Understanding Sphere Properties with a Rational Radius

This question asks us to consider a sphere where the radius is a rational number and determine if its surface area and volume are also rational numbers. To answer this, we need to recall the formulas for the surface area and volume of a sphere and understand the properties of rational and irrational numbers.

What are Rational and Irrational Numbers?

• A rational number is any number that can be expressed as the ratio or fraction \(\frac{p}{q}\) of two integers, where \(p\) is an integer and \(q\) is a non-zero integer. Examples include \(\frac{1}{2}, 3, -4, 0, \frac{22}{7}\).
• An irrational number is a number that cannot be expressed as a simple fraction \(\frac{p}{q}\). Their decimal representations are non-terminating and non-repeating. A famous example is \(\pi\).

A key property to remember is that the product of a non-zero rational number and an irrational number is always irrational.

Formulas for Sphere Surface Area and Volume

Let \(r\) be the radius of the sphere.

• The formula for the surface area (\(A\)) of a sphere is: \(A = 4 \pi r^2\).
• The formula for the volume (\(V\)) of a sphere is: \(V = \frac{4}{3} \pi r^3\).

We are given that the radius \(r\) is a rational number.

Analyzing Statement 1: Surface Area

Statement 1 says: "Its surface area is rational."

The surface area is \(A = 4 \pi r^2\). Since \(r\) is a rational number, let's say \(r = \frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\). For a real sphere, \(r\) must be positive, so \(p \neq 0\).

Substituting the rational radius into the formula:

\[A = 4 \pi \left( \frac{p}{q} \right)^2 = 4 \pi \frac{p^2}{q^2}\]

We can rewrite this as:

\[A = \left( \frac{4p^2}{q^2} \right) \pi\]

Here, \(\frac{4p^2}{q^2}\) is a number formed by integers \(4, p, q\). Since \(p \neq 0\) and \(q \neq 0\), \(p^2 \neq 0\) and \(q^2 \neq 0\). Therefore, \(\frac{4p^2}{q^2}\) is a non-zero rational number.

The surface area \(A\) is the product of a non-zero rational number (\(\frac{4p^2}{q^2}\)) and the number \(\pi\). We know that \(\pi\) is an irrational number.

According to the property mentioned earlier, the product of a non-zero rational number and an irrational number is irrational.

Thus, the surface area \(A\) is irrational.

Therefore, statement 1 is incorrect.

Analyzing Statement 2: Volume

Statement 2 says: "Its volume is rational."

The volume is \(V = \frac{4}{3} \pi r^3\). Again, since \(r\) is a rational number, let \(r = \frac{p}{q}\) where \(p\) and \(q\) are non-zero integers.

Substituting the rational radius into the formula:

\[V = \frac{4}{3} \pi \left( \frac{p}{q} \right)^3 = \frac{4}{3} \pi \frac{p^3}{q^3}\]

We can rewrite this as:

\[V = \left( \frac{4p^3}{3q^3} \right) \pi\]

Here, \(\frac{4p^3}{3q^3}\) is a number formed by integers \(4, 3, p, q\). Since \(p \neq 0\) and \(q \neq 0\), \(p^3 \neq 0\) and \(q^3 \neq 0\). Therefore, \(\frac{4p^3}{3q^3}\) is a non-zero rational number.

The volume \(V\) is the product of a non-zero rational number (\(\frac{4p^3}{3q^3}\)) and the number \(\pi\). We know that \(\pi\) is an irrational number.

The product of a non-zero rational number and an irrational number is irrational.

Thus, the volume \(V\) is irrational.

Therefore, statement 2 is incorrect.

Conclusion

Based on our analysis, both statement 1 (surface area is rational) and statement 2 (volume is rational) are incorrect when the radius of the sphere is a rational number. The presence of the irrational number \(\pi\) in the formulas for surface area and volume makes these quantities irrational whenever the radius is non-zero.

The correct answer is that neither statement 1 nor statement 2 is correct.

Revision Table: Sphere Calculations

Additional Information: Rationality and \(\pi\)

The number \(\pi\) is a fundamental mathematical constant representing the ratio of a circle's circumference to its diameter. It is a transcendental number, which is a type of irrational number. This means it is not a root of any non-zero polynomial equation with integer coefficients.

Because \(\pi\) is irrational, any expression that involves a non-zero rational multiple of \(\pi\) will also be irrational. This is why the surface area (\(4 r^2 \times \pi\)) and volume (\(\frac{4}{3} r^3 \times \pi\)) of a sphere with a non-zero rational radius (\(r\)) are always irrational. The terms \(4r^2\) and \(\frac{4}{3}r^3\) are rational when \(r\) is rational, but multiplying by \(\pi\) makes the result irrational.