0
$\begingroup$

It is a rule that:

A rational number can be expressed in $\frac{a}{b}$ form and an irrational number cannot be expressed in such form. It is even said that (in order to promote the $\frac{a}{b}$ form for expressing the rational numbers... that both, that is the numerator and the denominator should be in the form of integers).

So if a number can be expressed in $\frac{a}{b}$ form, then only can it be a rational number. And if a number cannot be expressed in $\frac{a}{b}$ form then it is irrational. Therefore, if a number is expressed in $\frac{a}{b}$ form, then it will be recurring or will terminate after few decimal digits. Let us take a number, $\frac{1}{2}=0.5$, it is a rational number, and both the numerator and the denominator are integers. Now let us take another number, $\frac{355}{113} = 3.1415929203539\dots $ it is not the symbol of recurrence. but what I want to convey is that it will go up to a number of decimal digits, and will start to re occur. Thus any number will start to re-occur after a few decimal digits. Suppose we take the number $\sqrt{2}$. Its value is taken to be approximately $1.41$. If we proceed this rooting we may get our answer until one crore or one extra decimal digits. But still it will end up somewhere. So that is just an assumption that it will not end. But it may end. so I am stuck on it. Can anybody help me out to clear my ideas?

$\endgroup$
5
$\begingroup$

We actually know for a fact that the decimal expansion of $\sqrt 2$ will not end and not be recurring. It is easy to see that if the decimal expansion of a number is reccuring, then the number can be written as $\frac{p}{q}$ for some integers $p$ and $q$.

On the other hands, it can be proven that $\sqrt 2$ cannot be written as $\frac pq$, meaning also that its decimal expansion cannot end and cannot be reccuring.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.