Question

Write the set of all positive integers whose cube is odd.

Answer 1

Hint: In this question, we have to form the set of all positive integers, whose cubes are odd. Let us consider an example, 1 is a positive integer, so its cube is ${{\left( 1 \right)}^{3}}=1\times 1\times 1=1$, which is an odd number, so it will be considered for the set. If we look at 2, then ${{\left( 2 \right)}^{3}}=2\times 2\times 2=8$, which is an even number, so it will not be considered. So, we will try to find any general pattern which can represent the positive integers whose cube is odd.

Complete step-by-step answer:

We have been asked in the question to write a set of all positive integers whose cube is odd. We know that positive numbers are those numbers that start from 0 to infinity. So, to find the positive integers whose cubes are odd, we will need to find the cube of these numbers and check if they are odd or not. We will consider the numbers for the set only if their cubes are odd. For that, we will find the cube of the positive numbers one by one from 1 to 6 and try to find if there is a pattern among them, so that we can represent them. So, let us start with 1, so we get,
${{\left( 1 \right)}^{3}}=1\times 1\times 1=1$, which is an odd number.
${{\left( 2 \right)}^{3}}=2\times 2\times 2=8$, which is an even number, so it will not be considered.
${{\left( 3 \right)}^{3}}=3\times 3\times 3=27$, which is an odd number.
${{\left( 4 \right)}^{3}}=4\times 4\times 4=64$, which is an even number, so it will not be considered.
${{\left( 5 \right)}^{3}}=5\times 5\times 5=125$, which is an odd number.
${{\left( 6 \right)}^{3}}=6\times 6\times 6=216$, which is an even number, so it will not be considered.
So, we can see from the above that the cube of every odd number is an odd number. We know that the odd numbers can be represented generally as (2n + 1), where n is a natural number.
Therefore we can represent the set of positive numbers whose cubes are odd as, \left\\{ 2n+1:n\in z,n>0 \right\\} in the set builder form and as, \left\\{ 1,3,5,7\ldots \ldots \infty \right\\} in roster form.

Note: The students usually get confused when they have to convert the odd numbers to a general form, because of which they are unable to form the set in the set builder form for this question. So, the students should be aware of the general representation of odd numbers, that is (2n + 1), so that they can write the set builder form easily.