MATH LESSONS
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LIST OF ARTICLES
Breaking Down Fractions
Divisibility Heuristics
Breaking Down Fractions
Divisibility Heuristics
1) Breaking Down Fractions
A majority of Americans admit an inability to perform operations on fractions. In this article, I am going to "break down fractions", and show you just how easy they are to calculate with.
First, I am not going to call them fractions. They are properly called Rational Numbers. Why? Because they are a ratio of two integers. We define the set of Rational Numbers as;
The set of all numbers of the form \(\frac{p}{q}\) such that p and \(q\) are both integers, except that \(q\) cannot be zero.
So, what is an integer? An integer is any number in the number set \(\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}\). The symbol \(\ldots\) is called an ellipse and means goes on forever. Why can \(q\) never be zero? I'm sure you have heard that division by zero is not possible, or is undefined, but why?
If the following statement is true, then the statement after it is also true. The first statement implies the second.
Suppose \(a, b,\) and \(c\) are any numbers other than zero, such that \(a \times b = c\). Then, \(a = c \div b = \frac{c}{b}\). This implication works in the other direction also. If \(a, b,\) and \(c\) are any numbers other than zero, then \(a = c \div b = \frac{c}{b}\) implies that \(a \times b = c\). This is where we get the fact families for multiplication and division.
Here is a concrete example. We know that \(3 \times 4 = 12\). But this means that \(3 = 12 \div 4 = \frac{12}{4}\). The other way around, we know that \(3 = 12 \div 4 = \frac{12}{4}\). If we multiply both sides of the equation by \(4\), we get that \(3 \times 4 = 12\).
Now, suppose we have two non-zero numbers \(a\) and \(b\), together with zero, and we claim that \(a = b \div 0 = \frac{b}{0}\). Then, it follows that \(a \times 0 = b\), where \(a\) and \(b\) are both non-zero. But \(a \times 0 = 0 \neq b\) no matter the value of the non-zero a. Then, \(a = b \div 0 = \frac{b}{0}\) must be false, which it is.
First, I am not going to call them fractions. They are properly called Rational Numbers. Why? Because they are a ratio of two integers. We define the set of Rational Numbers as;
The set of all numbers of the form \(\frac{p}{q}\) such that p and \(q\) are both integers, except that \(q\) cannot be zero.
So, what is an integer? An integer is any number in the number set \(\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}\). The symbol \(\ldots\) is called an ellipse and means goes on forever. Why can \(q\) never be zero? I'm sure you have heard that division by zero is not possible, or is undefined, but why?
If the following statement is true, then the statement after it is also true. The first statement implies the second.
Suppose \(a, b,\) and \(c\) are any numbers other than zero, such that \(a \times b = c\). Then, \(a = c \div b = \frac{c}{b}\). This implication works in the other direction also. If \(a, b,\) and \(c\) are any numbers other than zero, then \(a = c \div b = \frac{c}{b}\) implies that \(a \times b = c\). This is where we get the fact families for multiplication and division.
Here is a concrete example. We know that \(3 \times 4 = 12\). But this means that \(3 = 12 \div 4 = \frac{12}{4}\). The other way around, we know that \(3 = 12 \div 4 = \frac{12}{4}\). If we multiply both sides of the equation by \(4\), we get that \(3 \times 4 = 12\).
Now, suppose we have two non-zero numbers \(a\) and \(b\), together with zero, and we claim that \(a = b \div 0 = \frac{b}{0}\). Then, it follows that \(a \times 0 = b\), where \(a\) and \(b\) are both non-zero. But \(a \times 0 = 0 \neq b\) no matter the value of the non-zero a. Then, \(a = b \div 0 = \frac{b}{0}\) must be false, which it is.
-- More to come. Be patient. --
2) Divisibility Heuristics
Many people, mathematics teachers included, call these "divisibility rules". Really though, they are used to help a student determine if some integer divides another integer, and not to actually carry out a division operation. For that reason, I prefer the term "heuristic". There are many sites on the Interwebs where these heuristics can be found. Some of them may explain a particular heuristic in a different way than do I. If you can not follow my explanation, simply googlify "Divisibility Rules".
In some of the heuristics that follow, I will use a term or phrase that may be unfamiliar to many readers. In those cases, I will explain the term or phrase BEFORE I state the heuristic.
In some of the heuristics that follow, I will use a term or phrase that may be unfamiliar to many readers. In those cases, I will explain the term or phrase BEFORE I state the heuristic.
Divisibility by 2
Every integer with the numeral 0, 2, 4, 6, or 8 in the ones place is divisible by 2.
These are the even numbers. Some examples are 12, 280, 5794, and 135798.
These are the even numbers. Some examples are 12, 280, 5794, and 135798.
Divisibility by 3
Phrase Explained: Sum of the digits: The sum of the digits of an integer is found by adding the digits that comprise the integer. For example, the sum of the digits of 2,376,081 is (2 + 3 + 7 + 6 + 0 + 8 + 1) = 27, and the sum of the digits of 27 is (2 + 7) = 9. When using a sum of the digits heuristic, you may stop stage you wish, or you may carry out the summing process until you are left with a single digit integer.
Every integer whose digit sum is divisible by 3, 6, or 9 is itself divisible by 3.
For example, the sum of the digits of 217,251 is (2 + 1 + 7 + 2 + 5 + 1) = 18, and 18 is divisible by 3. Hence 217,251 is also divisible by three. The sum of the digits of 217,253 is (2 + 1 + 7 + 2 + 5 + 3) = 20. 20 is clearly not divisible by 3, so neither is 217,253.
For example, the sum of the digits of 217,251 is (2 + 1 + 7 + 2 + 5 + 1) = 18, and 18 is divisible by 3. Hence 217,251 is also divisible by three. The sum of the digits of 217,253 is (2 + 1 + 7 + 2 + 5 + 3) = 20. 20 is clearly not divisible by 3, so neither is 217,253.
Divisibility by 4
Every integer with a 2 or 6 in the ones place and an odd integer in the tens place is divisible by 4. Additionally, every integer with a 0, 4, or 8 in the ones place and an even integer in the tens place is also divisible by 4.
Let's look at a pattern, 12 and 16 are divisible by 4, but 22 and 26 are not. 10, 14, and 18 are not divisible by four, but 20, 24, and 28 are divisible by 4. This pattern continues for ever and ever, Ahem.
Let's look at a pattern, 12 and 16 are divisible by 4, but 22 and 26 are not. 10, 14, and 18 are not divisible by four, but 20, 24, and 28 are divisible by 4. This pattern continues for ever and ever, Ahem.
Alternate Rule
Multiply the number in the tens digit by two and add 2. If the result is divisible by 4, so is the original integer.
For example, 243,652 yields (5 x 2) + 2 = 10 + 2 = 12. 12 is divisible by 4, so 243,652 is as well. (This is equivalent to the rule given above.)
For example, 243,652 yields (5 x 2) + 2 = 10 + 2 = 12. 12 is divisible by 4, so 243,652 is as well. (This is equivalent to the rule given above.)
Divisibility by 5
Every integer with a 0 or 5 in the ones place is divisible by 5.
Divisibility by 6
Every integer divisible by BOTH 2 and 3 is also divisible by 6.
If an integer is divisible by 2, that means 2 is a factor of the integer. The same goes for 3. But 2 x 3 = 6. So an integer with 2 and 3 as a factor has 6 as a factor, also too as well.
If an integer is divisible by 2, that means 2 is a factor of the integer. The same goes for 3. But 2 x 3 = 6. So an integer with 2 and 3 as a factor has 6 as a factor, also too as well.
Divisibility by 7
WORKS ONLY WITH INTEGERS WITH 3 OR MORE DIGITS.
If the integer has exactly 3 digits, multiply the digit in the ones place by 2 and subtract the result from the integer formed by the first two digits in the original integer. If the result is divisible by 7, so is the original integer.
For example, 441 yields 44 - (1 x 2) = 42. Since 42 is divisible by 7 (42 = 7 x 6), so is 441.
For example, 441 yields 44 - (1 x 2) = 42. Since 42 is divisible by 7 (42 = 7 x 6), so is 441.
If the integer has more than 3 digits, alternate subtracting and adding periods from right to left. If this will yields an integer with exactly 3 digits, apply the rule above.
For example, 383,565,861 yields (861 - 565 + 383) = 296 + 383 = 679. Then 67 - (9 x 2) = 67 - 18 = 49 = 7 x 7.
For example, 383,565,861 yields (861 - 565 + 383) = 296 + 383 = 679. Then 67 - (9 x 2) = 67 - 18 = 49 = 7 x 7.
Alternate Rule
If the procedure above yields a 4 digit integer, apply the procedure to that number.
Divisibility by 8
If the digit in the hundreds place is even and the integer formed by the last 2 digits are divisible by 8, so is the original integer. Otherwise, if the digit in the hundreds place is odd, and the integer formed by the last 2 digits plus 4 is divisible by 8, so is the original integer.
Divisibility by 10
Every integer with a 0 in the ones place is divisible by 10.
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