I'm making a test, and there's a question: "Which one of the given expresions, takes value which is natural number". $a) \frac{17}{21} + \frac{12}{42}$ $b) \frac{\sqrt{64}}{\sqrt{8}}$ $c) \frac{\sqrt{75}}{\sqrt{3}}$ $g) -\frac{11}{12}-\frac{1}{12}$
The answer is $c$ however I don't get how to figure this out.
$\endgroup$3 Answers
$\begingroup$You have to compute all given numbers.
For c): $$\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{\frac{75}{3}}=\sqrt{25}=5$$
For d): Is negative, hence not natural.
For a): $\frac{17}{21}+\frac{12}{42}=\frac{34}{42}+\frac{12}{42}=\frac{46}{42}\in(1,2)$ hence not natural.
For b): $\frac{\sqrt{64}}{\sqrt{8}}=\sqrt{\frac{64}{8}}=\sqrt{8}=2\sqrt{2}$ -- irrational.
$\endgroup$ $\begingroup$You have to simplify the expressions. For $(a)$,
$$\frac{17}{21}+\frac{12}{42}=\frac{17}{21}+\frac6{21}=\frac{23}{21}\;,$$
and since $\frac{23}{21}$ is not even an integer, it certainly isn’t a natural number.
For $(b)$,
$$\frac{\sqrt{64}}{\sqrt8}=\sqrt{\frac{64}8}=\sqrt8=2\sqrt2\;;$$
again this isn’t even an integer. In fact, this one isn’t even a rational number.
If you simplify $(c)$ similarly, you get $\sqrt{25}=5$, which is a natural number.
Finally, $-\frac{11}{12}-\frac1{12}=-\frac{12}{12}=-1$; this is an integer, but it’s a negative integer, so it’s not a natural number.
$\endgroup$ $\begingroup$$$\frac{\sqrt{75}}{\sqrt{3}}=\frac{\sqrt{25\cdot 3}}{\sqrt{3}}=\frac{\sqrt{25}\cdot \sqrt{3}}{\sqrt{3}}= \sqrt{25}=5$$
$\endgroup$
