Question

List all the elements of the following set:
B=\left\\{ x:x=\dfrac{1}{2n-1},1\le n\le 5 \right\\}

Answer 1

Hint: In order to solve this question, we need to know few things about the set builder from a set of type B = { f(x): x = something, range}, then f(x) represents the function which will give us the elements of set B, x = something represents the condition through which we will find the possible value of x and range represents the possible value for which we will get x.

Complete step-by-step answer:

In this question, we have to list all the elements of the set B=\left\\{ x:x=\dfrac{1}{2n-1},1\le n\le 5 \right\\}. Here, we can say that f(x) = x, we will get the values of x by using the relation $x=\dfrac{1}{2n-1}$ and the values of n for which we will get the value of x is $1\le n\le 5$. Now, let us calculate the possible value of x for n = 1, 2, 3, 4, 5. So, we will get,
For n = 1, x = $\dfrac{1}{2\left( 1 \right)-1}=\dfrac{1}{1}$
For n = 2, x = $\dfrac{1}{2\left( 2 \right)-1}=\dfrac{1}{3}$
For n = 3, x = $\dfrac{1}{2\left( 3 \right)-1}=\dfrac{1}{5}$
For n = 4, x = $\dfrac{1}{2\left( 4 \right)-1}=\dfrac{1}{7}$
For n = 5, x = $\dfrac{1}{2\left( 5 \right)-1}=\dfrac{1}{9}$
And we know that f(x) = x. So, the possible value of f(x) are $1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9}$. Hence the elements of the set B are \left\\{ 1,\dfrac{1}{3},\dfrac{1}{5},\dfrac{1}{7},\dfrac{1}{9} \right\\}.

Note: The possible mistake one can do is by taking the values of n other than the whole number. Also, by assuming the function $f\left( x \right)= \dfrac{1}{2x-1}$ where $x\in N$ and $1\le x\le 5$. In this case, also, we will get the same answer which we got through the conventional method.