There are many ways to write numeric fractions.
Improper, proper, and mixed fractions are examples of different way of writing them.
Interpreting fractions are another concern to math learners.
Common mistakes are always made in mathematically operation involving fractions.
Example of math error having fractions:
5 1/3 x 1/2 = ?
Correct answer will be (16 / 3) x (1/2) = 16 / 6
Mistake: 5 x (1/3) x (1/2) = 5 / 6 !
What is the error? ==> Interpretation mistake.
5 1/3 is taken to be 5 x (1/3) ! ==> 5 and 1/3 are taken to be 2 separate numbers !
Actually 5 1/3 is a mixed number with 5 whole and 1/3 piece.
Common error is thinking 5 1/3 as 5 multiply by (1/3).
Learn to interpret fractions properly will solve this unnecessary mistake.
Note:
5 1/3 is converted to improper fraction by changing the numerator. The numerator is changed by having the denominator "3" of the (1/3) multiply to the "5" and adding the "1" of the (1/3).
Summary: 3 x 5 + 1 = 16 at the improper fraction numerator.
Learned how to read fractions? Its right interpretation will serve you good by making less mistakes while doing fractional manipulations.
.
Monday, 4 August 2008
Interpretation of Trigonometric Functions
What are the trigonometric operators or functions in mathematics?
They are the three basic operators, namely:
They represent ratio or number. They are related to angle and sides of a triangle.
How are they expressed mathematically?
Example:
cos Y = 0.707 with "Y" representing the angle in question.
In words, the meaning of the above example is "The cosine of angle Y is the number 0.707".
The word "cos" is performing a trigonometrical operation on the angle Y.
(This goes for the other trigonometic functions).
Moving forward with that understanding, it is clear that the "cos" has the same operational rank as the basic mathematical operators +, -, x, and /.
"cos" is not a variable!
"cos" cannot stand alone.
Cos A = 0.707 does not mean "cos" multiply "A" = 0.707.
What it really meant is angle A is trigonometrically operated using "cos" to reflect a number.
"Cos" and "A" has to be together to form a useful meaning on the maths function.
But note, "cos" can be rewritten as cos-1 to represent another trigonometric function of Inverse Cosine.
Example: cos A = 0.707 ===> A = cos-1 0.707
It means , in words, the inverse cosine of 0.707 is the angle A.
Here the operator "cos" can be taken separately and moved over to the opposite side of the equal sign to the number ratio 0.707 in order to get the numerical value of the angle A.
Here the emphasis should be taken that "cos-1", though works like a variable (in this case), is done to fulfill the function of "Inverse cosine" and not as a variable.
Common trigonometry mistake by students / maths learners is to treat the "cos" (or the others) as a variable to be operated upon.
Message:
"cos" is not a number and therefore cannot be a variable. It has to work hand-in-hand with an angle to be meaningful. Make sense?
I repeat, "cos" is a function to operate with an angle. It cannot stand alone. Clear ?
Good!
:)
.
They are the three basic operators, namely:
- Sine (written as "sin")
- Cosine (written as "cos")
- Tangent (written as "tan")
They represent ratio or number. They are related to angle and sides of a triangle.
How are they expressed mathematically?
Example:
cos Y = 0.707 with "Y" representing the angle in question.
In words, the meaning of the above example is "The cosine of angle Y is the number 0.707".
The word "cos" is performing a trigonometrical operation on the angle Y.
(This goes for the other trigonometic functions).
Moving forward with that understanding, it is clear that the "cos" has the same operational rank as the basic mathematical operators +, -, x, and /.
"cos" is not a variable!
"cos" cannot stand alone.
Cos A = 0.707 does not mean "cos" multiply "A" = 0.707.
What it really meant is angle A is trigonometrically operated using "cos" to reflect a number.
"Cos" and "A" has to be together to form a useful meaning on the maths function.
But note, "cos" can be rewritten as cos-1 to represent another trigonometric function of Inverse Cosine.
Example: cos A = 0.707 ===> A = cos-1 0.707
It means , in words, the inverse cosine of 0.707 is the angle A.
Here the operator "cos" can be taken separately and moved over to the opposite side of the equal sign to the number ratio 0.707 in order to get the numerical value of the angle A.
Here the emphasis should be taken that "cos-1", though works like a variable (in this case), is done to fulfill the function of "Inverse cosine" and not as a variable.
Common trigonometry mistake by students / maths learners is to treat the "cos" (or the others) as a variable to be operated upon.
Message:
"cos" is not a number and therefore cannot be a variable. It has to work hand-in-hand with an angle to be meaningful. Make sense?
I repeat, "cos" is a function to operate with an angle. It cannot stand alone. Clear ?
Good!
:)
.
Simultaneous Equations | Three Unknowns
Solving simultaneous equations can be done with many methods. If you understand the underlying concept of elimination, the process will be simple and fun to use.
In this post, the Elimination method to solve 3 unknowns is presented. In a previous post of this website, the steps to solve simultaneous equations having 2 unknowns using Elimination method can be reviewed by clicking this link.
The concept to solving simultaneous equations with 3 unknowns is the same, except the steps are more, since more equations and unknowns are present. Logical, right?
Elimination means the removal of variables or unknown in the process of solving. Study the approach or concept as it serves a good foundation which can be applied to any quantity of unknowns.
The strategy used in the Elimination method of solving is to choose a random unknown to be removed. This reduces the 3-unknown type of question to a simpler 2-unknown type question that can be easily dealt with. The details are presented below.
Letus begin with an example.
x + 2y + 3z = 10 --- (A)
2x + y + z = 9 --- (B)
3x + y + 2z = 13 --- (C)
Step 1:
Look at the 3 equations and decide which unknown is easy to remove. From equation (B) and (C), we can see that the unknown "y" can be easily eliminated by just subtracting the 2 mentioned equations.
(C) - (B): x + 0 + z = 4 --- (D) ==> unknown "y" disappeared !
Step 2:
Find another pair of equations with the purpose of removing the same unknown in step 1.
In this example, there is no 2 other combinations that allow us to remove the same unknown just by performing adding or subtracting. We need to do something first before we can move on in step 2.
Let's target the use of equation (A) and (B) to remove the "y" unknown. Note this is randomly selected.
We see that by multiplying equation (B) by 2, we get 4x + 2y + 2z = 18 ---(E).
The "2y" is now the same as that in equation (A).
Rewriting for simplification,
x + 2y + 3z = 10 --- (A)
4x + 2y + 2z = 18 ---(E)
(A) - (E): -3x + 0 + z = -8 --- (F)
Step 3:
Now we have obtained 2 new equations with only 2 unknowns.
This is the strategy to reduce the more complex 3-unknown question to the simpler 2-unknown question.
Rewriting the 2 new equations here:
x + 0 + Z = 4 --- (D)
-3x + 0 + z = -8 --- (F)
From here onwards, you can review the steps by clicking the link here. It teaches you how to proceed with solving 2-unknown simultaneous equation using the Elimination method.
Some common mistakes made:
The sign of the variables and numbers in the equation are not done correctly after subtraction of 2 equations. Example of this is can be seen using the above question.
If we subtract equation (F) from (D), we should get 4x + 0 = 12.
The common error is to do the x -3x instead of x -(-3x) = x + 3x. The "-" sign error !
To avoid this type of careless mistake, focus on the sign when doing subtraction of equations.
Just keep your mind clear, and constantly remind yourself of this "slip of the mind" mistake.
.
In this post, the Elimination method to solve 3 unknowns is presented. In a previous post of this website, the steps to solve simultaneous equations having 2 unknowns using Elimination method can be reviewed by clicking this link.
The concept to solving simultaneous equations with 3 unknowns is the same, except the steps are more, since more equations and unknowns are present. Logical, right?
Elimination means the removal of variables or unknown in the process of solving. Study the approach or concept as it serves a good foundation which can be applied to any quantity of unknowns.
The strategy used in the Elimination method of solving is to choose a random unknown to be removed. This reduces the 3-unknown type of question to a simpler 2-unknown type question that can be easily dealt with. The details are presented below.
Letus begin with an example.
x + 2y + 3z = 10 --- (A)
2x + y + z = 9 --- (B)
3x + y + 2z = 13 --- (C)
Step 1:
Look at the 3 equations and decide which unknown is easy to remove. From equation (B) and (C), we can see that the unknown "y" can be easily eliminated by just subtracting the 2 mentioned equations.
(C) - (B): x + 0 + z = 4 --- (D) ==> unknown "y" disappeared !
Step 2:
Find another pair of equations with the purpose of removing the same unknown in step 1.
In this example, there is no 2 other combinations that allow us to remove the same unknown just by performing adding or subtracting. We need to do something first before we can move on in step 2.
Let's target the use of equation (A) and (B) to remove the "y" unknown. Note this is randomly selected.
We see that by multiplying equation (B) by 2, we get 4x + 2y + 2z = 18 ---(E).
The "2y" is now the same as that in equation (A).
Rewriting for simplification,
x + 2y + 3z = 10 --- (A)
4x + 2y + 2z = 18 ---(E)
(A) - (E): -3x + 0 + z = -8 --- (F)
Step 3:
Now we have obtained 2 new equations with only 2 unknowns.
This is the strategy to reduce the more complex 3-unknown question to the simpler 2-unknown question.
Rewriting the 2 new equations here:
x + 0 + Z = 4 --- (D)
-3x + 0 + z = -8 --- (F)
From here onwards, you can review the steps by clicking the link here. It teaches you how to proceed with solving 2-unknown simultaneous equation using the Elimination method.
Some common mistakes made:
The sign of the variables and numbers in the equation are not done correctly after subtraction of 2 equations. Example of this is can be seen using the above question.
If we subtract equation (F) from (D), we should get 4x + 0 = 12.
The common error is to do the x -3x instead of x -(-3x) = x + 3x. The "-" sign error !
To avoid this type of careless mistake, focus on the sign when doing subtraction of equations.
Just keep your mind clear, and constantly remind yourself of this "slip of the mind" mistake.
.
Sunday, 3 August 2008
Exciting Patterns of Maths
Maths is an exciting subject.
Why so?
From the many patterns it forms, through its mathematical relation or expression, we can see that they tell us some hidden message, although this message is up to one's interpretation.
175 = 11 + 72 + 53
135 = 11 + 32 + 53
518 = 51 + 12 + 83
How about 596 ?
How about 598 ?
Below is another exciting set of patterns that we can see in maths:
9801 = 992 and 98 + 01 = 99
Look at the sequence of numbers for the above expressions, they are the SAME!
Numbers do make life exciting with its many hidden patterns. It is an area where the human mind can challenge itself to discover more unexplored relations.
Interesting maths, right?
.
Why so?
From the many patterns it forms, through its mathematical relation or expression, we can see that they tell us some hidden message, although this message is up to one's interpretation.
175 = 11 + 72 + 53
135 = 11 + 32 + 53
518 = 51 + 12 + 83
How about 596 ?
How about 598 ?
Below is another exciting set of patterns that we can see in maths:
9801 = 992 and 98 + 01 = 99
Look at the sequence of numbers for the above expressions, they are the SAME!
Numbers do make life exciting with its many hidden patterns. It is an area where the human mind can challenge itself to discover more unexplored relations.
Interesting maths, right?
.
Addition of Squared Numbers, Its Interesting Facts
Adding of squared numbers has many "side" stories besides the usual Pythagorean Theorem.
Looking at the below pattern of adding 2 square numbers., will tell us that it is our friendly Pythagorean Theorem.
32 + 42 = 52
How about if we double the equation of the Theorem, we will get the below:
2(32 + 42) = 2 (9 + 16) = 50
However, the resultant 50 can be expressed into 1 + 49, which is also 12 + 72.
Therefore we can easily say that,
2(32 + 42) = 12 + 72.
Message: Doubling an addition of 2 squared numbers results in another addition of 2 squared numbers.
Another example:
2(12 + 22) = 12 + 32
Why is this so?
What is the underlying principle of the finding?
To answer the questions, we need to understand the below.
What is the actual meaning of "squaring" a number?
It is to find the area within the square region bounded by the number as its side. Therefore the addition of 2 square numbers is simply the sum of 2 square areas.
We also know that addition of 2 square numbers is Pythagorean Theorem at work, and is actually the addition of 2 areas in a square. Pythagorean Theorem also states that this addition of 2 smaller square numbers results in a bigger square area.
Thus, by doubling the resulting bigger square area (as done at the beginning of this post), we still obtain another square area, but only double in size. A square area, no matter what, always has the area format of A x A, with A being its side unit.
By reversing the results of Pythagorean Theorem, starting from the bigger area, we can logically expect another set of square numbers having the same added size. This explains why doubling the sum of 2 squared numbers can form yet another equation of 2 squared numbers added up.
The knowledge presented here, has application in the engineering area in that surface area of an object can be re-sized by doubling or reducing by half and still keeping the same proportion relatively.
.
Looking at the below pattern of adding 2 square numbers., will tell us that it is our friendly Pythagorean Theorem.
32 + 42 = 52
How about if we double the equation of the Theorem, we will get the below:
2(32 + 42) = 2 (9 + 16) = 50
However, the resultant 50 can be expressed into 1 + 49, which is also 12 + 72.
Therefore we can easily say that,
2(32 + 42) = 12 + 72.
Message: Doubling an addition of 2 squared numbers results in another addition of 2 squared numbers.
Another example:
2(12 + 22) = 12 + 32
Why is this so?
What is the underlying principle of the finding?
To answer the questions, we need to understand the below.
What is the actual meaning of "squaring" a number?
It is to find the area within the square region bounded by the number as its side. Therefore the addition of 2 square numbers is simply the sum of 2 square areas.
We also know that addition of 2 square numbers is Pythagorean Theorem at work, and is actually the addition of 2 areas in a square. Pythagorean Theorem also states that this addition of 2 smaller square numbers results in a bigger square area.
Thus, by doubling the resulting bigger square area (as done at the beginning of this post), we still obtain another square area, but only double in size. A square area, no matter what, always has the area format of A x A, with A being its side unit.
By reversing the results of Pythagorean Theorem, starting from the bigger area, we can logically expect another set of square numbers having the same added size. This explains why doubling the sum of 2 squared numbers can form yet another equation of 2 squared numbers added up.
The knowledge presented here, has application in the engineering area in that surface area of an object can be re-sized by doubling or reducing by half and still keeping the same proportion relatively.
.
Mental Logarithm Easily Computed
In our daily life, we encounter many maths computing which we do mentally. Some are simple addition and subtraction. Some involves more complex operation. But with breaking-down strategies, we are abale to simplify this mental process to simpler steps for the mind to handle.
Here I will discuss on the mental computation with maths operation like the indices and logarithm, that we may come also in some daily situations.
Below site an example. Read and enjoy it.
Example: 10x = 8
To solve this example, we need to convert it into the logarithm form to move forward.
log 10x = log 8 ---(A)
Applying the Laws of Logarithm, log Yx ==> x log Y
Expression (A) above, become x (1) = x = log 8
However, we are stuck here! What is the answer for log 8? And mentally?
How to carry on with mental logarithm computation?
Mental logarithm computation can be easily processed if we can simply memorise a few number of the basic logarithm.
The basic numbers that we need to remember and suffice for common mental logarithm usage are:
With the 3 basic logarithm numbers known, we can list out the first few log to see its usefulness:
With the above list, most of the logarithm numbers can be easily computed mentally.
But do note, however, that logarithm of prime number cannot be done using above method.
Examples are log 11, log 17, etc.
However, we should not be deterred from this as they are the minority and the list above serves to cover most of the logarithm solutions which can be easily done mentally.
:)
.
Here I will discuss on the mental computation with maths operation like the indices and logarithm, that we may come also in some daily situations.
Below site an example. Read and enjoy it.
Example: 10x = 8
To solve this example, we need to convert it into the logarithm form to move forward.
log 10x = log 8 ---(A)
Applying the Laws of Logarithm, log Yx ==> x log Y
Expression (A) above, become x (1) = x = log 8
However, we are stuck here! What is the answer for log 8? And mentally?
How to carry on with mental logarithm computation?
Mental logarithm computation can be easily processed if we can simply memorise a few number of the basic logarithm.
The basic numbers that we need to remember and suffice for common mental logarithm usage are:
- log 2 = 0.301
- log 3 = 0.477
- log 7 = 0.845
With the 3 basic logarithm numbers known, we can list out the first few log to see its usefulness:
- log 1 = 0
- log 2 = 0.301
- log 3 = 0.477
- log 4 = log (2 x 2) = log 2 + log 2 = 0.301 + 0.301 = 0.602
- log 5 = log (10 / 2) = log 10 - log 2 = 1 - 0.301 = 0.699
- log 6 = log (2 x 3) = log 2 + log 3 = 0.301 + 0.477 = 0.778
- log 7 = 0.845
- log 8 = log (2 x 2 x 2) = log 2 + log 2 + log 2 = 0.301 x 3 = 0.903
- log 9 = log (3 x 3) = log 3 + log 3 = 0.477 + 0.477 = 0.954
- log 10 = 1
With the above list, most of the logarithm numbers can be easily computed mentally.
But do note, however, that logarithm of prime number cannot be done using above method.
Examples are log 11, log 17, etc.
However, we should not be deterred from this as they are the minority and the list above serves to cover most of the logarithm solutions which can be easily done mentally.
:)
.
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