So I decided to try out the book Basic Math & Pre-Algebra Workbook for Dummies
and so far I've been pretty happy with it. It's very basic and
understandable (I suppose that's why it's for "idiots"). I have had
trouble studying for the GMAT because I haven't known where to begin.
The Workbook for Dummies begins in a perfect place; it doesn't attempt
to teach you addition and subtraction, although it does go through long
division. I trudged through Part I, and I feel like I've picked up many
new details that I missed when I was learning the same stuff in
elementary school.
The first subject the book discussed was number systems. Apparently we use the Hindu-Arabic number system,
which basically uses decimal symbols to place value (assigning a digit a
greater or lesser value based on where the number appears) for the 10
digits we are familiar with: 0,1,2,3,4,5,6,7,8,9. Here is a better
summary: Hindu-Arabic numberation system.
0, unlike the other numbers, has no value. It simply acts as a placeholder.
There are two types of zeros:
Placeholders: These are on the right of one or more non-zero digits
Leading zeros:
These are to the left of a non-zero digit, and are extraneous. They are
only useful for corresponding numbers to alphabetical letters, such as
01 for A and 11 for K, or for holding a place for additional numbers
(think of a car odometer, for example)
The most common type of placeholder in the Hindu-Arabic number system uses exponential powers of 10, namely:
Let's use the number 234 (Two hundred and thirty four) as an example:
Millions Hundred Thousands Ten Thousands Thousands Hundreds Tens Ones
2 3 4
234 = 10² * 2 + 10¹ * 3 + 10⁰ * 4 = 100*2 + 10*3 + 1*4 = 200 + 30 + 4
However, it the placeholder need not be 10 in general. The Binary System, for instance, uses only 0's and 1's (powers of 2).
Here is the same number, 234, written in Binary:
Powers of Two:
One Hundred Twenty Eight Sixty Fours Thirty Twos Sixteens Eights Fours Twos Ones
1 1 1
0 1 0 1 0
Start subtracting the highest power of 2 from the total number:
(Try writing it out yourself. I've found that solving something yourself at your own pace really helps)
234 - 128 = 106 <--"Remainder"
Take the amount that remains and subtract the next highest power of 2:
106 - 64 = 42
And again:
42 - 32 = 10
And again:
10 - 8 = 2
And finally, we are left with zero:
2 - 2 = 0
So, here is the same number, 234, calculated with exponential powers of 2
234
= 2⁷ * 1 + 2⁶ * 1 + 2⁵ *1 + 2⁴ * 0 + 2³ * 1 + 2² * 0 + 2¹ * 1 + 2⁰ * 0 =
128*1 + 64*1 + 32*1 + 16*0 + 8*1 + 4*0 + 2*1 + 1*0 = 128 + 64 + 32 + 8 +
2
Oh sure, you say, it got solved so nice and neatly because she used an even number and the exponents were in powers of two.
Okay then, let's try 233
Basically it's the same stuff as above:
One Hundred Twenty Eight Sixty Fours Thirty Twos Sixteens Eights Fours Twos Ones
1 1 1
0 1 0 0 1
233-128 = 105
105-64 = 41
41-32 = 9
9-8 = 1
1-1 = 0
We still get left with a zero. NO remainder.
And so we can conclude that 233 in binary is 11101001, while 234 in binary is 11101010
It gets a little more complicated when you use a
different base (because it's less familiar); let's use the power of 5,
for example:
One Hundred and Twenty Fives Twenty Fives Fives Ones
1 4 1 4
234 - 125 = 109
109 - (25 * 4) = 9
9 - 5 = 4
4 - 4 = 0
234 = 5³ * 1 + 5² * 4 + 5¹ * 1 + 5⁰ * 4 = 125*1 + 25*4 + 5*1 + 1*4 = 125 + 100 + 5 + 4
Kind of cool, isn't it?
I know, I got into exponents right away. But you see how easy it is, right? They don't look so scary when they're just placeholders, do they?
Don't be afraid to try it!
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