that any infinite geometric series where the first term (which is a1) is finite and the term multiplied is finite and smaller than 1, will converge.

The formula for sum of n parts of a geometric series can be established (so here it is just given) as a1(λ^n -1)/(λ-1). If λ<1, n infinite, the numerator becomes -1, the denominator stays as -1+λ. Both are finite if (as we set up) λ is finite, and we also set up that a1 is finite so finite(finite)=finite.

Eg 0.9repeat=0.9(0.1^n -1)/(0.1-1)=1.

 

[AdrienIer]

Yeah that's a normal way of putting it. Once you have the forumla you go to the limit and you're good.
As to 0.99999... the problem is that some people are vibe-math-ing it and claim that because it looks like it's less than 1 then it is strictly less than 1. Without ever explaining what number it is with a serious definition. But any serious math person would agree with you.

 

[Kyriakos]

A decade ago - before I bothered to reread secondary education math - I was among them So I can recall that my issue then was not identifying the number 0.9repeat as set (thinking it was ongoing). But any sum of parts by definition is set and there is nothing ongoing.
For infinite repeating decimal numbers, in particular, there is also the early middeschool approach of setting the number as x and then (depending on how many different decimals there are in the repeating group) having a multiple of x. Eg 0.9rep=x=>10x=9.9rep=>9x=9=>x=1.
Of course constructions (like these two for 0.9rep) are not insight. Insight is external to them.

 

[Kyriakos]

Is it fair to say that plane geometry relies on the most fundamental level (ie apart from the postulates, eg Euclid's) on definitions of rotation - and thus on central symmetry? For example, opposite angles are formed by the intersection of two lines, have a common vertex and their sides are opposite halflines with that vertex as sole common point, thus they have central symmetry with it as center. But given it is very common (and useful, of course) to present progressions of theories in geometry which can arrive at other approachable theorems through rotation in a lot more round-about (unintended pun) ways (eg not present either sum of triangle's angles as 2 right angles by using a parallel line to the base which has the external to the base vertex in it=>and not using this leads to the external angle of a triangle not being quickly identified as the sum of the two that aren't its supplementary).
Asking because all sorts of progressions of theorems do highlight properties which (while no doubt equally important as any other) obviously wouldn't be that sought after in specific problems if a different progression was used - and yet my sense currently is that central rotation is incomparably the fastest route and more importantly appears to be on a fundamental level ubiquitous even if only tacitly a part of the other routes.
So I am primarily asking because it irks me a bit that even in progressions which explicitly don't wish to refer to that approach to get a faster result (fewer steps in the current proof), parts of it are still there (like the opposite angles congruence).

To make more clear what I am talking about, I will give a very specific (and obv simple) example: you can prove that an external angle is larger than any of the two non-supplementary to it internal ones of the triangle, without explicitly using properties of rotation (180 degrees being the sum of internal angles isn't mentioned at all), but you would still be tacitly using them (for opposite angle congruence).
Or is there a way to provide geometric proofs for this - and similar - without at all using rotation?

It also irks me that in such approaches some elements are by and large taken for granted anyway (inevitably) due to visual presentation - for example that angle b1<angle b, which while evident, and also provable if one wishes to do it, still makes the observer perhaps too involved in the actual system. I am sure, of course, that there are non-geometric approaches which do not make use of what is visually evident geometrically and thus other properties become apparent much like they do when parts of the rotation properties are not taken for granted.

 

[Samson]

Simple question: The lowest common factor of any set of numbers is one, right?

This is in reference to this question, from a usually good source:

3. Find the lowest common factor of 88, 50, 65.

 

[Kyriakos]

I reread the question (numbers), and there aren't any other common factors anyway as 88 is not divisible by 5 and 65 isn't by 2.
So I assume the question seeks to have the pupil notice parts of the presented factorization approach in practice.

 

[Samson]

I reread the question (numbers), and there aren't any other common factors anyway as 88 is not divisible by 5 and 65 isn't by 2.
So I assume the question seeks to have the pupil notice parts of the presented factorization approach in practice.

The question as written would not require that. For any integer values of x and y, the lowest common factor is one so I would say that is the end of the question, there is no need to look at any factorisation. That is really the question though, if presented with such a question in the exam do you just write 1 and carry on?

 

[Kyriakos]

The question as written would not require that. For any integer values of x and y, the lowest common factor is one so I would say that is the end of the question, there is no need to look at any factorisation. That is really the question though, if presented with such a question in the exam do you just write 1 and carry on?

Well, yes? 1 is an issue only if you are asked to find common prime factors
Simply due to the definition excluding 1, so as to have every number be a unique product of primes to exponents.
If this program you linked to has already gone through that, then this question may just be examining if it was understood.