Reinventing mathematics for schools

I found this today. It’s worth a read.

Author: Gary S. Stager


Here’s an extract:

Conrad Wolfram estimates that 20,000 student lifetimes are wasted each year by school children engaged in mechanical (pencil and worksheet) calculations.
Expressed another way, we are spending twelve years educating kids to be a poor facsimile of a $2 calculator.
Forty years after the advent of cheap portable calculators, we are still debating whether children should be allowed to use one.

We are allowing education policy and curriculum to be shaped by the mathematical superstitions of Trump voters.
Educators need to take mathematics back and let Pearson keep “math.”

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Filed under education, math

Saving American Higher Education, but how?

I found the following article ” The Coddling of the American Mind” while rooting about, after finding what I was looking for.

Read it, or at least some of it, as it is quite long. The fear of being upset is screwing the college sector, and my guess is that the UK won’t be far behind. “You can’t say that. We shouldn’t be reading that. Someone will be traumatised”, and so on. All must be protected. It is the death of critical thinking.


Filed under college, critical thinking, education

Political correctness in K-12 math

This is definitely worth a read.

Here is a quote:

“High schools focus on elementary applications of advanced mathematics whereas most people really make more use of sophisticated applications of elementary mathematics. This accounts for much of the disconnect noted above, as well as the common complaint from employers that graduates don’t know any math. Many who master high school mathematics cannot think clearly about percentages or ratios.”

And here is the link:

Link found on f(t)’s blog

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Filed under algebra, abstract, teaching, education, math, CCSS, Common Core, K-12, standards, tests, depth

Bloggy young nerd women


Read the linked post from Michelle G. ” Picture yourself as a stereotypical male, about the question of intelligence, gender, and stereotype threat.”
The psychological investigations described were a real eye-opener to me. It is particularly relevant to MATH

Originally posted on mathbabe:

Today I want to share with you all two recent blog posts from nerdy young women. And who doesn’t love nerdy young women!?

The first is by Michelle G, an M.I.T. student, class of 2018, who is majoring in “14,” which is M.I.T. code for economics. She wrote a blogpost called Picture yourself as a stereotypical male, about the question of intelligence, gender, and stereotype threat. I thought I knew all about that stuff but I learned quite a bit from her post. p.s. Larry Summers, I hope you read this.

The second is by Meena Boppana, a Harvard math major who has guest blogged here on mathbabe before. This post is called The Making of A Girl Mathlete, and it describes her experience winning math competitions, often as the only girl. Even though I personally think math contests kind of suck, I appreciate how much Meena…

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Halving a triangle, follow-up number three, bisecting an angle

While looking at the bisection of area formula (see previous posts)
a’b’ = (1/2)ab
where a’ and b’ are the distances from the vertex of points on the sides of the triangle, and a and b are the lengths of the sides I remembered another formula about triangles, the bisection of the angle formula, with a” and b” being the lengths of the two parts into which the opposite side is divided, namely
a”/b” = a/b

These are like Cuba and Puerto Rico, “Two wings of the same bird”, Jose Marti (in Spanish)
Neither involves the angle itself, and so is very general. I decided that there must be a connection, and after a futile look for some duality in the situation I suddenly saw the connection, in simple algebraic terms:
a’b’ = a’/(1/b’)
and so a triangle with sides a’ and 1/b’ will give a”/b” = a’b’
and then a”/b” = (1/2)ab as well.

This is the construction for the halving a triangle.
half triangle and bisector0

This is the extended construction for the bisector. The opposite side is in brown.
half triangle and bisector1
and this is a gif showing how as the point a’ (E) is moved the brown line (OE, opposite side) moves parallel to itself, thus preserving the value of the ratio a”/b”

gif halving bisecting

And in case you got this far, some light relief. Bullfrog eats dog food, this morning. The dish is 10 inches across.
bullfrog dogdish

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Filed under algebra, bisecting, geometry, halving

Not more CCSSM horrors, just some glaring omissions.

Glaring omissions to me, that is.

The obsession with Al Gebra and manipulations has used up loads of time which could have been spent on

1. Parameters.
The sudden appearance of the word “parameter” in High School :
“Interpret expressions for functions in terms of the situation they model. 5. Interpret the parameters in a linear or exponential function in terms of a context.”
The idea of a parameter is basic to the study of functions and relationships. At the start the equation y = mx + b has four letters in it. x and y are variables. What on earth are m and b? Numbers? Fixed numbers? Variable numbers, but not as variable as variables? No, they are parameters for the line. For a given line they are fixed, but for different lines one or both are different.
(When I was at school we, that is the kids, used to call them “variable constants”)

2. Parametric representation of curves and relationships.
For example a circle. With parameter θ a point (x,y) on the unit circle is described by x = cos(θ), y = sin(θ)
and a parabola, parameter a, point on curve given by x = a, y = a2
and for a lot of curves the only neat way.
It also allows for ease in programming graphics of curves.

3. Polar coordinates. The ONLY mention of the word “polar” is with regard to representation of complex numbers. With no way of simple plotting them ?????
How about the function representation of a circle as r = 2 ??

There are others!

It was admitted at the time of development of the CCSSM that too much time was spent on K-8, and HS math was a rough job – so why can it not be modified ???????


Filed under CCSSM, Common Core, high school, omissions

Halving a triangle, follow-up number two, pursuing the hyperbola

Halving the triangle, any triangle, led to the equation XY = 2 as the condition on the points on two sides of the triangle, distant X and Y from the vertex.
The envelope of this set of lines turned out to be a hyperbola.
But XY = 2 defines a hyperbola – what is the connection ?

I took xy = 1 for the condition, on a standard xy grid, and wrote it as representing a function x —-> y, namely y = 1/x
The two points of interest are then (x,0) on the x-axis and (0,1/x) on the y-axis.
We need the equation of the line joining these two points, so first of all we have to see that our x, above, is telling us which line we are talking about, and so it is a parameter for the line.
We had better give it a different name, say p.
Now we can find the equation of the line in x,y form, using (p,0) and (0,1/p) for the two points:
(y – 0)/(1/p – 0) = (x – p)/(0 – p)
which is easier to read as yp = -x/p + 1, and easier to process as yp2 = -x + p

Now comes the fun bit !
To find the envelope of a set of straight lines we have to find the points of intersection of adjacent lines (really? adjacent?). To do this we have to find the partial derivative (derivative treating almost everything as constant) of the line equation with respect to the parameter p. A later post will reveal all about this mystifying procedure).
So do it and get  2yp = 1

And then eliminate p from the two equations, the line one and the derived one:
From the derived equation we get p = 1/(2y), so substituting in the line equation gives 1 = 2xy
This is the equation of the envelope, and written in functional form it is
y = 1/(2x), or (1/2)(1/x)
Yes ! Another rectangular hyperbola, with the same asymptotes.
(write it as xy = 1/2 if you like)

Now I thought “What will this process do with y = x2 ?”
So off I go, and to cut a long story short I found the following:
For y = x2 the envelope was y = (-1/4)x2, also a multiple of the original, with factor -1/4
parabola by axa track point
parabola by axa track point and line

Some surprise at this point, so I did it for 1/x2 and for x3
Similar results: Same function, with different factors.
Try it yourself ! ! ! ! ! ! !

This was too much ! No stopping ! Must find the general case ! (y = xk)
Skipping the now familiar details (left to the reader, in time honoured fashion) I found the following:

Original equation: y = xk

Equation of envelope: y = xk multiplied by -(k-1)k-1/kk

which I did think was quite neat.
The next post will be the last follow-up to the triangle halving.

Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click

and to download the .doc instructions file basics.doc


Filed under algebra, calculus, construction, envelope, geometrical

Halving a triangle, follow-up number one, ellipse

The previous post is “Analytic (coordinate) geometry has its good points, but elegance is not one of them”, in which a formula for any line cutting a triangle in half was found. The envelope of the cutting line for one section (of three) of the triangle was found to be always a hyperbola, which got me thinking “How do I get an ellipse?”. Clearly not by cutting a triangle in half, which involved taking two points A’ and B’on adjacent sides of the triangle, and making the product of their distances from the point of intersection of the sides equal to half the product of the lengths of the two sides.

So we cannot take two distances along two lines from the same point, lets try two distances from separate points on two lines, and keep the product of the distances constant. Magic:
envelope ellipse cropped
The base line is AF. Line DL is set parallel to AF. B and G are the two points of interest, where AB and DG are the two distances, and the envelope of the line BG is an ellipse which touches AF and DL. The really interesting thing about this is that the lines do not have to be parallel, and that as the points A and F are placed nearer and nearer to the point of intersection the ellipse becomes more and more hyperbolic at the nearby end.

The next post will be a different follow-up to the triangle halving.

Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click

and to download the .doc instructions file basics.doc


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Filed under conic sections, construction, geometrical

More on √2 – Common Crappiness Simply Seen (CCSS)

It’s strange how one can read something many times and miss the complete stupidity of it, in math at any rate.
This is from the CCSSM Grade 8:
The Number System 8.NS (Grade 8)
Know that there are numbers that are not rational, and approximate
them by rational numbers.
2. Use rational approximations of irrational numbers to compare the size
of irrational numbers, locate them approximately on a number line
diagram, and estimate the value of expressions (e.g., π2). For example,
by truncating the decimal expansion of √2, show that √2 is between 1 and
2, then between 1.4 and 1.5, and explain how to continue on to get better

I need an approximation to √2. Just get me the decimal expansion, please. Oh, and I need it to 73 decimal places.

Do I have to explain to the authors of this garbage that the only way I am going to get anywhere with √2 is by a process of successive approximation, NOT THE OTHER WAY ROUND ! !

And just try doing this for pi.

“I know that there are irrational numbers”. “How do you know that?”. “Because my teacher told me”.

And where will I encounter π2 ? Or “estimate the value of pi-e”.

And when we get to High School we find:

Use properties of rational and irrational numbers.
3. Explain why the sum or product of two rational numbers is rational;
that the sum of a rational number and an irrational number is irrational;
and that the product of a nonzero rational number and an irrational
number is irrational.

I find real difficulties explaining the last point.

I am not proposing that we go as far as Cauchy Sequences or Dedekind cuts, but if they cannot do a better job than this the topic is best stopped at “√2 is irrational and here’s why”. How many students can prove that √2+√3 is irrational?


Filed under arithmetic, confusion, irrational numbers

Analytic (coordinate) geometry has its good points, but elegance is not one of them.

This all started with a post by Maya Quinn ( on problem solving. An oblong piece has been removed from an oblong cake (why? who did that?).The problem was to cut the remainder of the cake into two equal parts. Lots of solutions, one in particular was very imaginative.
This led me to another problem – what if the cut-out piece was triangular?

In view of one of the solutions to the first problem I decided that the triangle should be chopped in half. Not as simple as chopping a rectangle in half !
envelope half a cake

So, with Polya at my side (he’s been there since 1962) I decided that an equilateral triangle would be a reasonable starting point, as at least it was obvious that there were a few lines through the centroid doing the job (the medians), so some generality was still around.
Here is the equilateral triangle, nicely resting on the x axis, side length 2, top point on the y axis.
half equilateral 1

The vertical median bisects the triangle, so I described the general bisector GH by the distances x=DG and y=EH

In order to find out more about the bisector lines I first found a relationship between x and y, which was y=2x/(1+x), based on the area calculations:

Height of equilateral triangle is √3, base is 2, so area is √3

Height of BGH triangle is (2-y)/2 * √3, base is 1+x, so area is (1/2)*(1+x)*(2-y)/2*√3

and this is to be half of √3

So (1/2)*(1+x)*(2-y) must be equal to 1, and this leads to  y=2x/(1+x)
This allowed me to find the coordinates of the point H and locate it correctly on the line.

I then joined the points and by moving G the line moved, and I tracked it, shown in green below.

Small notational irritation: In the diagram below G is now C and H is now K
triangle area bisector detail
The visible curve is called the envelope of the lines, and it (obviously) touches the medians.

The complete envelope of the bisecting lines consists of two more sections, making a “concave”triangle with the centroid in the middle.

At this point I figured that a change of variable was in order, and looking at the y=2x/(1+x) equation, and at the diagram it looked like the distances of the two points from the left hand vertex would be helpful:

This produced X=1+x and Y=2-y, and to my surprise the resulting equation was Y=2/X, or XY=2.

The significance of this last equation escaped me at this point.

So I found the coordinates of the two points C and K in a coordinate system with origin at point A, in terms of the new variables X and Y, found the line joining them, rewrote in terms of a parameter P, used a bit of calculus to get the envelope (I’ll do a post on this later), and it had the second degree equation

√3/4 = y2 – √3y – √3xy + 3x

which is a hyperbola !!! (see C. Smith “Conic Sections”)

At this point I stopped thinking about halving a triangle, and looked at the full envelope for one of the three sections, and got this:
envelope hyperbola equilateral 2
A complete hyperbola, and, not only that, its asymptotes appear to be sides of the equilateral triangle. (Which when you think about it is not completely unreasonable !).

Then, thinking about doing shear operations on the picture, and with the X, Y variables, and ratios of lengths of segments on the same line being unchanged, I constructed the whole lot on an arbitrary triangle and did the envelope:
envelope hyperbola

The triangle is ADF and the halving line is BG. Yes ! Same result !

At which point I saw a bit of light, and thought that “XY=2” does not involve angles at all.

Ooops, there’s  another formula for the area of a triangle: a*b*sin(C)/2

Then straightaway all was revealed. You can draw the picture !
1. Triangle ACB, point X on side AC, point Y on side BC, area=0.5*AC*BC*sin(ACB)
2. Triangle XCY, area 0.5*XC*YC*sin(XCY), but angles ACB and XCY are the same, so if we require the area of triangle XCY to be half the area of triangle ACB we get 0.5*XC*YC*sin(XCY)=0.5*0.5*AC*BC*sin(ACB)

which reduces to (XC/AC)*(YC/BC)=0.5, and this does not involve the angle.

This corresponds to our earlier XY=2 since the equilateral triangle had side length 2

Pure speculation suggests that this result may have some connection with the way that the angle bisector of an angle in a triangle divides the opposite side in the ratio of the two adjacent sides.
The geometric diagrams were all constructed with the web based program


Filed under envelope, geometry, math, triangle