Answer
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Hint: For solving this question we will take $\left( 2n+1 \right)$ and $\left( 2k+1 \right)$ as two odd integers where $n$ and $k$ are from the set of whole numbers. After that, we find the product $\left( 2n+1 \right)\left( 2k+1 \right)$ and then, we will try to prove the result by checking whether the result is divisible by 2 or not. Then, we will give the final answer easily.
Complete step-by-step solution -
Given:
We have to find whether the product of two odd numbers is even or odd.
Now, before we proceed we should know that any positive integer which is divisible by 2 is an even integer and which are not divisible by 2 are odd integers.
Now, let $\left( 2n+1 \right)$ and $\left( 2k+1 \right)$ are two odd numbers where $n$ and $k$ are from the set of whole numbers.
Now, we will multiply $\left( 2n+1 \right)$ and $\left( 2k+1 \right)$ . Then,
$\begin{align}
& \left( 2n+1 \right)\left( 2k+1 \right)=2n\times 2k+2n+2k+1 \\
& \Rightarrow \left( 2n+1 \right)\left( 2k+1 \right)=4nk+2n+2k+1 \\
\end{align}$
Now, in the above equation, we will take 2 as common from the terms $4nk$ , $2n$ and $2k$ . Then,
$\left( 2n+1 \right)\left( 2k+1 \right)=2\left( 2nk+n+k \right)+1$
Now, from the above result, we conclude that the product of any two odd integers will be of the form $2\left( 2nk+n+k \right)+1$ , where $n$ and $k$ are from the set of whole numbers.
Now, as per our assumption $n$ and $k$ are from the set of whole numbers. So, $2nk+n+k$ will also be an integer.
Now, let $2nk+n+k=A$ , where $A$ is any positive integer. Then,
$\begin{align}
& \left( 2n+1 \right)\left( 2k+1 \right)=2\left( 2nk+n+k \right)+1 \\
& \Rightarrow \left( 2n+1 \right)\left( 2k+1 \right)=2A+1 \\
\end{align}$
Now, from the above result, we conclude that the product of any two odd integers will be of the form $2A+1$ , where $A$ is any positive integer. So, we can say that $2A$ will be an even integer. Then, we conclude that the product of any two odd integers will be the sum of an even integer and 1.
Now, as we know that the sum of an even and odd integer is always odd. And from the above result, we conclude that the product of any two odd integers will be the sum of an even integer and 1.
Thus, from the above result, we conclude that the product of two odd integers will always be an odd integer.
Hence, (b) will be the correct option.
Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to find the correct answer. Moreover, we should remember this answer as a result and for checking our answer, we should take numerical examples for a better understanding of the concept of the number system.
Complete step-by-step solution -
Given:
We have to find whether the product of two odd numbers is even or odd.
Now, before we proceed we should know that any positive integer which is divisible by 2 is an even integer and which are not divisible by 2 are odd integers.
Now, let $\left( 2n+1 \right)$ and $\left( 2k+1 \right)$ are two odd numbers where $n$ and $k$ are from the set of whole numbers.
Now, we will multiply $\left( 2n+1 \right)$ and $\left( 2k+1 \right)$ . Then,
$\begin{align}
& \left( 2n+1 \right)\left( 2k+1 \right)=2n\times 2k+2n+2k+1 \\
& \Rightarrow \left( 2n+1 \right)\left( 2k+1 \right)=4nk+2n+2k+1 \\
\end{align}$
Now, in the above equation, we will take 2 as common from the terms $4nk$ , $2n$ and $2k$ . Then,
$\left( 2n+1 \right)\left( 2k+1 \right)=2\left( 2nk+n+k \right)+1$
Now, from the above result, we conclude that the product of any two odd integers will be of the form $2\left( 2nk+n+k \right)+1$ , where $n$ and $k$ are from the set of whole numbers.
Now, as per our assumption $n$ and $k$ are from the set of whole numbers. So, $2nk+n+k$ will also be an integer.
Now, let $2nk+n+k=A$ , where $A$ is any positive integer. Then,
$\begin{align}
& \left( 2n+1 \right)\left( 2k+1 \right)=2\left( 2nk+n+k \right)+1 \\
& \Rightarrow \left( 2n+1 \right)\left( 2k+1 \right)=2A+1 \\
\end{align}$
Now, from the above result, we conclude that the product of any two odd integers will be of the form $2A+1$ , where $A$ is any positive integer. So, we can say that $2A$ will be an even integer. Then, we conclude that the product of any two odd integers will be the sum of an even integer and 1.
Now, as we know that the sum of an even and odd integer is always odd. And from the above result, we conclude that the product of any two odd integers will be the sum of an even integer and 1.
Thus, from the above result, we conclude that the product of two odd integers will always be an odd integer.
Hence, (b) will be the correct option.
Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to find the correct answer. Moreover, we should remember this answer as a result and for checking our answer, we should take numerical examples for a better understanding of the concept of the number system.
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