Can rational numbers have decimals?

Question

I had a question in my exam paper - Which of the following is not a rational number?

a) $\sqrt{25}$

b) $\sqrt{45}$

c) $\sqrt\frac{256}{225}$

d) $\frac{3}{4}$

The answer to this is b. Now, $\sqrt{45} \approx 6.708$. Can someone explain why this not rational? Is it about decimal points?

A rational number is any number that can be expressed in the form of $\frac{p}{q}$, where $p,q$ are integers and $q\neq 0$.

So $\frac{3}{2}$ qualifies as a rational number right? But, in decimal form, $\frac{3}{2}$ is $1.5$ which has decimals. I thought integers don't have decimals, so 1.5 shouldn't be a rational number!

Can someone clear up my mind? Simple terms please :)

Regards.

Answer 1

You are correct, a rational number is one that can be expressed in the form $m/n$, where $m,n$ are integers (and $n\ne0$ - also we usually assume they have no common factors).

So 1.5 is a rational number because it can be expressed as 3/2.

Indeed, any finite decimal can be expressed in the form $m/n$ (think about what 0.123 means). Some infinite decimals can too, eg 0.333...

But $\sqrt5$ cannot. There is a standard proof for that. Are you familiar with it?

[Following your question below]

5 is a prime. Suppose we had $\sqrt5=\frac{a}{b}$. We can take out any common factors, so that $a,b$ are relatively prime. Then $a^2=5b^2$. But that implies that 5 divides $a^2$. Since 5 is prime, it must also divide $a$. Suppose $a=5c$. But now we can divide through by 5 to get $5c^2=b^2$. So exactly the same argument shows that $b$ is divisible by 5. Contradiction.

Answer 2

First off, your definition of rational numbers is correct. So $3/2$ is a rational number, but in the definition it does not say that $p/q$ should be an integer, only that both of $p$ and $q$ should. Rational numbers can have decimals and even an infinite decimals, BUT any rational number's decimals will have a repeating pattern at some point whether it be like $$ \frac23 = 0.666... $$ or $$\frac{92}{111000} = 0.000\hspace{2px}828\hspace{2px}828\hspace{2px}828... $$ or $$\frac32 = 1.500 \hspace{2px} 000 \hspace{2px} 000...$$ The reason why $\sqrt{45}$ is not rational is not because it has decimals. We have that $$\sqrt{45} = \sqrt{5\cdot 9} = \sqrt{5}\sqrt{9} = 3\sqrt{5},$$ so $\sqrt{45}$ must be irrational if $\sqrt{5}$ is irrational. In fact it is known that $\sqrt{p}$ for all primes $p$ is irrational such as $p=5$, see e.g. this.

Therefore we can say that $\sqrt{45}$ is irrational.

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To see why the product of a (non-zero) rational number and an irrational number is irrational, see this.

Answer 3

$$\sqrt{45} = \sqrt{9\cdot 5} =3\sqrt{5}, $$ but $\sqrt{5}$ isn't rational.

Short proof: if $\sqrt{5}=\dfrac{a}{b}$, $a,b\in\mathbb{N}$, and $GCD(a,b)=1$, then

$$ 5 = \dfrac{a^2}{b^2}, $$ $$ a^2 = 5b^2, $$

so $~~5|a^2$ $~~\Rightarrow~~$ $5|a$ $~~\Rightarrow~~$ $5|b^2$ $~~\Rightarrow~~$ $5|b$ $~~\Rightarrow~~$ $GCD(a,b)=5$. Contradiction.

Answer 4

Whether a number is a rational number does not depend on how it is expressed, but on whether it is possible to write it as $\frac pq$ where $p,q$ are integers and $q\neq 0$.

So any terminating decimal, for example $r=abc.defg$ is a rational number, because $10000r=abcdefg$ and $r=\frac {abcdefg}{10000}$

Also any decimal which eventually repeats represents a rational number, say $r=abc.defgefgefgefg \dots$.

Here we can look at $$10000r-10r=abcdefg.efgefgefg \dots -abcd.efgefgefg \dots=abcdefg-abcd$$

so that $9990r=abcdefg-abcd$

and $r=\frac {abcdefg-abcd}{9990}$

In fact any rational number will either terminate or repeat - think of doing long division to find the decimal expansion. Eventually you will get to remainder $0$, in which case the division ends, or you will eventually arrive at a remainder you have encountered before, in which case the pattern will repeat.

I've sketched this out because you might like to explore these ideas a bit for yourself.

Answer 5

As shown in this answer, if the solution of $$ x^2-45=0\tag{1} $$ is rational, then it is an integer. $x=6$ is too small and $x=7$ is too big, so there is no integer that satisfies $(1)$. Thus, there is no rational number that satisfies $(1)$.

That is, $\sqrt{45}$ is irrational.

Answer 6

3 out of the 4 options on that exam can be written in $\frac{p}{q}$ form, where p and q are integers. Note that integers can not have decimals. For example:

a) $\sqrt{25} = 5 = \frac{5}{1}$

b) $\sqrt{45} = \frac{?}{?}$

c) $\sqrt{\frac{256}{225}} = \frac{\sqrt{256}}{\sqrt{225}} = \frac{16}{15}$

d) $\frac{3}{4}$

So a, c, and d can be written in $\frac{p}{q}$ form (5,1,16,15,3,4 are all integers) but b cannot. If you think b is rational, then fill in the two question marks with integers and you will find that you cannot.

Answer 7

A rational has to be able to be shown as a fraction with integers on the numerator and denominator.

Some decimals can be written as fractions (examples below)

$3=\frac{3}{1}$

$1.5=\frac{3}{2}$

$1.66666...=\frac{5}{3}$

The problem arises in the fact that $\sqrt {45} =3 \sqrt{5}$ and $3$ is rational, but $\sqrt{5}$ can't be written in any fraction with integers on the top and bottom.

It cant actually be written as a decimal, as you wrote, it isnt equal to 6.708, its close to 6.708, and there is no write-able decimal that it is equal to.

Answer 8

Now, $\sqrt{45} \approx 6.708$. Can someone explain why this not rational? Is it about decimal points?

Your answer to this is below: $\sqrt{45}$ is not rational because it cannot be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers.

This also means it can’t be expressed in decimal form with terminating or repeating digits.

$\frac{3}{2}$ qualifies as a rational number right?

Yes. It is in the form $\frac{p}{q}$, and $3$ is an integer and $2$ is an integer.

But, in decimal form, $\frac{3}{2}$ is $1.5$ which has decimals. I thought integers don't have decimals, so 1.5 shouldn't be a rational number!

The problem here is that an integer is not the same thing as a rational number. $\frac{3}{2}$ is rational because it’s written in the form $\frac{p}{q}$, and both the $p$ and the $q$ are integers ($3$ and $2$). We’re not saying that $\frac{3}{2}$ is an integer. We’re saying that the numerator and the denominator are integers, making $\frac{3}{2}$ rational.

All integers are rational—but not all rational numbers are integers. $3$ is an integer and $2$ is an integer. So $\frac{3}{2}$ is rational but it happens not to be an integer.