Problem of the Day


Let N represent a real number. Does the decimal representation of \sqrt{N} have a finite number of non-zero digits to the right of the decimal point?

(1) N is not an integer.

(2) N>0

Reveal Answer

Answer

E. Statements 1 and 2 together are not sufficient.

See the Solution

Solution

[latexpage]

The question essentially asks whether the decimal representation of $\sqrt{N}$ ends or goes on forever. A number can have an infinite decimal representation in one of two ways, it could be an irrational number like $\pi$ or $\sqrt{2}$ with a decimal representation that goes on forever without any particular pattern, or it could be a rational number like $\frac{1}{3}=0.333333333…$ that just repeats the same number infinitely many times.

Statement (1): INSUFFICIENT. If $N$ is not an integer, it could be $N=\frac{1}{2}$ or it could be $N=\frac{9}{25}$, In the former case, $\sqrt{N}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$. This is an irrational number and hence has an infinite number of non-zero digits in its decimal representation. In the latter case,  $\sqrt{N}=\sqrt{\frac{9}{25}}=\frac{3}{5}=0.6$, which has a finite number of non-zero digits in its decimal representation.

Statement (2) INSUFFICIENT. The same two examples that we used to prove Statement (1) insufficient also apply here.

Statements (1) and (2): INSUFFICIENT. The same two examples also prove that both statements together are insufficient.


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