Question

prove algebraically that the recurring decimal 0.25 has the value 23÷90

Answer 1

The proof for the recurring decimal $0.2\bar5$ is given by the fraction $\frac{23}{90}$ is detailed below.

Proof:

let us consider the variable x such that :

$x=0.2\bar5$ (this is the )

$0.2\bar5$ can also be written as 0.255555...

Now we will find the value of 10x.

$10x = 10 \times 0.25\bar5 = 2.5555..=2.\bar5$ this is the second equation.

Subtracting the first equation from the second equation we get:

$10x-x=2.5\bar5-0.2\bar5\\\\or, 9x = 2.3\\\\or, x= \frac{2.3}{9}\\\\or, x =\frac{23}{90}$

A repeating decimal is unending and the exact value of the fraction cannot be written in the form of a decimal. Fractions such as $\frac{1}{6}$ gives us the decimal form 0.16666... which continues to perpetuity.

So to denote this as a recurring decimal we use the bar or dot symbol above the number which is repeating such as:

$\frac{1}{6}=0.1\bar6$

Therefore the recurring decimal $0.2\bar5$ is given by the  $\frac{23}{90}$ which is a fraction.

Learn more about recurring decimals at:

https://brainly.in/question/1730292

https://brainly.in/question/48573743

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