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Prove that the sum of two odd numbers is even.

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Last updated date: 24th Feb 2024
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IVSAT 2024
Answer
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Hint: To solve this question, we need to know the basic theory related to the number system. as we know a number is an arithmetic value that is used to express the quantity of an object and helps in making calculations. Here, we will have a look at the definition, examples, and properties of even and odd numbers in detail.

Complete step-by-step solution:
As we know, an even integer is defined as an integer which is the product of two and an integer while an odd integer is defined as an integer which is not even.
Let x and y be two odd numbers.
Then $x=2m+1$ for some natural number m and $y=2n+1$ for some natural number n.
Let’s take out their sum.
Thus $x+y=2m+1+2n+1=2(m+n+1)$
Here, with the addition 2 in there.
Therefore, $x+y$ is divisible by 2 and is even.
Here, we showed that if we add two odd numbers, the sum is always an even number.
To better convince ourselves that the given statement has some truth to it. We can test the statement with a few examples.
Now,
Let 1 and 3 be two odd numbers.
As we know, the sum of these two (1 and 3) is 4.
Which is divisible by 2 and as we know any number that can be exactly divided by 2 is called an even number.
Therefore, we conclude that the given statement is true.
Hence, We prove that the sum of two odd numbers is even.

Note: The sets of even and odd numbers can be expressed as follows,
Even = ${2k : k \in Z}$
Odd = ${2k + 1 : k \in Z}$
A formal definition of an even number is an integer of the form $n = 2k$, where k is an integer. An odd number is defined as an integer of the form $n = 2k + 1$. This classification applies only to integers. Non-integer numbers like $\dfrac{1}{2}$, 4.201, or infinity are neither even nor odd.