Given that the decimal expansion of irrational numbers is non-terminating, non-recurring, and is irrational, what can we conclude about the decimal expansion of ?

Question

Homi · Accepted Answer

Understand the Question

The decimal expansion of any irrational number is non-terminating (never ends) and non-recurring (never repeats).
Since $\sqrt{17}$ is given to be an irrational number, by definition, its decimal expansion must share these exact properties.

Step 1 · Deduce the Decimal Expansion of $\sqrt{17}$

By definition, any irrational number has a decimal expansion that is both non-terminating and non-recurring. Diagram 1

\begin{aligned} \text{If a number is irrational} &\implies \text{Its decimal expansion is non-terminating} \\[0.6em] \text{If a number is irrational} &\implies \text{Its decimal expansion is non-recurring} \\[0.6em] \text{Given that } \sqrt{17} &\text{ is irrational} \end{aligned}

Therefore, the decimal expansion of $\sqrt{17}$ is non-terminating and non-recurring.

Answer

The decimal expansion of $\sqrt{17}$ is non-terminating and non-recurring.

Common Mistakes

Rational vs. Irrational Expansions: Rational numbers have terminating or repeating decimal expansions, whereas irrational numbers are strictly non-terminating and non-recurring.
Unnecessary Computation: Attempting long-division square root extraction instead of using simple logical deduction from the given premise.

Question 3

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