$\begingroup$(a) A natural number,
(b) A rational number but not a natural number,
(c) An irrational number not exceeding 6,
(d) An irrational number exceeding 6.
Please help with this, i can't manage to simplify it. Thanks in advance.
$\endgroup$91 Answer
$\begingroup$Let us assume $$\sqrt{13+3\sqrt{\frac{23}{3}}} +\sqrt{13-3\sqrt{\frac{23}{3}}}=x $$ Squaring and simplifying gives $$26+20=x^2$$ which gives $$x=+\sqrt{46}$$
Since $6= \sqrt{36}<\sqrt{46}<\sqrt{49}=7$
The result is $(D)$
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