Math Examples of Properties of Real Numbers

We first find the multiplicative inverse of a real number Now to get the multiplicative identity of any real number, we multiply 7 and its multiplicative inverse i.e. 1/7. We get, = 7 * (1/7), = 1. So we find that multiplicative identity of 7 is 1.

We first find the additive inverse of a real number 2 is -2. Now to get the additive identity of any real number, we add 2 and its additive inverse. We get, = 2 + ( -2 ) = 0 So we find that additive identity of 2 is 0.

We observe that if we multiply 1 with the given number i.e. 3.4, then the product is the original real number. So, we get 3.4 * 1 = 3.4 Moreover we also know that if the multiplicative inverse of any real number is multiplied by the original number, then we get the multiplicative identity as the result. Let us try it for the given real number 3.4 First we write 3.4 as 34/10 We observe that its multiplicative inverse is 10 / 34 Now multiplying both real Numbers we get: ( 34/10) * (10/34) = 1 So we conclude that 1 is the multiplicative identity for 3.4 If we try it for any other real number , we get the same answer. So 1 is the multiplicative identity for all the Real Numbers.

Here we see that if zero ( 0) is added to 6/5, then ( 6 / 5) + 0 = 6 / 5 More over we can also check that if the additive inverse of any real number is added to the given number, the result we get is additive identity of the number. Let us check the same: We see that the additive inverse of 6 / 5 is -6 / 5. So, (6/ 5) + (-6/5) = (6/ 5) - (6/5) = 0. So it is the additive identity of 6/5 or we conclude that 0 is the additive identity for any real number.

We have the pair of real Numbers 2 / 5 and 2. To find the real numbers between these two numbers, we first add the two numbers and then divide them by 2 Sp we get, = ( 2 / 5 ) + 2, = ( 2 / 5 ) + ( 2 /1 ). Taking L.C.M of 5 and 1 as 5, we first make the denominator same For this, we multiply the numerator and the denominator of 2 by 5, we get, = ( 2 / 5 ) + 2*5/5 == ( 2 / 5 ) + ( 10 / 5 ) = ( 2 + 10 ) / 5 = 12 / 5 Now dividing the above number by 2, we get the real number which lies between = ( 2 / 5 ) and 2 = ( 12 / 5 ) ÷ 2 = (6 / 5 ) Now if we proceed in the same way we observe that there exist a real number between 2/5 and 6 / 5. Also there exist a real number between 6/5 and 2. This check can go up to an endless process and we will find a new number every time. Thus this list is endless and so there are infinite numbers between the given two numbers. This property of real numbers, where we can find infinite real numbers between any two real numbers is called the Density Property of Real Numbers.

If we look at these two Numbers as Natural Numbers, we observe that there exist no number between two numbers, but it we take these numbers as real numbers, We find the number between 1 and 2 by = ( 1 + 2 ) /2, = 3/2. = 1.5, so we observe that 1.5 is a real number which exist between 1 and 2. Now in the same way we again check how many real numbers exist between 1 and 1.5. For this again we apply the same formula and get, = ( 1 + 1.5 ) /2, = 2.5 / 2 = 1.25. Thus 1.25 is a real number between 1 and 1.5, so it also lies between 1 and 2 also. In the same way we can proceed and see that this chain is endless. Thus we come to a conclusion that there are infinite real numbers between two real numbers. We also observe that these numbers cannot be counted, So we come to a conclusion that real numbers are dense, So it satisfies the Density Property of Real Numbers.

For solving this real number expression by distributive property we need to follow the steps given below:

Step 1: Write the given expression,

 y= 5(x +1) + 2(y+1),

Step 2: Now we simplify the given real Numbers expression,

y = 5x +5 +2y +2,

y =   5x +2y +7.

Step 1: Write the given real Numbers expression,

y=  2(5 + 3)

Step 2: Now we simplify the given Real Numbers expression,

y = 10 + 6

y = 16

Step 1: Write the given real number expression,

 y= 8p²– 2(3p² +1)+ p,

Step 2: Now, we simplify the above expression,

y = 8p² – 6p² -2 +p,

y = 2p² + p -2.

Step 1: Write the given equation,

y= 5s² + 4s(s+1),

Step 2: Now, we simplify the given real Numbers equation,

y = 5s² + 4s² +4s,

y = 9s² + 4s.

Step 1: Write the given expression,

y= (5r+1) –(r-4) –r,

Step 2: Now, we simplify the given real Numbers expression,

y = 5r +1 – r + 4 -r,

y =   3r + 5.

For solving the above expression by distributive property we need to follow the following steps:

Step 1: Write the given expression,

y= = (n-1)- n(n+4) +n,

Step 2: Now, we simplify the given real Numbers expression,

y = n - 1 – n² – 4n + n,

y = -n² -2n -1.

Here, we are solving this question by using Associative Property but final answer don’t get affected 

=>   (5+4) + 10 =5+(4 + 10),

=>  9 +10 =5 + 14,

=>19 = 19.

In above example we can see that by changing the Position of brackets we can perform the calculation without affecting the final answer.

In this question, given expression is a Combination of real Numbers in which we are using different operators. Now, we use Associative Property to change the Position of the brackets, which doesn't make any change in the final answer.    

Suppose x=1,y =2, a=3 then,

 => 5(1) +(7*2 + 4*3) =(5(1) + 7*2) + 4*3,

 =>   31 = 31.

So answer is 5x +(7y +4a).

Apply commutative law on the given problem, we get:

5*(12) = 60 and 6*(13)=78,

We can rewrite our question as:

60 +78,

Now, apply Commutative Property:

a+b = b+a

Here 60 is the given value for a and 78 is the given value for b.

We will take left hand side:

60 +78

=138

Take right hand side,

78+60,

=138.

Here, both the left hand side and right hand side are equal.