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.

Observe that $25=5^2$ and $\frac{256}{225} = \frac{2^8}{9\times 25} = \frac{16^2}{3^2\times 5^2}$, so now do you see why $a)$ and $c)$ are rational numbers? — user37238, Sep 08 '14 at 13:16
That shows options, but doesn't prove that b) is irrational. a, c and d maybe rational, but that doesn't prevent b from being rational. — GreatBlitz, Sep 08 '14 at 13:17
For $n=p/q$ to be rational, as you've correctly stated, both $p$ and $q$ have to be integers. But that does not imply that $p/q$ must be an integer as well. — rubik, Sep 08 '14 at 13:18
because sqrt(5) is not a rational, because you cannot represent sqrt(5) in the form p/q. — njzk2, Sep 08 '14 at 16:50
In $\sqrt{45} \approx 6.708$ the left side is not rational but the right side is. Approximate equality does not preserve rationality; indeed to know whether a number is rational it is of no use whatsoever to know approximations to it. On the other hand any number written with decimals (necessarily finitely many of them) is rational. But not every rational number is (or rather can be) written with finitely many decimals. — Marc van Leeuwen, Sep 08 '14 at 16:50
It's rather easy to prove (by contradiction, usually) that, if $p$ and $q$ are coprime $\sqrt{p/q}$ is rational iff $p=n^2$ and $q=m^2$ for some integers $n,m$. — Jean-Claude Arbaut, Sep 08 '14 at 17:45
$6.708$ is rational but $\sqrt{45}$ is not. Given an irrational number, you can find a rational approximation as close as you need. — Henry, Sep 08 '14 at 18:18

score 23 · Answer 1 · edited Apr 13 '17 at 12:21

23

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.

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

edited Apr 13 '17 at 12:21

Community

1

answered Sep 08 '14 at 13:08

Eff

12,989

Explain why sqrt(5) is irrational? – GreatBlitz Sep 08 '14 at 13:23
5

@GreatBlitz The square root of any prime number is irrational. – Eff Sep 08 '14 at 13:41
@user112061 Great answer! Just for completeness, I've added a little note at the bottom. – beep-boop Sep 08 '14 at 17:44
3

In fact, the square root of a non-square integer is never rational (and the square root of a rational number is never rational unless the numerator and denominator, in reduced form, are both squares). This is easy enough to show in the contrapositive: the square of every rational number $a/b$ can be written as $a^2/b^2$, where $a^2$ and $b^2$ share no common factors, and $a^2/b^2$ is therefore in reduced form. – Ilmari Karonen Sep 09 '14 at 00:41

score 17 · Answer 2 · edited Sep 09 '14 at 09:25

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.

score 5 · Answer 3 · edited Apr 13 '17 at 12:20

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.

edited Apr 13 '17 at 12:20

Community

1

answered Sep 08 '14 at 15:24

robjohn

345,667

1

You're stating that $x$ must be an integer in this answer, but that's not true. $x^2-2.25=0$ works just fine ($x=1.5$). $\sqrt{45}$ may be irrational, but that doesn't mean that $\sqrt{x}$ must be an integer. – Brian S Sep 08 '14 at 19:12
@BrianS: From the cited answer, the polynomial needs to be not only monic, but also all its coefficients need to be integers: "any rational number $x$ which satisfies $(2)$ with integer $c_k$ must be an integer". However, $2.25$ is not an integer. – robjohn Sep 08 '14 at 21:48

score 4 · Answer 4 · answered Sep 08 '14 at 13:34

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.

score 3 · Answer 5 · edited Sep 09 '14 at 00:51

3

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.

edited Sep 09 '14 at 00:51

robjohn

345,667

answered Sep 08 '14 at 13:06

almagest

18,380

No, I am not. Can you explain it?
EDIT: How is 0.3333... rational?
– GreatBlitz Sep 08 '14 at 13:20
Because it can also be expressed as 1/3. You might think that an infinite decimal cannot be a rational number, but it can. 0.333... means $\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\dots$ and that infinite series can easily be summed to give 1/3. – almagest Sep 08 '14 at 13:26
Basically this? http://www.math.utah.edu/~pa/math/q1.html – GreatBlitz Sep 08 '14 at 13:29
Yes. Note that it only works for a prime. So $\sqrt{n}$ is irrational if for some prime $p$, the highest power of $p$ dividing $n$ is an odd power. So $45=3^25$. The three is irrelevant, but the 5 makes $\sqrt{45}$ irrational. – almagest Sep 08 '14 at 13:41
You missed a $ to close off your MathJaX at $n\ne0$ – Brian S Sep 08 '14 at 19:07
Oops. Sorry. Did you fix it? I cannot find it! – almagest Sep 08 '14 at 19:19
@almagest: I fixed it. Your answer and Oleg567's are very similar in idea and time, so you both deserve +1. – robjohn Sep 09 '14 at 08:12

score 3 · Answer 6 · answered Sep 08 '14 at 13:10

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.

answered Sep 08 '14 at 13:10

Oleg567

17,295

why was this answer deleted? – robjohn Sep 09 '14 at 00:52
@robjohn, I deleted it, because thought that exactly the same (classic) proof is for $\sqrt{2}$, somewhere in Wiki or other widely known resources. – Oleg567 Sep 09 '14 at 07:51
1

+1 It is a standard proof, but those are useful. – robjohn Sep 09 '14 at 08:10
@robjohn, thank you :) – Oleg567 Sep 09 '14 at 08:27
Additional exercise: Show why this proof fails for $\sqrt4$, hint: somewhere in Greece, an old man found what was to be called the fundamental theorem of arithmetic. – Bourbaki is ALIVE Sep 09 '14 at 17:13
@Bourbaki is ALIVE, if consider $\sqrt{4}$, then $a^2=4b^2$ $~~\Rightarrow~~$ $2|a^2$ (note, not $4|a^2$, as in the case of $\sqrt{5}$). So, there are no contradictions in the case of $\sqrt{4}$. – Oleg567 Sep 10 '14 at 16:30

score 2 · Answer 7 · answered Sep 08 '14 at 15:34

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.

score 1 · Answer 8 · answered Sep 08 '14 at 17:19

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.

Can rational numbers have decimals?

8 Answers8