Types of numbers

To answer this, we need to remember how the set of real numbers $\mathbb{R}$ is arranged. In The number system, real numbers are separated into two clear categories that do not overlap.

In other words, $\mathbb{R}$ is split into two disjoint sets:

1. Rational numbers ($\mathbb{Q}$)
2. Irrational numbers

Because every real number must be either rational or irrational, a real number that is not irrational must belong to the other group.

$\mathbb{R}=\mathbb{Q}\cup \text{Irrational numbers}$

Since these two parts of the partition share no elements, anything in $\mathbb{R}$ that is not irrational must be in $\mathbb{Q}$.

Now consider why some choices don’t work. Even though integers and natural numbers are not irrational, they don’t include all non-irrational real numbers.

• Option B: Integers $\mathbb{Z}$ are inside $\mathbb{Q}$, but they leave out numbers like $\frac{1}{2}$, which is also not irrational.
• Option C: Natural numbers $\mathbb{N}$ are an even narrower set, containing only counting numbers (and sometimes zero), so they miss many non-irrational values.

Option D is also removed because it describes irrational numbers, while the question asks for numbers that are not irrational.

So, using the definition of how $\mathbb{R}$ is split and eliminating the other options, the set of all real numbers that are not irrational is the set of rational numbers $\mathbb{Q}$.

To solve this, focus on the structure of the real numbers $\mathbb{R}$. In The number system, we learn that each real number is either rational (can be expressed as a fraction) or irrational (cannot be expressed as a fraction).

The question is asking for the set of all numbers that are not irrational within $\mathbb{R}$.

Option D can be crossed out immediately. If we want numbers that are not irrational, then the set of irrational numbers is exactly the set we do not want.

Check Option C: Natural numbers $\mathbb{N}$ (often $0,1,2,\mathrm{3...}$).

Natural numbers are not irrational, but they are far from the full collection. For instance, $-5$ and $0.5$ are also not irrational, yet neither is in $\mathbb{N}$. Since Option C is only a subset, it must be eliminated.

Now look at Option B: Integers $\mathbb{Z}$ ($...,-2,-1,0,1,2,...$).

Integers are also not irrational, but they still don’t cover every non-irrational real number. Values like $\frac{2}{3}$ or $1.25$ are not irrational, but they are not integers. So Option B is also eliminated.

After eliminating the other options, Option A remains: the rational numbers $\mathbb{Q}$.

By definition, $\mathbb{R}$ is made from rational and irrational numbers. So if a real number is not irrational, it must be rational. This includes natural numbers, integers, fractions, and terminating/repeating decimals—everything in $\mathbb{R}$ that isn’t irrational.

To answer this, we need the conditions from The number system topic. We want a number $x$ that meets both:

1. $x$ is an integer ($x\in \mathbb{Z}$).
2. $x$ is not a natural number ($x\notin \mathbb{N}$).

The question tells us the natural numbers are: $\mathbb{N}=\{1,2,3,\dots \}$

Integers include negative whole numbers, zero, and positive whole numbers: $\mathbb{Z}=\{\dots ,-2,-1,0,1,2,\dots \}$

Consider $0.5$.

It is a decimal and not a whole number. Integers are only whole numbers (and their negatives, plus zero). So $0.5\notin \mathbb{Z}$, meaning it does not satisfy the first requirement.

Now look at $\pi$.

$\pi$ is an irrational number (about $3.14159\dots$). Since it is not a whole number, it is not an integer. Therefore, $\pi \notin \mathbb{Z}$, so it cannot be the answer.

Next, evaluate $\sqrt{4}$.

Simplify: $\sqrt{4}=2$

The number $2$ is an integer ($2\in \mathbb{Z}$), but it is also in the natural numbers because $\mathbb{N}=\{1,2,3,\dots \}$. The question needs a value that is not in $\mathbb{N}$, so this option does not work.

Finally, check $-8$:

1. Integer? Yes. It is a negative whole number, so $-8\in \mathbb{Z}$.
2. Natural number? No. Natural numbers begin at $1$ ($1,2,3,\dots$), so $-8\notin \mathbb{N}$.

So this option matches both requirements.

To solve this, remember the meanings of the sets from The number system:

• Integers ($\mathbb{Z}$): whole numbers including negatives, zero, and positives (e.g., $\dots ,-2,-1,0,1,2,\dots$).
• Natural numbers ($\mathbb{N}$): by the convention given, these are the counting numbers $\{1,2,3,\dots \}$.

We need a number that is in $\mathbb{Z}$ but not in $\mathbb{N}$.

First, the number must be an integer ($\mathbb{Z}$). Check the options:

• Option B: $0.5$ is a decimal number, so it is not a whole number.
• Option C: $\pi$ is irrational (about $3.14159\dots$), so it is also not a whole number.

Because neither is an integer, neither can be correct.

Next, the correct choice must not be in $\mathbb{N}$. From what remains:

• Option D: $\sqrt{4}$ simplifies to $2$.
• Since $2$ is a positive counting number, it is in $\mathbb{N}$.

So Option D fails the second condition.

That leaves Option A: $-8$.

1. $-8$ is an integer, so $-8\in \mathbb{Z}$.
2. Natural numbers start at $1$ using the given convention, so $-8\notin \mathbb{N}$.

Therefore, Option A is the correct answer.

To decide this, start by remembering the meaning of a rational number from The Number System. A number is rational if it can be written as:

$\frac{p}{q}$

where $p$ and $q$ are integers and $q\ne 0$. So, if every terminating decimal can always be expressed in this fraction form, then the statement must be true.

Use the example in the statement. The decimal $0.375$ ends and has three digits after the decimal point. That means we can place it over $1000$:

$0.375=\frac{375}{1000}$

And since $1000=10^3$, the denominator is a power of $10$, exactly as claimed.

In general, if a terminating decimal stops after $n$ decimal places, it can always be written as:

$\frac{\text{integer digits}}{10^n}$

Both the numerator (the digits as an integer) and the denominator $10^n$ are integers, so the decimal fits the form $\frac{p}{q}$. Therefore, every terminating decimal is rational, no matter how long or short it is.

Now consider the other options:

• Option B cannot be right because the fraction conversion works consistently, so the statement isn’t false.
• Option C cannot be right because numbers bigger than $1$ still work (for instance, $2.5=\frac{25}{10}$).
• Option D cannot be right because even decimals with many places still convert (for instance, $0.00001=\frac{1}{10^5}$).

Because any terminating decimal can be written as a fraction with denominator $10^n$, the statement is correct.

Begin by unpacking the statement. A terminating decimal is one that ends. A rational number is any number that can be written as $\frac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$.

The statement gives this example: $0.375=\frac{375}{1000}=\frac{375}{10^3}$

Since $10^3$ is a power of 10, this shows the example is rational. The question is whether this reasoning always works, or whether the extra limits suggested by options C and D could be true.

Option C suggests the idea works only for decimals between $0$ and $1$. Test that by using a terminating decimal larger than $1$, such as $12.5$.

$\begin{array}{rl}12.5& =\frac{125}{10}\\ & =\frac{125}{10^1}\end{array}$

This still has a denominator that is a power of $10$, so the rule is not limited to the $0$ to $1$ interval. That makes Option C incorrect.

Option D suggests the idea works only when there are fewer than three decimal places. Test this with a terminating decimal that has five decimal places, like $0.12345$.

$\begin{array}{rl}0.12345& =\frac{12345}{100000}\\ & =\frac{12345}{10^5}\end{array}$

This also produces a denominator that is a power of $10$, so the number of decimal places doesn’t break the rule. Therefore, Option D is incorrect.

Since the claim works for a larger number and for a longer terminating decimal, the reasoning is clearly general. Any terminating decimal with $n$ digits after the decimal point can be written as:

$\text{Decimal}=\frac{\text{Integer}}{10^n}$

That is always a fraction $\frac{p}{q}$ with integers, so it represents a rational number. So the statement cannot be false.

After removing the options that try to add false restrictions (C and D) and seeing that the conversion always works, the statement is true.