I know I’m doing things back-to-front –
arse-about - as we say in Oz (and possibly elsewhere) but, considering all the
esoteric mathematics I produce on this blog, I thought I should try and explain
some basics.

As I mentioned earlier this year in a post on
‘analogy’, mathematics is a cumulative endeavour and you can’t understand
calculus, for example, if you don’t know algebra. I’ve come across more than a
few highly intelligent people, of both sexes, who struggle with maths (or math as
Americans call it) and the sight of an equation stops them in their tracks.

Mathematics is one of those topics where
the gap, between what you are expected to know and what you actually learn, can
grow as you progress through school, mainly because you were stumped by
algebra. You know: the day you were suddenly faced with numbers being replaced
by letters; and things like counting, adding, subtracting, dividing,
multiplying, fractions and even decimals suddenly seemed irrelevant. In other
words, everything you’d learned about mathematics, which was firmly grounded in
numbers – something you’d learned almost as soon as you could talk – suddenly seemed
useless. Even Carl Jung, according to his autobiography, stopped understanding
maths the day he had to deal with ‘x’. In fact, his wife, Emma, had a better
understanding of physics than Jung did.

But for those who jump this hurdle,
seemingly effortlessly, ‘x’ is a liberator in the same way that the imaginary
number

*i*is perceived by those who appreciate its multi-purposefulness. In both cases, we can do a lot more than we could before, and that is why algebra is a stepping-stone to higher mathematics.
Fundamentally, mathematics is not so much
about numbers as the relationship between numbers, and algebra allows us to see
the relationships without the numbers, and that’s the conceptual hurdle one has
to overcome.

I’ll give a very simple example that
everyone should know: Pythagoras’s triangle.

I don’t even have to draw it, I only have
to state it: a

^{2}+ b^{2}= c^{2}; and you should know what I’m talking about. But a picture is worth innumerable words.
The point is that we can use actual integers, called Pythagorean triples, that obey this relationship; the smallest
being 5

^{2}= 4^{2}+ 3^{2}. Do the math as you Americans like to say.
But the truth is that this relationship
applies to all Pythagorean triangles, irrespective of their size, length of
sides and units of measurement. The only criteria being that the triangle is
‘flat’, or Euclidean (is not on a curved surface) and contains one right angle
(90

^{o}).
By using letters, we have stated a
mathematical truth, a universal law that applies right across the universe.
Pythagoras’s triangle was discovered well before Pythagoras (circa 500BC) by
the Egyptians, Babylonians and the Chinese, and possibly other cultures as
well.

Most of the mathematics, that I do,
involves the manipulation of algebraic equations, including a lot of the stuff
I describe on this blog. If you know how to manipulate equations, you can do a
lot of mathematics, but if you don’t, you can’t do any.

A lot of people are taught BIDMAS, which
gives the priority of working out an equation: Brackets, Indices, Division,
Multiplication, Addition and Subtraction. To be honest, I’ve never come across
a mathematician who uses it.

On the other hand, a lot of maths books
talk about the commutative law, the associative law and the distributive law as
the fundaments of algebra.

There is a commutative law for addition and
a commutative law for multiplication, which are both simple and basic.

A + B = B + A and A x B = B x
A (that’s it)

Obviously there is no commutative law for
subtraction or division.

A – B ≠ B – A and A/B ≠ B/A (pretty obvious)

There are some areas of mathematics where
this rule doesn’t apply, like matrices, but we won’t go there.

The associative law also applies to
addition and multiplication.

So A + (B + C) = (A + B) + C and A x (B x C) = (A x B) x C

It effectively says that it doesn’t matter
what order you perform these operations you’ll get the same result, and,
obviously, you can extend this to any length of numbers, because any addition
or multiplication creates a new number that can then be added or multiplied to
any other number or string of numbers.

But the most important rule to understand
is the distributive law because it combines addition and multiplication and can
be extended to include subtraction and division (if you know what you're doing). The distributive law lies at
the heart of algebra.

A(B + C) = AB + AC and A(B + C) ≠ AB + C (where AB = A x B)

And this is where brackets come in under
BIDMAS. In other words, if you do what’s in the brackets first you’ll be okay.
But you can also eliminate the brackets and get the same answer if you follow
the distributive rule.

But we can extend this: 1/A(B - C) = B/A -
C/A (where B/A = B ÷ A)

And
-A(B – C) = CA – BA because
(-1)

^{2}= 1, so a minus times a minus equals a plus.
If 1/A(B + C) = B/A + C/A then (B + C)/A =
B/A + C/A

And
A/C + B/D = (DA + BC)/DC

To appreciate this do the converse:

(DA + BC)/DC = ~~D~~A/~~D~~C + B~~C~~/D~~C~~
= A/C + B/D

But the most important technique one can
learn is how to change the subject of an equation. If we go back to
Pythagoras’s equation:

a

^{2}+ b^{2}= c^{2}what’s b = ?
The very simple rule is that whatever you
do to one side of an equation you must do to the other side. So if you take
something away from one side you must take it away from the other side and if
you multiply or divide one side by something you must do the same on the other
side.

So, given the above example, the first
thing we want to do is isolate b

^{2}. Which means we take a^{2}from the LHS and also the RHS (left hand side and right hand side).
So b

^{2}= c^{2}– a^{2}
And to get b from b

^{2}we take the square root of b^{2}, which means we take the square root of the RHS.
So b = √(c

^{2}– a^{2})
Note b ≠ c – a

^{ }because √(c^{2}– a^{2}) ≠ √c^{2}- √a^{2}
In the same way that (a + b)

^{2}≠ a^{2}+ b^{2}
In fact (a + b)

^{2}= (a + b)(a + b)
And applying the distributive law: (a +
b)(a + b) = a(a + b) + b(a + b)

Which expands to a

^{2}+ ab + ba + b^{2}= a^{2}+ 2ab + b^{2}
But (a + b)(a – b) = a

^{2}– b^{2}(work it out for yourself)
An equation by definition (and by name)
means that something equals something. To maintain the

**equality**whatever you do on one side must be done on the other side, and that’s basically the most important rule of all. So if you take the square root or a logarithm or whatever of a single quantity on one side you must take the square root or logarithm or whatever of everything on the other side. Which means you put brackets around everything first and apply the distributive law if possible, and, if not, leave it in brackets like I did with the example of Pythagoras’s equation.
Final Example: A/B = C + D What’s B = ?

Invert both sides: B/A = 1/(C + D)

Multiply both sides by A: B = A/(C + D) (Easy)

Note: A/(C + D) ≠ A/C + A/D

To understand why, take the inverse: (C + D)/A = C/A + D/A

And take the inverse again: A/(C + D) = 1/(C/A + D/A) ≠ A/C + A/D

## 2 comments:

"Obviously there is no commutative law for subtraction or division."

Plausibly that's because subtraction and division aren't true operations. Really, subtraction is just the addition of a negative number and division is just multiplication by a number between 0 and 1.

Hi Eli,

Yes, and, basically that's the way it is in algebra: a number can be positive or negative and it can be inverted or not, but it's important to keep track of those attributes.

In fact, if I was teaching this, I'd make that point about the associative law: you can do them in any order you want as long as the negative numbers stay negative and the inverted ones stay inverted.

I think algebra is a conceptual hurdle for a lot of people and I don't think BIDMAS helps at all. It assumes you're ignorant and, if you depend on it, it will keep you ignorant.

Regards, Paul.

Post a Comment