Category Archives: algebra

Complex Numbers via Rigid Motions

https://howardat58.files.wordpress.com/2015/02/complex-numbers-by-rotations.doc

Complex numbers via rigid motions
Just a bit mathematical !

I wrote this in response to a post by Michael Pershan:
http://rationalexpressions.blogspot.com/2015/02/could-this-introduce-kids-to-complex.html?

The way I have presented it is showing how mathematicians think. Get an idea, try it out, if it appears to work then attempt to produce a logical and mathematically sound derivation.
(This last part I have not included)
The idea is that wherever you have operations on things, and one operation can be followed by another of the same type, then you can consider the combinations of the operations separately from the things being operated on. The result is a new type of algebra, in this case the algebra of rotations.
Read on . . .

Rotations around the origin.
angle 180 deg or pi
Y = -y, and X = -x —> coordinate transformation
so (1,0) goes to (-1,0) and (-1,0) goes to (1,0)
Let us call this transformation H (for a half turn)

angle 90 deg or pi/2
Y = x, and X = -y
so (1,0) goes to (0,1) and (-1,0) goes to (0,-1)
and (0,1) goes to (-1,0) and (0,-1) goes to (1,0)
Let us call this transformation Q (for a quarter turn)

Then H(x,y) = (-x,-y)
and Q(x,y) = (-y,x)

Applying H twice we have H(H(x,y)) = (x,y) and if we are bold we can write HH(x,y) = (x,y)
and then HH = I, where I is the identity or do nothing transformation.
In the same way we find QQ = H

Now I is like multiplying the coodinates by 1
and H is like multiplying the coordinates by -1
This is not too outrageous, as a dilation can be seen as a multiplication of the coordinates by a number <> 1

So, continuing into uncharted territory,
we have H squared = 1 (fits with (-1)*(-1) = 1
and Q squared = -1 (fits with QQ = H, at least)

So what is Q ?
Let us suppose that it is some sort of a number, definitely not a normal one,
and let its value be called k.
What we can be fairly sure of is that k does not multiply each of the coordinates.
This appears to be meaningful only for the normal numbers.

Now the “number” k describes a rotation of 90, so we would expect that the square root of k to describe a rotation of 45

At this point it helps if the reader is familiar with extending the rational numbers by the introduction of the square root of 2 (a surd, although this jargon seems to have disappeared).

Let us assume that sqrt(k) is a simple combination of a normal number and a multiple of k:
sqrt(k) = a + bk
Then k = sqr(a) + sqr(b)*sqr(k) + 2abk, and sqr(k) = -1
which gives k = sqr(a)-sqr(b) + 2abk and then (2ab-1)k = sqr(a) – sqr(b)

From this, since k is not a normal number, 2ab = 1 and sqr(a) = sqr(b)
which gives a = b and then a = b = 1/root(2)

Now we have a “number” representing a 45 degree rotation. namely
(1/root(2)*(1 + k)

If we plot this and the other rotation numbers as points on a coordinate axis grid with ordinary numbers horizontally and k numbers vertically we see that all the points are on the unit circle, at positions corresponding to the rotation angles they describe.

OMG there must be something in this ! ! !

The continuation is left to the reader (as in some Victorian novels)

ps. root() and sqrt() are square root functions, and sqr() is the squaring function .

pps. Diagrams may be drawn at your leisure !

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Filed under abstract, algebra, education, geometry, operations, teaching

Another Common Core Math Horror

I thought I had found them all, but NO.

Subtraction. Read this
————-
Kindergarten
Operations and Algebraic Thinking
• Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from.
————-
What has subtraction got to do with taking apart ???
(The examples are all of the form 9 = 3 + 6 and so on).

Also there is NO mention at all of subtraction as a way of finding the difference between two numbers, or of finding the larger of two numbers (anywhere).

While I am in critical mode I offer two more, less awful, horrors from Grade 1:

“To add 2 + 6 + 4,…”  and  “For example, subtract 10 – 8″.

The poor symbols are clearly in great pain at this point. Just read aloud exactly what is written…..

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The Distributive Law, again !

The formal statement of the distributive law should read as follows:

If a, b, c and d are numbers, or algebraic expressions (same thing really) and b = c + d then ab = ac + ad

It is a by-product of the law that it tells you how to expand an expression with a bracketed factor.

In any case, what’s the big deal ?

gif distributive law

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The angle bisector theorem and a locus

The theorem states that if we bisect an angle of a triangle then the two parts of the side cut by the bisector are in the same ratio as the two other sides of the triangle. So I thought, what if we move the point, G in the gif below, so that GB always bisects the angle AGC, then where does it go ? Or, more mathematically, what is its locus. Geometry failed me at this point so I did a coordinate geometry version of the ratios and found that the locus was a circle ( a little surprised !). The algebra is fairly simple.

gif bisector locus

What was a real surprise was that if we take the origin of measurements to be the other point where the circle cross the line ABC, then point C is at the harmonic mean of points A and B. There is YOUR problem from this post.

Getting the construction to show all this was tricky. It is shown below.

locus angle bisector point full

I did the construction on my geometry program, Geostruct (see my Software page).

Line bb passes through A. Point D is on line bb. Point E is on this line and also on the line through C parallel to line cc joining B and D, So AD and DE are in the same ratio as AC and CB. When point D is moved the ratios are unchanged (feature of the program). So the point of interest is the intersection of a circle centre A, radius AD, and a circle centre C and radius DE (= FC)

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Completing the (four sided) square

completing the square

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Radius of the inscribed circle of a right angled triangle

incircle radius with text

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Pythagoras, triples, 3,4,5, a calculator.

How to generate Pythagorean triples (example: 3,4,5), well one way at least.

Starting with (x + y)2 = x2 + y2 + 2xy and (x – y)2 = x2 + y2 – 2xy

we can write the difference of two squares

(x + y)2  –  (x – y)2 = 4xy

and if we write  x = A2 and y = B2 the right hand side is a square as well.

Thus:

(A2  +  B2) 2 – (A2 – B2) 2 = 4A2 B2 = (2AB) 2

which can be written as

(A2  +  B2) 2 = (A2 – B2) 2 + (2AB) 2

the Pythagoras form.

Now just put in some integers for A and B

2 and 1 gives 3,4,5

Conjecture1: This process generates ALL the Pythagorean triples.

Conjecture2: Every odd number belongs to some  Pythagorean triple.

Have fun…….

My next post will be about finding the radius of the inscribed circle in a right angled triangle…..

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“Observe and make use of structure”. Observe would be a start. A tale from the chalkface.

Here’s my little story:
It was a class of day release students on a Higher National Certificate course in engineering. I reached a point in one class with a relationship between p and q, p = kq, with k a constant. “What’s its graph look like”, I asked. Deathly silence. “Ok, let’s try x and y”. Result y = kx. Same question, same response. “Well, what about y = 3x ?”. Same question, same response. So I wrote y = 3x + 2. Their eyes lit up, and they unanimously shouted “It’s a straight line!”.

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CCSS SMP 7 Look for and make use of structure. Sums of powers of the natural numbers

The following is unreadable. Use the browser zoom or click the picture, or download the .doc file from https://howardat58.files.wordpress.com/2014/10/sums-of-powers.doc
sums of powers pic version

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Subtraction in algebra – let’s use algebra !

I have seen some heavy handed ways of explaining the identity

a – (b + c) = a – b – c

Let us use algebra. Give the left hand side a name, say d .Then

a – (b + c) = d

This is an equation, so add (b+c) to each side and get

a = d + (b + c), then a = d + b + c as the parentheses are now superfluous.

Now subtract  b  from each side

a – b = d + c

Now subtract  c  from each side

a – b – c = d

so  a – (b + c) =  a – b – c

or is this too simple ? Look, no messing with p – q = p + -(q) stuff,

and no appeal to the famous distributive law.

You can do this, and other stuff, with numbers as well.

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