Question

Find the positive number smaller than \[\dfrac{1}{{{2}^{1000}}}\] and justify your answer.

Answer 1

Hint: We solve this problem by finding the value for which the number \[\dfrac{1}{{{2}^{1000}}}\] nearly equals to.
We find the number that \[\dfrac{1}{{{2}^{1000}}}\] is tending to so that we can find the positive numbers that are less than that tending number keeping in mind that the positive number is an integer.
For finding the number for which the \[\dfrac{1}{{{2}^{1000}}}\] is tending, we find the each and every power of \[\dfrac{1}{2}\] to find the general value of \[\dfrac{1}{{{2}^{1000}}}\]
We use one formula of exponents that is
\[\dfrac{1}{{{x}^{a}}}={{\left( \dfrac{1}{x} \right)}^{a}}\]

Complete step-by-step solution
We are asked to find the positive number that is less than \[\dfrac{1}{{{2}^{1000}}}\]
We know that the value of \[\dfrac{1}{2}\] in decimal form is 0.5
Now let us find the square of \[\dfrac{1}{2}\] then we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}={{\left( 0.5 \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}=0.25 \\
\end{align}\]
Now, let us find the cube of \[\dfrac{1}{2}\] then we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{3}}={{\left( 0.5 \right)}^{3}} \\
 & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{3}}=0.125 \\
\end{align}\]
Now, let us find the value of \[{{4}^{th}}\]power of \[\dfrac{1}{2}\] then we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{4}}={{\left( 0.5 \right)}^{4}} \\
 & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{4}}=0.0625 \\
\end{align}\]
Similarly, let us find the \[{{5}^{th}}\] power of \[\dfrac{1}{2}\] then we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{5}}={{\left( 0.5 \right)}^{5}} \\
 & \Rightarrow {{\left( \dfrac{1}{2} \right)}^{5}}=0.03125 \\
\end{align}\]
Here, we can see that as we increase the power of \[\dfrac{1}{2}\] it is more and more reaching 0
So, we can say that the \[{{1000}^{th}}\] power of \[\dfrac{1}{2}\] is nearly equal to 0 that is
\[\Rightarrow {{\left( \dfrac{1}{2} \right)}^{1000}}\simeq 0\]
We know that the formula of exponents that is
\[\dfrac{1}{{{x}^{a}}}={{\left( \dfrac{1}{x} \right)}^{a}}\]
By using this formula we get
\[\Rightarrow \dfrac{1}{{{2}^{1000}}}\simeq 0\]
Here, we can see that the value of is nearly equal to 0
We know that there will be no positive number that is less than 0
So, we can conclude that the there is no number that is less than \[\dfrac{1}{{{2}^{1000}}}\]

Note: Students may make mistakes in understanding the question.
We are asked to find the positive number that is less than \[\dfrac{1}{{{2}^{1000}}}\]
Here the positive number represents the integer because we are not mentioned the type of number we need to find. If it is mentioned that to find the positive rational number then we have an infinite number of rational numbers that are less than \[\dfrac{1}{{{2}^{1000}}}\]
The rational numbers that are less than \[\dfrac{1}{{{2}^{1000}}}\] can be obtained by increasing the power of 2 in \[\dfrac{1}{{{2}^{1000}}}\] that is the numbers \[\dfrac{1}{{{2}^{1001}}},\dfrac{1}{{{2}^{1002}}},.....\] all are less than \[\dfrac{1}{{{2}^{1000}}}\]
But when we have not mentioned the type of number to be found then we assume it as the integer.