I had 5 apples, and then my brother gave me 2 more apples.

I now have 7 apples.

The process is

**Without the pictures: **

*Starting state (5 apples)*

* Action (add 2 apples)(and bunch them up !)*

* Final state (7 apples)*

With a single sentence or equation, with the result

Start with 5, add 2, and resulting state is 7

Or the equation with apples

Start with 5 apples, add 2 more apples, and result is 7 apples

**Subtraction**

I have 7 apples now, and then my brother takes 3 of them away.

I end up with 4 apples .

The process is

**Without the pictures:**

*Starting state (7 apples)*

* Action (subtract 3 apples)*

* Final state (4 apples)*

With a single sentence or equation, and the result

Start with 7, subtract 3, and resulting state is 4

Or the equation with apples

Start with 7 apples, subtract 3 apples, and result is 4 apples

**Both of these equations are read from left to right**

**— – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – — – **

If I start with 1 and then add 1 I get 2. This doesn’t seem to be a problem at all.

If I start with 2 and then add 1 I get 3. This doesn’t seem to be too big a problem.

If I start with 5 and then add 3 I get 8. This starts to get interesting because starting with 2 and adding 5 has exactly the same final state.

It is all to do with counting.

If I count a pile of bricks and there are ten of them I can jumble the bricks around and count them again. I will still have ten of them. The count is “how many” and nothing else.

Consequently the rearrangement has no counting effect, and so any collection of numbers with plus signs and a total of ten can be rearranged, to give the same total of ten.

For example:

**start 2, add 3, add 4, add 1** is the same total as **start 2, add 4, add 1, add 3** (result ten)

We can also swap the start number and the add number with an example:

**start 3, add 5** is **start 4, add 4** and then **start 5, add 3**. Same total.

And finally, with **add** we can combine two adds to get **add 3, add 2 = add 5**

The apple example:

*Starting state (7 apples)*

* Action (subtract 3 apples)*

* Final state (4 apples)*

The starting state*, or start number, is 7*

The action is

and the final state is

**— – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – — – **

Have a look at the following:

**9 and −2 and −4**

The final state is ** 3, **but so is

Consequently the two subtractions can be swapped.

It’s even better than this: The adds and the subtracts can be put in any order, with the same final state.

Try

* start with 9, add 2, subtract 3, subtract 5, add 4, subtract 2* (final state 5)

or rearranged

**— – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – — – **

*Operations and Algebraic Thinking, CCSS grade 1 (italics)*

* Represent and solve problems involving addition and subtraction.*

* 1. ………*

* 2. ………Understand and apply properties of operations and the relationship*

* between addition and subtraction.*

* 3. Apply properties of operations as strategies to add and subtract.*

* Examples:*

* If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of*

* addition.) To add 2 + 6 + 4, the second two numbers can be added to make*

* a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)*

In mathematics there are numerous **algebraic** structures each consisting of a set of elements equipped with an **operation** that combines any two elements to form a third element. (Wikipedia, adapted, 19th century math)

So we can find an example of this where any two elements in the set of natural numbers can be combined with “+”, with the usual interpretation. (7 + 4 = 11).

More specifically we can use the “commutative property” of addition, a feature of the natural number system, to tell us that 7 + 4 equals 4 + 7 , for example.

**BUT** take a look at the above example, with different numbers:

**7 − 6 + 4 **

Swap **4** and **6**

**7 − 4****+ 6**

So we have to avoid the negative operator.

The conclusion is that the commutative rule applies only when **all** the operators are plus operators.

*Table 3. The properties of operations. Here a, b and c stand for arbitrary numbers in a given number system. The **properties of operations apply to the rational number system, the real number system, and the complex number **system. (CCSS)*

*Commutative property of addition is a+ b = b + a*

This is a definition. It states that “a complicated name” is what we know already, and can actually prove by counting. **So it is a complete waste of time for K-7 or further.**

Not only that, but the actions **add 3** and **subtract 4** are the original 17th and 18th century statements, and are **proper algebra**.

**I FOUND THIS ACCOUNT OF THE CHAIN RULE FROM SAM SHAH**

** https://samjshah.com/2018/02/08/there-might-be-light-at-the-end-of-the-chain-rule-tunnel-maybe/**

**The partial text is as follows:**

Find two points close to each other, like (x,g(f(x)) and (x+0.001,g(f(x+0.001)).

Find the slope between those two points: {g(f(x+0.001)-g(f(x))}/{(x+0.001)-x}.

There we go. An approximation for the derivative! (We can use limits to write the exact expression for the derivative if we want.)

But that doesn’t help us understand that {d/dx}[g(f(x)]=g'(f(x))f'(x) on any level. They seem disconnected!

But I’m on my way there. I’m following things in this way: x >>> f >>> g

Check out this thing I whipped up after school today. The diagram on top does x >>> f and the diagram on the bottom does f >>> g. The diagram on the right does both. It shows how two initial inputs (in this case, 3 and 3.001) change as they go through the functions f and g.

At the very bottom, you see the heart of this.

It has {δg}/{δf} times {δf}/{δx}={δg}/{δx}

(END OF SAM SHAH’S BIT)

**Proof of the chain rule.**

** We can be more exact and use the derivatives, and show that the formula is true for Y = G(F(X)) with F(X) = X ^{ N} and G(F) = F^{ M}**

**SOME DEFINITIONS:**

** A function is a process which converts an input into a corresponding output.**

** In symbols, input –> f –> output**

**Examples:**

** The input is transformed (converted) into the output,**

** usually by a formula or an expression using the values of the input:**

** output = 2(input) + 5**

** or you can write this as output = 2 times input + 5**

**The input and the output are both expressions, which can be**

** a: a number, or**

** b: a single variable, x or y or z … , or**

** c: a more complicated expression**

**The commonest form for the input-output relationship for a function f**

** is, as an example, f(x) = 3x + 4, where x is an input**

** and the corresponding output is 3x + 4**

** f is the label of the function,**

** and f(x) is the expression whose value is 3x + 4**

** f(x) = 3x + 4 is then an equation**

** The equation can be seen for example as f(8) = 3 x 8 + 4,**

** or f(y) =3y + 4 using y as the input,**

** or f(z ^{2} + 5z + 7) = 3(z^{2} + 5z + 7) + 4 using an expression.**

**In its most simple formulation the input is not present and the equation is simply f = 3y +4, where the input**

** is identified as the ‘y’.**

** Now if an equation has a single variable on the left and an expression on the right then**

** a) it can be interpreted as a function (functional form) with f(y) = 3y + 4, and**

** b) the expression on the right can be substituted for the variable on the left.**

**Example**

** Let g(A) = A + 2 be a function g with output A + 2**

** Then it can be identified with the equation g = A + 2**

** Let g have the input x ^{2} – 4x + 3**

https://samjshah.com/2018/02/08/there-might-be-light-at-the-end-of-the-chain-rule-tunnel-maybe/

**The partial text is as follows:**

Find two points close to each other, like (x,g(f(x)) and (x+0.001,g(f(x+0.001)).

Find the slope between those two points: {g(f(x+0.001)-g(f(x))}/{(x+0.001)-x}.

There we go. An approximation for the derivative! (We can use limits to write the exact expression for the derivative if we want.)

But that doesn’t help us understand that {d/dx}[g(f(x)]=g'(f(x))f'(x) on any level. They seem disconnected!

But I’m on my way there. I’m following things in this way: x >>> f >>> g

Check out this thing I whipped up after school today. The diagram on top does x \rightarrow f and the diagram on the bottom does f \rightarrow g. The diagram on the right does both. It shows how two initial inputs (in this case, 3 and 3.001) change as they go through the functions f and g.

At the very bottom, you see the heart of this.

It has {δg}/{δf} times {δf}/{δx}={δg}/{δx}

We can be more exact and use the derivatives, and show that the formula is true for y = g(f(x)) with f(x) = x

The direct solution for y’ is the derivative of x

The formula solution is mf

and finally mnx

A function is a process which converts an input into a corresponding output.

In symbols, input –> f –> output

The input is transformed (converted) into the output,

usually by a formula or an expression using the values of the input:

output = 2(input) + 5

or you can write this as output = 2 x input + 5

a: a number, or

b: a single variable, x or y or z … , or

c: a more complicated expression

is, as an example, f(x) = 3x + 4, where x is an input

and the corresponding output is 3x + 4

and f(x) is the expression whose value is 3x + 4

f(x) = 3x + 4 is then an equation

The equation can be seen for example as f(8) = 3 x 8 + 4,

or f(y) =3y + 4 using y as the input,

or f(z

is identified as the ‘y’.

**Now if an equation has a single variable on the left and an expression on the right then**

** a) it can be interpreted as a function (functional form) with f(y) = 3y + 4, and**

** b) the expression on the right can be substituted for the variable on the left.**

** Example**

** Let g(A) = A + 2 be a function g with output A + 2**

** Then it can be identified with the equation g = A + 2**

** Let g have the input x ^{2} – 4x + 3**

It’s well documented that many school students and adults alike are less than fond of mathematics. It tends to be a theme I discuss on here. I’ve singled out over-emphasis on speed of processes,…]]>

I like “Add a zero”, but lying in maths is wonderfully stupid. Read on !!!!!

It’s well documented that many school students and adults alike are less than fond of mathematics. It tends to be a theme I discuss on here. I’ve singled out over-emphasis on speed of processes, misguided attempts at trying to convince students it’s entirely relevant to their daily lives, and of course, not explaining things, as factors. These are all ways in which we, the teachers, are sometimes subconsciously influencing things – but it’s certainly not all our fault though. We live in a country where, typically, it’s a badge of honour to be crap at maths and still miraculously live a normal life. Our bloated curriculum and imbalanced subject hierarchy don’t exactly help either. I was reading a maths book for trainee teachers the other day and I came across a familiarly painful explanation of the column method of subtraction, stating that you ‘cannot subtract three from…

View original post 409 more words

Just a right angled triangle.

First, the definition of a parabola from the focus and directrix.

Pick a line, the directrix, and a point (B) not on that line (the focus):

Find the line at right angles, passing through a point (C) on that line.

Now find the line from B to C, and the midpoint of BC, which will be D.

Find the line at right angles to BC from D, and the intersection of this line and the vertical line, E, is a point on the parabola.

As point C is moved the parabola is traced out.

The picture is completed with the line BE. Check it!

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The circle has radius 1 and centre at the origin.

The line is x = a

Now 1/a is the inverse of a, so a * 1/a = 1, and is fixed.

The line z from (1/a,0) to (x,y) is orthogonal to the radial line, so r/z = Y/a and the two triangles are similar

and r/(1/a) = a/R

Hence rR = a * (1/a) = 1, and both conclusions are true.

The point (x,y) follows a circular path, and rR = 1

]]>Do read it.

It’s quite short!

]]>

I have used dots as they are easier to create.

The quadratic is viewed initially as the “standard form”, and then rebuilt dynamically line by line into the “square plus a bit over” form, as shown in the following sequence:

The odd valued coefficient of x in the original expression can appear as a row and a column of half-dots, or half squares in the area model form.

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