Is 4.56(with bar on top of 6) a rational or irrational number?

[mathven]

There are two parts that I am confused with, so any kind help will be greatly appreciated!
1. Can 4.56(with bar on top of 6) be expressed as a fraction? (If it can be, then it's a rational number right?)
2. Does the "56" in ".566666..." makes the number non-recurring? (Because an irrational number is non-terminating and non-recurring)

 

[lookagain]

1. Can 4.56(with bar on top of 6) be expressed as a fraction? (If it can be, then it's a rational number right?)

Because of what you know/stated in number 2) about irrational numbers, the complement would mean that the decimal numbers
of certain rational numbers can terminate or blocks of them can regularly repeat (after some point) for other rational numbers.

Regardless, you can answer the question in your parentheses first. If you determine that it is a rational number, then you
will know it can be expressed as a fraction. Knowing what optional forms of the fraction, or how to get there, is not my
point at this time.

2. Does the "56" in ".566666..." makes the number non-recurring? (Because an irrational number is non-terminating and non-recurring)

You could be asking about just the digit 5. The three dots shows that the sixes repeat forever, or that they are "recurring."

 

[stapel]

There are two parts that I am confused with, so any kind help will be greatly appreciated!
1. Can 4.56(with bar on top of 6) be expressed as a fraction? (If it can be, then it's a rational number right?)

If you haven't yet been shown the trick for these, here it is:

Given a decimal expansion such as $0.333\overline{3}$ (specifically, a decimal expansion with a one-digit repetition, and no other digits significant digits coming before the repeated part), you can assign the expansion a name (generally, $x$), multiply this by $10$, subtract, and then divide:

$\qquad x = 0.33333...$
$\qquad 10x = 3.33333...$

$\qquad \begin{array}{r} 10x = 3.33333\ldots \\ \underline{\phantom{10}x = 0.33333\ldots} \\ 9x = 3\phantom{.33333\ldots}\end{array}$

$\qquad \dfrac{9x}{9} = \dfrac{3}{9}$

$\qquad x = \dfrac{1}{3}$

In your case, you have $4.5666\overline{6}$. There is just one digit in the repeated part, so multiply by $10$. This will give you:

$\qquad \begin{array}{r} 10x = 45.6666\ldots \\ \underline{\phantom{10}x = \phantom{5}4.5666\ldots} \\ 9x = 41.1\phantom{666\ldots}\end{array}$

Where does this lead? In particular, does it lead to a fraction?

2. Does the "56" in ".566666..." makes the number non-recurring? (Because an irrational number is non-terminating and non-recurring)

All that matters is that, eventually (in this case, right after the "5"), you get the same digit repeating forever. And that "dot, dot, dot" at the end of the "6666" means "forever after, in this manner". So the sixes never end. It's non-terminating, but it is certain not non-recurring.

 

[mario99]

If you haven't yet been shown the trick for these, here it is:

Given a decimal expansion such as $0.333\overline{3}$ (specifically, a decimal expansion with a one-digit repetition, and no other digits significant digits coming before the repeated part), you can assign the expansion a name (generally, $x$), multiply this by $10$, subtract, and then divide:

$\qquad x = 0.33333...$
$\qquad 10x = 3.33333...$

$\qquad \begin{array}{r} 10x = 3.33333\ldots \\ \underline{\phantom{10}x = 0.33333\ldots} \\ 9x = 3\phantom{.33333\ldots}\end{array}$

$\qquad \dfrac{9x}{9} = \dfrac{3}{9}$

$\qquad x = \dfrac{1}{3}$

In your case, you have $4.5666\overline{6}$. There is just one digit in the repeated part, so multiply by $10$. This will give you:

$\qquad \begin{array}{r} 10x = 45.6666\ldots \\ \underline{\phantom{10}x = \phantom{5}4.5666\ldots} \\ 9x = 41.1\phantom{666\ldots}\end{array}$

Where does this lead? In particular, does it lead to a fraction?



All that matters is that, eventually (in this case, right after the "5"), you get the same digit repeating forever. And that "dot, dot, dot" at the end of the "6666" means "forever after, in this manner". So the sixes never end. It's non-terminating, but it is certain not non-recurring.

Hey stapel.

I will teach you a better trick. You just need to divide by $9$.

$\displaystyle \frac{1}{9} = 0.111\overline{1}$


$\displaystyle \frac{2}{9} = 0.222\overline{2}$


$\displaystyle \frac{3}{9} = 0.333\overline{3}$


$\displaystyle \frac{4}{9} = 0.444\overline{4}$


$\displaystyle \frac{5}{9} = 0.555\overline{5}$


$\displaystyle \frac{6}{9} = 0.666\overline{6}$


$\displaystyle \frac{7}{9} = 0.777\overline{7}$


$\displaystyle \frac{8}{9} = 0.888\overline{8}$


I will leave it for you as an exercise to find the fraction of $0.999\overline{9}$


Now suppose that I want the number starts with two zeros, $\displaystyle 0.0$

You just need to divide the fraction by $10$.

$\displaystyle \frac{\frac{6}{9}}{10} = 0.0666\overline{6}$

Do you want three zeros, 0.00? Just divide by $100$

$\displaystyle \frac{\frac{6}{9}}{100} = 0.00666\overline{6}$

Do you want four zeros, 0.000? Divide by $1000$ and so on.

The OP number is:

$\displaystyle 4.5666\overline{6} = 4 + \frac{1}{2} + \frac{\frac{6}{9}}{10} = \frac{137}{30}$

It looks rational to me

 

[stapel]

Hey stapel.

I will teach you a better trick. You just need to divide by $9$.

How will your method work when the repeated portion contains two or more digits, such as $0.\overline{142857}$?

 

[mario99]

How will your method work when the repeated portion contains two or more digits, such as $0.\overline{142857}$?

Well same as before

Just divide by $9$, not one $9$, but six $999999$


$\displaystyle 0.\overline{142857} = \frac{142857}{999999}$


Bonus:

$\displaystyle 0.\overline{14} = \frac{14}{99}$


$\displaystyle 0.\overline{142} = \frac{142}{999}$


$\displaystyle 0.\overline{1428} = \frac{1428}{9999}$


$\displaystyle 0.\overline{14285} = \frac{14285}{99999}$