Number Theory Equality of Radicals

Question

Suppose I have an equality $9\sqrt{5} = \sqrt{26} + \sqrt{3}$ (I know this is a false equation, but bear with me). Does the fact that these do not simplify to 0 = 0 imply that these two expressions are not equal? I know you could simply approximate the square roots, but I wanted a quick rule to determine equality. To me it seems like they wouldn't be equal, intuitively. Can someone please give me a rule for this that is supported by (I'm assuming) basic number theory and explain the property? Thank you.

Answer 1

In this case you can use : $$ 2 < \sqrt{5} < 3 \implies 18 < 9\sqrt{5} < 27 $$

$$\ 5 < \sqrt{26} <6, \ 1<\sqrt{3}<2 \implies 6<\sqrt{26}+ \sqrt{3}<8$$

to see they aren't equal, in general this is a simple check to use, however you can easily see it could fail.

Answer 2

I don't think there is a "quick" rule for such problems in general, since radicals encompass a wide range of types of numbers. Consider this nested radical equality, which I found with a web search: $$ \sqrt{5} + \sqrt{22+2\sqrt{5}} = \sqrt{11+2\sqrt{29}} + \sqrt{16 - 2\sqrt{29} + 2\sqrt{55-10\sqrt{29}}}. $$ An approach to such problems other than feeding it into a computer will probably have to account for idiosyncratic properties of the particular numbers involved.