Section 1.1
§1 Converting from decimal to binary
Remember, in decimal the base of each place value is 10. If you want to break it down, in the decimal
system the place values look like the following:
_
_
5
10
10
_
4
_
3
10
10
_
2
_
1
100
10
.
So a number like 3,012 in the decimal system is equal to  3 1000    0 100   110    2 1 . Note
that we multiply each quantity times its base value.
In binary, the base of each place value is 2. The place values look like the following:
_
_
_
_
_
_
_
6
5
4
3
2
1
20
2
2
2
2
2
2
. Note that 26  64 , etc…
Example: Convert the decimal number 45 to a binary number.
So what we need to do is find the largest place value below the number in question, which in this case is
45. Hence, the largest place value below 45 is 32, which is 25 . So put a 1 there.
1
_
_
_
_
_
5
4
3
2
1
20
2
2
2
2
2
.
Next, we subtract 32 from 45 to get 13. So, we need to find the next largest base value below 13. That
base value is 8, which is 23 . So put a 1 there. Note that we put a 0 for the place value of 16.
1
0
1
_
_
_
5
4
3
2
1
20
2
2
2
2
2
Next, we subtract 8 from 13. This is 5. The next largest place value below 5 is 4. So put a 1 there.
1
0
1
1
_
_
5
4
3
2
1
20
2
2
2
2
2
You should see that the final answer is 101101. Note that if we multiply each place value times its
quantity, we get 1 32    0 16   1 8   1 4    0  2   11  45
PRACTICE
1) Convert the decimal number 59 to binary.
§2 Converting from decimal to hexadecimal
Example: Convert the decimal number 141 to a hexadecimal number.
In hexa, the base of each place value is 16. The place values look like the following:
_
_
_
_
_
_
_
166 165 164 163 162 161 160
Most hexadecimal numbers are not larger than two digits. So we only really need the first two places:
_
_
1
16
160
So, since our decimal number in question is 141, we need to find out how many times 16 goes into 141
without going over. Let’s make a list:
16  0  0
16  8  128
16 1  16
16  2  32
16  3  48
16  4  64
16  5  80
16  6  96
16  7  112
16  9  144
16 10  160
16 11  176
16 12  192
16 13  206
16 14  224
16 15  240
Hence, we see that 16 goes into 141 8 times without going over. So put a 8 in that place.
8
_
1
16
160
Since 16 times 8 = 128, the remainder is 13. In hexa, the numbers bigger than 9 are represented by
letters.
A  10, B  11, C  12, D  13, E  14, F  15
Hence, 13 corresponds to D. So the decimal number 141 is represented by 8D in hexa. If we expand this,
we see that we get  8 16   13 1  141
PRACTICE
2) Convert the decimal number 187 to hexadecimal.
§3 Estimation and Rounding
These terms actually do not mean the same thing! When we are asked to estimate, it can be done to any
place value. For the most part, when we estimate, we usually round to the leading value. This is called
Leading Value Estimation. The leading value is simply the first number. For example, if you are asked to
estimate the total cost of a $4.79 cup of coffee and $2.39 muffin, we would round each value only to its
leading digit. Notice you were not asked which place value to estimate to. So, we would estimate the
cost of the coffee to be $5 and the cost of the muffin to be $2 for a total cost of $7.
Rounding is a different story. When we round, you are told which place value to round to. The steps are
as follows:
1) First determine the round-off digit, which is the digit in the specified place value.
2) If the first digit to the right of the round-off digit is greater than 5, do not change the round-off digit.
All digits between the decimal point and the round-off digit should become zeros. Any digit to the right
of the decimal point simply goes away.
3) If the first digit to the right of the round-off digit is less than 5, increase the round-off digit by 1. Apply
the same steps as above the to the remaining digits.
For example, say we want to round the number 1704.6916. If I ask you to estimate this number we
would use leading digit estimation. The round-off digit is simply the first (or left-most) digit, which in this
case is 1. So we need to look at the first digit to the right of the round-off digit. Since it is greater than 5,
we increase the round-off digit by 1. All numbers between the round-off digit and the decimal become
zeros, while all digits to the right of the decimal point simply drop. So our answer is 2000.
Say we want to round the digit to the nearest tens. The round-off digit is the zero in the tens place. Look
at the digit to the right – it’s a 4. Hence we do not change the round-off digit. Our final answer is 1700.
Say we want to round to the nearest hundredth. The round-off digit is the 9 to the right of the decimal
point. Look at the digit to its right – it’s a 1. Since it is not greater than 5, we leave the round-off digit as
is. Since we are rounding to a place value to the right of the decimal point, any value to the right of the
decimal point drops. The answer is then 1704.69.
PRACTICE
3) Round 5482.493 to nearest a) hundreds
b) ones
c) hundredths
§4 Order of Operations
Let’s say I ask you to calculate the answer for
the answer 13. Which one is correct?
3  5  2 . You may get the answer 16, or you may get
Well, it turns out they both can’t be correct. The key is, there must be some correct ORDER to perform
the operations. They are as follows:
1) Parentheses (or grouping symbols such as braces or brackets)
2) Exponents
3) Multiplication or division, as they appear going from left to right
4) Addition or subtraction, as they appear going from left to right.
The easy way to remember the order is to use the acronym PEMDAS, or Please Excuse My Dear Aunt
Sally.
It’s important to note that we perform the multiplication and division as we see them going from left to
right. It does not mean we always do multiplication before division. The same thing applies with the
addition and subtraction.
For example, say we want to simplify the expression 12  (7  3)
2
 2 . It’s a good idea to do one
operation at a time and make sure you write everything out!
The first operation to perform is the parentheses. The expression simplifies to 12  (4)
2
 2 . The next
operation to perform is the exponent. The expression becomes 12  16  2. What would you do next?
Make sure you are able to get the correct answer, which should be 4!
NOTE: The grouping symbols get a little confusing sometimes. For example, how would you interpret
10  3(2) ? In this case, the parentheses means we multiply. So we actually have to multiply the 3 and
2 first. You should end up with 10  6, which equals 4.
PRACTICE
4) Simplify
8.3  8.8  2.2  1.5
5) Simplify
8  25  7 
6) Simplify
5   3  1  15  9 
2