Examples of Real Numbers

Scientific Notations are generally used to represent the Numbers in mathematics that seem to be very insignificant or very big. This type of number will possess just 2 factors out which first one appears in between numbers 1 and 10. Second factor is usually written in form of an exponent of 10. For example, 1012 means exponent of 10 is 12. This is what we call as scientific representation of any number. We are calling two numbers as factors because numbers that are multiplied to each other to get a product are called as factors only.

Let us consider an example to understand the concept of converting any number into its scientific form: We have to represent 0 Point 0035 times 10 to the power of negative 2 in scientific notation. Remember first factor has to lie between 1 and 10. This makes necessary for us to keep on shifting the decimal point until we get a digit lying between 1 and 10. Here, in our example the number is 0.0035. We can see that digit that lies between 1 and 10 is lying to the right of decimal at thousands place. So, at least 3 times the decimal point has to be shifted to get desired factor. That is we have to multiply 0.0035 by 103 to get 3 in the units place to the left side of decimal. We get:
0.0035 * 103 / 103 = 3.5 * 10^-3,
Also number has to be multiplied by 10-2 and then final scientific notation will be decided. On multiplying 10-2 by 3.5 * 10-3, we get following result:
3.5 * 10^{- (3 + 2)} = 3.5 * 10^-5,
So, second factor is 10^-5 and scientific notation for desired number is given as:
3.5 * 10^-5.

In mathematical word problems involvement of operations like addition, subtraction, multiplication and division seem to be much in use. Word problems are basically expressed in the form of statements. These statements are essentially not mathematical expressions or Numbers. We ourselves have to understand complete word problem properly and then extracting the necessary information that can be used for evaluation purpose. Next part of word problems is the solution. In a word problem all dimensions for various quantities have to be handled in a proper way. If conversions are required, do them purposefully. Let us consider an example of such problems to understand their working in Math properly.
Suppose we have following example word problem:

Q. The sum of two numbers is 15 if one of the numbers is X, the other would be best represented by which of the following? Options are:
A.      X + 15
B.      X – 15
C.      15 – X
D.     X
Solution: Word problem says that numbers when added give result 15 and one number out of two is X. Let us assume that second number is given as Y. According to the word problem, only information we are available with is their sum, i.e.
X + Y = 15,
Taking 'X' to another side as it is also a known quantity; we get value of 'Y' as:
Y = 15 – X,

Thus correct representation of value of 'Y' is given by option C i.e. 15 – X. Thus we see, to find the value of one missing variable in the problem we need just relevant information. Similarly, for word problems involving more than 1 variables, we need that information.

We first arrange the product in form of column i.e hundreds, tens and ones and then proceed: 
    Hundreds       tens           ones
       4                   6               3
      X                    1               2
____________________________
                                                    We first multiply 463 by 2, we get
   
 
 Hundreds       tens           ones
       4                   6               3
      X                    1               2
____________________________
    8                   12                6.    Now we find that there is 12 on tens, so we retain 2 on tens place and carry 1 to hundreds place.
 
    Hundreds       tens           ones
       4                   6               3
      X                    1               2
____________________________
     9                    2                6    now we take 463 by 1 which is at tens place, by putting X at ones place
  46                     3                X
___________________________
 5 5                    5                6     We add both the product and get  the result

 So 463 * 12 = 5556.

We first arrange the product in form of column i.e. hundreds, tens and ones and then proceed:
             Hundreds    tens  ones
                   3            6       7
                  X                      3
____________________________ On multiplying by 3 we get
                                         21
  Here we get 21 at ones place so we retain 1 and carry 2 at tens place. This 2 is added to 18 at tens ( 6 * 3 ) + 2 = 20
 
          Hundreds    tens  ones
                   3            6       7
                  X                      3
____________________________ 
                            20          1
--------------------------------------------------
Here we get 20 at tens place so we retain 0 and carry 2 at hundreds place. This 2 is added to ( 3 X 3 ) = 9 + 2 = 11
 
 
          Hundreds    tens  ones
                   3            6       7
                  X                      3
____________________________ 
                  11          0          1
-------------------------------------------------
 So answer is 1101. 

We know  by zero property of multiplication of the Numbers, that any number multiplied by 0, the result is always 0
                 So 180 X 0 = 0 

We know by Identity Property of Multiplication, that when any number is multiplied by 1, the product is the number itself, so
=> 356 * 1 = 356. 

First we write the Numbers in their proper places and get:
          hundreds            tens     ones
              3                       8          5
              X                       2          0
________________________________
             0                        0          0   we first multiply all the digits by 0, which gives 0 at all the places.
 Now we multiply 385 by 2 . Before this  we put X at ones place.  so we get
 
        hundreds            tens     ones
              3                       8          5
              X                       2          0
________________________________
             0                        0          0 
 
             6                       16         X  This 16 is at tens place, so we retain 6 at tens place and carry 1 to hundreds place
_______________________________
 
 
hundreds            tens     ones
              3                       8          5
              X                       2          0
________________________________
           1+ 0                     0          0 
 
             6                       6         X  
________________________________
             7                       6          0 
________________________________

So, answer is 760.  

In such questions, we first multiply 135 by 9 and then multiply it by 100, i.e. we put 2 zeroes with the answer on the right hand side.
 Lets see how to proceed:
   hundreds     tens      ones
     1                 3            5
      X                              9
________________________
                                  45, we find ( 5 X 9 = 45 ), we write 5 at tens place and carry 4 to be added to tens place.
 
 hundreds     tens      ones
     1                 3            5
      X                              9
________________________
                    (27 +4)      5 
 hundreds     tens      ones
     1                 3            5
      X                              9
________________________
                       31           5
_________________________
Now from 31, 1 is left at tens place and 3 is carried over to hundreds place
 hundreds     tens      ones
     1                 3            5
      X                              9
________________________
    (9+3)             1         5 
 ________________________
Adding 9 + 3 = 12 , so we get
 
 hundreds     tens      ones
     1                 3            5
      X                              9
________________________
    12               1            5  
________________________
  
Now we multiply 1215 with by 100, we get:
 = 1215 X 100,
 = 121500.

In such problems, when we have 300 * 20,
So, it can be written as 3 * 2 * 1000,
= 6 * 1000,
= 6000. 

We find the sum very complicated and rather difficult to solve. But remember when any number is multiplied by 0, the result is always 0. This property of multiplication is called the property of multiplication with 0. So we get the result of this problem without actually solving the question,
= 0. 

We know that any number multiplied by 1 gives the same number as the result. This is called the Identity Property of Multiplication.
Thus, when 840 * 1 = 840.

 Here we first multiply 240 by 3, which is at ones place and then multiply 240 by 20 (as 2 is at tens place, then we add both the digits to get the desired result).
             Hundreds   tens   ones
               2                4        0
                                X         3
 ___________________________
              7               2          0 (we get 7   as 3 * 2 = 6 add 1, which is carried from tens place)
Now 
 
      Hundreds   tens   ones
          2                4        0
                          X         2
 ___________________________
         4               8          0
 Thus 240 X 20 = 4800
 
Now to get the result of 240 X 23, we add 4800 + 720 and get,
  
      4  8  0  0
  +      7  2  0
    ___________
      5  5  2   0  
_______________

We first arrange the product in form of column of tens and ones and then proceed: 
    Tens      Ones
    2             3
   X              5
____________
              15, we write 5 in ones place and take 1 as a carry over, so it becomes
 
    Tens      Ones
    2             3
   X              5
____________
   11          5    

Here we multiply 2 * 5 = 10 and add 1 to it, so at ones place we get 5 and at tens place we get 11.
So  we have 23 X 5 = 115.

In such cases, we need not to apply column method to get the solution to this.
As any number is multiplied by 1, the result is same number. This property is called the Identity Property of Multiplication.
So we get:
= 6009 * 1,
= 6009.

In such problems we can write them as 
345 * 2 * 100,
So first we solve 345 * 2
Write the digits as per their place value and then multiply:
 
 
        Hundreds    Tens    Ones
            3               4         5
            X                          2
      ______________________
         6                 8         10, from 10, 0 retains the ones value and 1 is carried over to tens place,
  We get,
    Hundreds    Tens    Ones
            3               4         5
            X                          2
      ______________________
         6                 9           0
      ______________________
 Now we multiply 690 by 100,
 = 690 * 100,
 = 69000.