If two numbers a and b are odd, then prove that their sum a + b is even
In the early days, the ancestors did not know counting and used to represent different objects by symbols and letters. Later they used numbers to represent things and that leads to the formation of the number system. Today if one can count the things then that could be possible because of the number system. Further number systems are categorized into different sub-parts like decimal number system, binary number system, octal number system, hexadecimal number system. The decimal number system contains different types of numbers based on their characteristics like natural numbers, whole numbers, integers, rational numbers, irrational numbers, etc. In whole numbers, there are even and odd numbers, let’s learn about even numbers,
All the numbers apart from even numbers in the category of whole numbers are odd numbers. Odd numbers are the ones that cannot be divided by 2, for instance, 5, 7, 9, etc. all these numbers leave a number 1 as a remainder when divided by 2.
If a number is completely divisible by 2, leaving 0 as remainder then that number is known as an even number. Even number is the subset of integers. Successive +2 or -2 to the even number is also an even number. Examples of even numbers are -6, -8, 10, 6, etc. 0 is also an even number because when 0 is divided by 2, the remainder is 0.
If two numbers a and b are odd then prove that their sum is even.
If ‘a’ and ‘b’ are odd numbers it means that the remainder of a ÷ 2 will be one and similarly b is also an odd number so the remainder of b ÷ 2 will also be one.
The remainder of (a ÷ 2) = 1
The remainder of (b ÷ 2) = 1
Now to prove that remainder of (a + b) is also one, divide it by 2 then we can say that a + b is also an odd number.
Divide (a + b) by 2.
= (a + b) ÷ 2
By using distributive property, (x + y) ÷ z = (x ÷ z) + (y ÷ z)
= (a ÷ 2) + (b ÷ 2)
As discussed the remainder of (a ÷ 2) and the remainder of (b÷2) is zero. So it can be concluded that the remainder of (a ÷ 2) +(b ÷ 2) will be 1 + 1, i.e. 2, and since 2 is itself completely divisible by 2, the final remainder will again be 0.
Divide a + b by 2, the remainder is 0 and it fulfills the condition of an even number so can say that a + b is an even number.
Question 1: If 4 and 6 is an even number then prove that their sum will also be an even number.
Divide a number by 2 and get 0 as remainder then that number is known as an even number.
Divide 4 by 2, we got 2 as the quotient and 0 as the remainder.
Divide 6 by 2 we got 3 as the quotient and 0 zero as the remainder.
The summation of 4 and 6 is 10.
Divide 10 by 2, it is gotten 5 as quotient and 0 as remainder. So it can be concluded that 10 is also an even number.
Hence proved, the summation of two even numbers is also an even number.
Question 2: Prove that sum of -6 and 2 is an even number.
Divide a number by 2 and got 0 as remainder then that number is known as an even number.
Divide -6 by 2 we got -3 as quotient and 0 as remainder.
Divide 2 by 2, we got 1 as the quotient and 0 as the remainder.
Sum of -6 and 2,
= (-6) + (2)
= -6 + 2
When we divide -4 by 2, we got 0 as the remainder so -4 is an even number.