Tips on Estimation

In maths, we do come across topics on estimation.

This topics relate to our daily living very closely and is truly practical. 
Example is when we go shopping and starts to count the expenses to be paid. 

But mathematical estimation has one key concern.
To what extent or how accurate does one wish to?

There is no right or wrong to an answer when dealing with estimation. After all it is an ESTIMATED answer.

The basic requirement is thus to get as close to the true answer as possible.

Let's start with one simple example to demonstrate the concept.

Example:
y = 0.501 + square root(3.89)
What is y without using calculator ?

Answer 1:
y = 1 + square root (4)
y = 1 + 2 = 3

Answer 2:
y = 0.5 + square root (4)
y = 0.5 + 2 = 2.5

You can now see that both answers is close to the actual answer of 2.4733.
However, it is the gap or extent of the difference you wish for.

If possible, LOOK carefully at the numbers and give a best estimation closest to ability to compute the answer.

Here, from the above, you will notice the first term of 0.501 decides the outcome.
Estimate this 0.501 to what numeric value?

Think further and you will find that 0.501 to 1 will give you a bigger difference compared to 0.501 to 0.5.

If you can mentally handle 0.5 as the estimated value for computation, go for it since this should be closer to the final outcome.

Thus in conclusion, do look a bit closer to the numbers presented in your problem and simply do a brief calculation of the differences between estimated and raw data. Then you are one step closer to getting a good estimation.

Estimation, finally, boils down to how far you are to the actual answer. Nothing difficult.

Interesting? Any more suggestions?

.

Basic Decimal Conversion

`
Every thing falls back to basic.

If the fundamentals are weak, any maths learners will have a hard time moving forward in their maths learning journey.

Let me quote an example.

How do we change 4.75 to fraction.

We can use 475 / 100 and reduce it through long division.  This will give 4 and 3/4.

However, if we know that 4.75 is actually consisting of 4 add to 0.75, the conversion will be simpler.

4 and 0.75 means 4 + (3/4) which leads directly to 4 whole and 3/4.  Same as answer of above.
(NOTE:  0.75 is a quarter which equates to 3/4).

Simple isn't it?

Maths is interesting.

:-)

Number of Answers | Common mistake

Maths can be tricky when you are not careful.
This is not to frighten you, though.

This post is just to remind you of the wonderful aspect of maths in covering all areas.

Below is an example of what I meant.

Let's take the quadratic eqaution solving as a starting point

x² = 5x

x  = 5x  /  x   = 5  (Answer)

At first, this looks pretty fine. The answer, when substituted back, produces match of equation.

But this is actually not complete.

Those doing quadratic equation will know 2nd order (x2) equation evaluates to 2 answsers.
The answers may be the same though.

Now, if we approach it using another method, let's see the different.

x² - 5x = 0

==> x (x - 5) = 0   , after factorising

==>  x = 0  and  (x - 5) = 0
==>  x = 0  and   x = 5

There are two answers now.

We had the x = 5 initially, but what about this new x = 0.

We have missed out on the x = 0 with the first mehtod. It looks OK then.

What happen?
It may be due to lack of experience handling this form of maths question.

The concept in solving quadratic equation is actually not limited to second order.
The hidden message is depending on the order, the number of answers will follow suit.

What I meant is :
2nd order gives 2 answers,
3rd order gives 3 answers,
4th order gives 4 answers, etc.

It is this verry message that maths learner should capture. Otherwise you will be tricked to give only one answer which leads you to "mistakes" of being incomplete.

I agree that this is tricky, but within reasonable argument.
If a student practice hard (and smart), he will not fall prey to this type of simple math problem.

Do not get con again.

Enjoy maths. It's fun and interesting.

:D

Area Displacement Theory

.
Maths is not all about calculation.

There are always more to it than meet the eyes.

This is especially true when you are doing geometrical questions where you are involved with area, perimeter and so on.

Displacement theory or its equivalent is always done without the knowledge of many people.

What is this theory about?

Let's look at one example below.

In the diagram above, you will see a path (white coloured) going across a blue platform.
If you are asked to find the area of this path, what can you do to obtain this area?

If no data of dimension is given, it is definitely not possible.

Now if the width of the path and the vertical length of the blue platform is given, can you compute the answer?

Again , this need a bit of thinking.

Displacement theory kicks in here. Look at the diagram on the right.

It is the displaced or closed up portion of the blue platform that does the trick.
Here you will notice the dashed line forming a white rectangluar area on the right-most side of the white blue platform.

Are you able to find the area of this white rectangular piece?
The width of this rectangle piece is ACTUAL the width of the white path!

You should now be able to calculate the area of this rectangular piece since the path width and length of the rectangular block is known or deduced now.

How this is possibe is through the "hidden" clue or step of closing up the path revealing the simpler rectangular area that any decent maths student can calculate.

Hence, maths is wonderful in that it tests you not only about applcations of maths tools, but your other "intelligence".

Having known displacement theory here, I believe you are really for the Math Challenge 23.
Go there and answer the question, and be quick before others grap the position one ... 

:-D
.

Percentage | Common Mistake

*
Percentage problems can be tricky at times when you are careless.

Let's us look at one maths word problem related to it.

Question:
There are 3 persons, John, Mary and Jane.
John is richer than Mary by 10%, and Mary is richer than Jane by 20%.
Is John richer than Jane by (10 + 20)% = 30% ?

Most students, upon quick thinking, will acknowledge that 30% is the correct answer.

Is it so?

To verify the answer, let us assume that Jane has $1000.
As such, Mary will have (100+20)% of $1000 = 1.2 x $1000 = $1200.
John is then 1.1 x $1200 = $1320 richer than Jane ==> By 32%.

If 30% is correct, we should get 1.3 X $1000 = $1300.

The latter number (dollar) is not the same as the first worked out solution.
Why?

Mistake in understanding what is percentage:
To assume that John is 10% + 20% richer than Jane is incorrect.
This is due to the fact that percentage has to take a common reference for this to be correct.

In the word problem, the percentages of comparison are not to a common reference.
The first one is to Mary, while the next is to Jane.

These made the denominator of the ratio different.
Thus adding the percentage up is a mistake, and an easy one too!

.

Logarithm Operation Explained (2)

'
Logarithm operator exists as a mathematical tool to allow us to convert a number to another form in terms of base and power.

Can you make 0.5 in terms of 10^y or 5^y ?

The above is a maths question that can use logarithm to solve.

The application requires the Power rule that states that log_kD^m is equivalent to mlog_kD.
With that knowledge, to convert the above question of 0.5 to various base, we simply log the 0.5 to its respecive base.

Let's go for the base 10.

Original:   0.5 = 10^y

Performing "log" on both sides:
log₁₀ 0.5 = log₁₀10^y 
-0.301      = y  log₁₀10
-0.301      = y

or another way to put the outcome is 0.5 = 10^-0.301

We have managed to convert the original number of 0.5 to one with base 10 and a power (index) of "-0.301".

We can also similarly do the same for a base of 5. =>  0.5 = 5^y
Here we just "log" to base 5.

log₅ (0.5) = log₅ 5^y = y
-0.431       = y

Thus 0.5 = 5^-0.431

From the 2 examples above, you can see the wonders of having logarithm as a conversion tool.
Once you understand this principles, you will appreciate logarithm.

Cheers!
:D

Solving Maths Visually

.
Maths is interesting!
There are many exciting concepts and techniques one can apply.
Read on ...

There are many ways to solve a maths question.
Normally it involves taking many steps with related sequence.

However, there are also simple ways to handle a maths problem.

One such solving method is simply through visual steps.
This has no working at all.

What do I mean?

Let's look at an example.

Example:
Determine the angle B from the diagram below. Angle A is 40 degree.

Solution:

Angle B is same as angle A = 40 degree.

There is no working at all. Just the visual determination.

Concept of this trigonometrical question in geometry:
When 2 straight lines cross and meet at an angle to each other, the angle opposite to any one is the same.

As such, angle C is also equal to angle D.

This is visual maths.

Interesting?

:-D

Maths Symbols versus Real life Applications| Teaching style

'
Is maths interesting and is easy to learn or teach?

There are always argument over how to really capture maths students' attention and make them understand the principles and concepts in maths.

One school of thought is to approach the real life case studies.
Here actual applications are taught to make aware the usefulness of maths.
Problem-based case studies are planned into the curriculum to allow the learners to learn maths.

Another approach that is conventional is to pump in concepts and techniques using symbols and hard formulae.
Here students learn maths consists of symbols and their true meaning in the workings.
No real life indication is mentioned or just skimmed through. The focus is purely the use of maths tools to solve questions.

What are the advantages and disadvantages?

1) For real life approach, there are chances that the learners may couple their maths learning to only that particular application.
If train speed is mentioned or used, the students may only understand that the maths tools apply to train and not areoplane. The scope of aplication is a factor and serve to be a disadvantage.
The advantage is that the learners can relate to the usefulness of maths and will pay more attention and feel more fulfilled.

2) The advantage of pure symbolic approach in the conventional teaching method is that scope of the maths tools are wide. No tying down of the maths techniques to any specific area enables fredom of use.
The disadvantage, of course, is that the students may take a longer time to see that actual useulness of the maths tools and principles.

So what is the best method to learn maths?

I suppose the answer lies with the type of students and the topics to be taught.
No one way is best.
It has to be customised to suit the students or majority of them
Flexibility is thus the best methods and getting a good maths teacher who can read the minds of the students is the better choice.

If the classroom teacher is not up to expectation or has some constraints, it will be good to look for private tutors to brush up the maths learning. Note, classroom teaching does not cater for individual needs and this is a fact.

Happy maths learning.
Maths is interesting!
Don't forget it.

:-)

Simple Tricky Maths Pattern

.
Maths forms interesting patterns if you care to look around.

One such example is highlighted below:

12  11  10  09  08  07  06   <== row 1  (decreasing by 1)
01  02  03  04  05  06  07   <== row 2  (increasing by 1)

If you add the numbers in row 1 to that in row 2, you will get, surprisingly, the same answer throughout!

13  13  13  13  13  13  13

Why is it so?

It is a simple trick to the unwaries, actually.

Maths is not that difficult, to start off.
The answers to the above additions seem to be accidental in having the same number.
It is actually not accidental if you think abit.

The answers were intended to be 13 to begin with.

13 = 12 + 1
13 = 11 + 2
13 = 10 + 3  ....  and so on.

Though the additions seems magical with one row in ascending mode while the other row in descending mode, it is the visual maths trick that corrupts and confuses the mind.

If your principles of mathematics is good and strong, you will break through this simple trick in no time.

To excite young minds, this is a good one to try on them.

Happy playing with maths.
Maths is interesting!

:D

Application of Boolean OR operation

'
Besides the Boolean AND operation and its application, Boolean OR operation also serves a special application.

For the AND operation, which is equivalent to the maths multiplication, the use is for it to pull any untied input to a system to zero state.

For the OR operation, the concept is similar, except that now it is the maths addition.
Instead of pulling the untied input to the system to zero, you can set it to the other Boolean state, which is the "1".

In base 2 number system (Boolean), if the input is 000101, and you need the data to be all 1, just OR the input with 111111.

What you will get after this OR (adding) operation is 111111.

This is so due to the fact that 1 + 0 = 1.

Here you will see that option to set the untied input to 0 or 1 can be done using either the AND or OR operation of the Boolean system.

This is maths, if you are aware.
This is the application of maths after understanding the principles of maths operation.

Any comment?

:D

Boolean OR operation | Special Maths Addition

'
Boolean means 2-state operation.

OR means any one state that is valid will result in a positive outcome.

Example: When James or Mary come, the show will be start.

In maths, this translates to ADD, except only 2 states can occur (that is, on or off only).
1 = on, 0 = off (or vice versa).

What do I mean?

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 1

From the above addition, you will notice that as long as one state is ON, the result will be ON.
The operation is that of an "add".

Note: Since the operation is Boolean, it cannot go above 1 or 2 and above cannot exist. Only base 2 number (0, 1) can happen.

This field of maths is known as Boolean Algebra, a special maths operation used in digital electronics.

Caution:
If you are using this Boolean process, make it clear that the number base system is 2, otherwise 1 + 1 = 2!

1 + 1 = 2 in Boolean means that you go over the ceiling!

Another interesting maths concept, right?

Maths is interesting!

:D

Application of Boolean AND operation

'
Boolean, though, of 2 states, can still be useful.

The concepts of multiplication in this AND operation can be used to perform cancellation of unknowns.

Anything times zero gives zero.
This is a common knowledge.

Here we are talking about basic fundamental maths. Nothing difficult.

Cancellation of unknowns in the AND operation means that we are able to mask out the unwanted or redefine the unknowns to a definite state.
Here the unknown become a zero, a known state or condition.

A more specific application is in the masking of unconnected inputs to a processing system. If the inputs are scanned and compared for status updates, the inputs have to be accurate or of a known status. Otherwise, comparison results become meaningless.

Here the Boolean multiplication (AND) forces the unconnected input to a zero. This is then an accurate input status for comparison.

Here, maths is applied to technical application, and its usefulness become apparent.
This is the power of understanding maths , and is exactly what makes maths interesting.

.

What is Percentage?

'
Many people do understand what percentage is about.

If you see a sales offer with 50% discount, you will know that it is cheaper by half.
If the offer is with 40% less, you will also realise that is a good deal since it is almost half the original price.

But what is this percentage in detail?

A mark of 4 / 5 is reflected as a fraction.

A learned maths person will understand that it is (4 /5) x 100 = 80%

If you tell a student that he achieved 50 marks. Is that enough?

The information will not be enough as the total score is not known (unless using the default told before hand).

Thus a mark of 50 upon 50, and a mark of 50 upon 100 means different story altogether.

I believe you will agree totally!

A percentage will always reflect better information since 80% means 80 / 100.
The base of 100 is taken as the default.

The actual base number is immaterial in this matter.

An information of 50 marks compared to 50% showed the power of using percentage.

For absolute marks, you need to tell the total marks to form a complete message.
Using percentage, the fraction part of the calculation can be ignored.

Percentage and fraction are related. But percentage used a common base number (100) to commonise the value.

For comparison sake, percentage, thus stand a more proper way to tell the result.

Example:

Is it easy to tell the closeness between 49 / 56 and 46 / 56,
or is it easier to tell between 78% and 81% ?

The answer is obvious, I hope.

Maths is interesting, and mastering simple concept makes maths learning even more interesting.
Happy learning.

Cheers! :D

Meaning of Minus

'
To the ignorant, a "-" or minus is a strange symbol.

Is this more so when it is expressed as "-6", "-5 km", "-$3", etc.

What is then the true meaning of this "minus" sign?

Answer: It indicates a reversal of action of intention.

Maths is a useful tool to help explain this concept.

Example:

When a car moves forward by 5 km, it moves +5 km. ( By default, no sign means positive)
When it reverses by 5 km, it moves -5 km. The direction of movement reverses!

When a person gain $3, he has +$3 in his pocket.
When he loses $3, he has -$3.

From the 2 examples above, you can see that the "minus" sign is a reversal to the default.

If you "reverse" and then "reverse" again, you find yourself in the positive direction.
(-1) x (-1) = + 1

If you reverse 5 times, 5 x (-1) = -5 . You are facing the reverse to the default starting direction.

Now take note of this coming information.
If you reverse the car by 5m and another 5m, you reversed in total (-5) + (-5) = -10m.

If you reverse the car by 5m, followed by changing the direction and moving by another 5m, you moved (-5) + (+5) = 0m.
(Reverse direction followed by forward direction).

Does all these "reversing" cause a daze in you?

Do not despair.
Message is "When there is a reversal, put a minus in front of the number". Simply that!

I owe you $5 ==> -$5.
I gained $5 but lost $3 ==> + $5 + (- $3) = $2 (Action followed by another using "+").

Understand?

Hope this post on minus sign reduces your anxiety about this little maths symbol.

Cheers :-)

.

Power Law of Logarithm Explained

Logarithm study has a few formulae that are important and key to solving math questions.

By remembering them , you will be in line to solve logarithmic problems and, maybe, fast too.

However, what if you forget them?
Does it mean that you are not able to solve the question regardless of speed?

Do not despair.
As long as you are able to manage the 2 basic laws in logarithm, you are safe.

The product law and the quotient law are must for any students.

Why do I say that?

Let us take the Power Law and do a review.

n log x = log X ⁿ

Why is it so? What if you forget this law? Any problem?

These are the very queries any new learners exposed to logarithm will ask.

First allow me to go through the product law of logarithm.
log (XY) = log X + log Y

Here you see that the product of "X" and "Y" in logarithmic operation, becomes a "sum"of the individual logarithms.

4 log Y = log Y + log Y + log Y + log Y (adding up 4 of the log Y)

Using product law, you know that these 4 terms can be combined to log (Y x Y x Y x Y).

log (Y x Y x Y x Y) = log Y⁴

Now, you see, through the product law, you are able to equate the 4 log Y into log Y⁴ ,
meaning, 4 log Y = log Y⁴.

You see that you did not utilise the Power Law here,and yet is able to form this formula!
Amazing isn't it.

What is the message here?
The message is that, when you have the basic understanding in logarithmic principles, you will be able to twist and turn any given problems to come out a solution.

You had used the basic product law to discover this unique Power Law.
It is similar to other laws and also can be expanded to cover other maths topics too.

Do enjoy maths.
Do discover more exciting twists it presents wwith a bit of thinking.

Happy learning :-)

.

Time Calculation May Not Be Easy

.
Counting from 1 to 100 is normal for anyone. Just increment by 1. It's that simple.

But counting time may be another story for young math students.

Why is this so?

There is the seconds, the mintes, and the hours to handle. They differ by the umber 60.

The "carry over" through addition is the number SIX!

1 min 30 sec added by 40 sec gives .......

1 min 70 sec?

Here the 70 sec includes 60 sec + 10 sec. You need to understand that for time, 60 sec means one whole minute.
Thus, 70 sec = 60 sec + 10 sec ===> 70 sec = 1 min + 10 sec.

Therefore, 1 min 30 sec add 40 sec gives ==> 2 min 10 sec.

Compare this to adding 30 by 40. Answer = 70. No more analysis.

Time base on 1 hour = 60 min, 1 min = 60 sec concept.

To test true understanding of addition (and subtraction), time is a good gauge and tool to assess learners.

Test it on young kids today to annoy them... *#^&@!
Cheers!

.

Constants and Variables

In maths, you will encounter many "letters". This "letters" are the symbols used to form equations and mathematical expressions.

("letters" here may mean symbols like the theta, beta, etc)

Some of these symbols are constants and some represent variables.

Knowing what are constants and variables is crucial for mastery of maths learning.

What are constants?
Constants, as the word literally means, are items that have number that never change.

What are variables?
Variables are symbols that changes in value.

Example A:
y = 3 x + 2

y and x are variables, and 3 and 2 are obviously constants.


Example B:
log x = 5^y

x and y are variables , and 5 is constant.

Example C:
y = mx + c ( for straight line equation)

y and x are variables, and m and c are constants.

This may be confusing to some maths students when they start plotting graphs.
Here, the straight line is continuously moving with the value of x and y.

So why is the "m" identified as constant?

You need to know that "m" represents the "GRADIENT" of the line.
The line has the same slope at any value of x and y.
Thus "m" is a constant.
This is a typical concept that commonly goes wrong.

Therefore when you really understand what changes are considered "variables" and those that remain stable are known as "constants", you are in line for good maths study!

.

Complex Number Explained (2)

Complex number consists of real and imaginary terms to cover the concern that not all real number can satisfy any maths question.

An example was provided in this post (click the link for information).

What is then the different between this Real and Imaginary term?

In graphical form, the Real term is presented in the horizontal axis whereas the Imaginary term is along the vertical axis.

This is purposely so to has no impact of the Imaginary term on the Real term.
(Think in term of the cosine aspect of a pure Imaginary axis).
The overlapping portion of the Imaginary vertical part is ZERO on the Real axis.

See the diagram below for understanding.

The concept of this diagram is to allow learners visually see that the Real and Imaginary terms are unique in themselves and have no link to each other.

But another issue appears.

What is it?

Looking at the diagram, you will see that the complex number defined as Z = a + ib,
where "a" is the real term and "ib" is the imaginary,
will create directional value.

Some solution to specific maths problem requires the complete a + ib format.

Thus the introduction of complex number to offset the impossibility of solving equation using only real numbers, forces the angular dimension into the answer.

With this angular dimension coming into the mathematically picture, the principles of quadrant, as in the trigonometry studies, will be utilized to identify the various answers.

Complex number is then made "complex" mainly due to this directional information added to the normal Real numbers.

Do not be frighten off by this new addition, as, if you know very well how it comes about, you will welcome it. This imaginary term helps you solve many interesting maths equation that normal working cannot.

Love this complex number.

.

Simple Factor Multiplication

Multiplying is simple.

What is 4 x 3?
Answer is 4 x 3 = 12.

Simple?
Sure it is.

How about y(y - 1)?
Answer is y² - y.

Again simple? Sure.

But how about (y + 1)(2y + 3)?
Many of you may find this simple and basic.

But you may still come across some who did not grasp this factor multiplication.
Mistake still occur for this maths operation involving factors.

What is the mistake commonly seen?

(y + 1)(2y + 3) is given as (y)(2y) + (1)(3).
First term multiply by first term, second one multiply with the second one. That's all.
This is incorrect mathematically.

This is a misconception of what multiplication does.

Let me explain.
(y + 1)(2y + 3) can be interpreted as (y)(2y + 3) plus (1)(2y + 3).
This is key to this form of maths operation.

The second term (2y + 3) is multiplied by the first term "y" of the first factor (y + 1).
(2y + 3) is next multiplied by the second term "1" of the first factor.
The result of these two operations are then added up, since it is y add 1 (as reflected in the first factor).

The correct answer is then:
(y + 1)(2y + 3)
= (y)(2y) + (y)(3) + (1)(2y) + (1)(3)
= 2y² + 3y + 2y + 3
= 2y² + 5y + 3

Learn from the mistake, and do not repeat it.
This is the basic concept in learning from mistakes. They are our teacher.

Remember, maths is interesting!
A twist can be destructive or constructive.
That is where maths is special and challenging.

Cheers! :-)
.

Special Product A^2 - B^2 and A^2 + B^2

The product format A² - B² is a special form of algebraic expression.

It is equal to (A + B)(A - B).

Many other expressions, like the cos2A = cos²A - sin²B,
can also be expressed in the special product form, that is,

cos2A = (cos A + sin A)(cos A - sin A).

But, how about A² + B² ?

Can it be expressed in the (A + B)(A - B) format?

Why not.

However we need to deviate a bit from the norm.

We need to know the imaginary "i" in complex number system.
Click here for a review to it.

Since i² = -1, we can make use of this property for the special product A² + B².

Here it goes....

A² + B² = A² - (i²)(B²)

Using the basic of Indices, it can be modified to,

A²-(iB)²

This becomes, therefore, (A + iB)(A - iB).

In summary,

A² - B² = (A + B)(A - B), and

A² + B² = (A + iB)(A - iB).

The principles of the special product still holds regardless of the addition or subtraction operation between A square and B square.

Student: What an interesting twist is mathematics in this matter. Anything seems to be simple if we know the technique!

.