What can the maximum number of digits be in the recurring block of digits in the decimal expansion of $$\frac{1}{17}$$? Perform the division to check your answer.

Asked by Vishal kumar | 2 years ago |  285

##### Solution :-

We need to find the number of digits in the recurring block of $$\frac{1}{17}$$.

Let us perform the long division to get the recurring block of $$\frac{1}{17}$$.

We need to divide 1 by 17, to get We can observe that while dividing 1 by 17 we get 16 number of digits in the repeating block of decimal expansion which will continue to be 1 after carrying out 16 continuous divisions.

Therefore, we conclude that $$\frac{1}{17}$$ = 0.0588235294117647..... or $$\frac{1}{17}$$ = $$0.\overline{0588235294117647}$$, which is a non-terminating and recurring decimal.

Answered by Vishal kumar | 2 years ago

### Related Questions

#### Visualise the representation of 5.37̅ on the number line upto 5 decimal places, that is upto 5.37777.

Visualise the representation of $$5.3\overline{7}$$ on the number line upto 5 decimal places, that is upto 5.37777.

#### Visualise 2.665 on the number line, using successive magnification.

Visualise 2.665 on the number line, using successive magnification.

#### Find whether the following statements are true or false:

Find whether the following statements are true or false:

(i) Every real number is either rational or irrational.

(ii) π is an irrational number.

(iii) Irrational numbers cannot be represented by points on the number line.

Represent $$\sqrt{10.5}$$  on the real number line.
Represent $$\sqrt{9.4}$$  on the real number line.