if yes,why in rational number can first simplify then decision about rational and irrational , but in Rational expressions cannot simplify and must decision befor simplifying about rational and irrational expressions?
Your number is (3 sqrt(2)) / sqrt(2) = 3, and is a rational number indeed. I don't know exactly how to interpret the rest of the question. If r is a positive rational number and p is some positive real number, then sqrt(r^2 p) / sqrt(p) is always rational, being equal r. Possibly your question refers to situtions in which sqrt(c) is not uniquely determined, as for c negative real number or complex non-real number. In those situations a discussion is necessary. Also, in general expressions the discussion is necesary, because the denominator must be different from 0, and so on.
Your number is (3 sqrt(2)) / sqrt(2) = 3, and is a rational number indeed. I don't know exactly how to interpret the rest of the question. If r is a positive rational number and p is some positive real number, then sqrt(r^2 p) / sqrt(p) is always rational, being equal r. Possibly your question refers to situtions in which sqrt(c) is not uniquely determined, as for c negative real number or complex non-real number. In those situations a discussion is necessary. Also, in general expressions the discussion is necesary, because the denominator must be different from 0, and so on.
I, too, cannot quite understand the second part of the question. But I can guess at it and would conclude that an answer would incorporate the following observations.
the word 'rational' does not simply refer to a 'ratio-resembling' expression A/B - where A and B may be anything at all
the square root sign, or operator, or multivariate function (however you wish to think of it) is associative over division and multiplication.
I.e. both√A*√B = √(A* B) and - in particular - √A/ √B = √(A/ B)
Which is, I suspect, why "can first simplify then decision".
The answer to the first part of the question is given in the answers above, and thus I won't rehash them here. Consider the more general problem as follows, quoted from Robert B. Ash's "A Primer of Abstract Mathematics: "If n and k are positive integers but the kth root of n is not an integer, show that the kth root of n is irrational." The proof appears in the back of the book. Ash gives the following hint: "If the kth root of n equals a/b. where a and b are integers. them a^k = b^k n. Use the Unique Factorization Theorem to produce a contradiction." Here the Unique Factorization Theorem implies that any positive integer has a unique prime factorization. The answer to the problem appears verbatim in the back of Ash's book as follows: "If p^e appears in the prime factorization of a, then by the Unique Factorization Theorem, p^(ke) must appear in the prime factorization of a^k. Thus all exponents in the prime factorization of a^k (and similarly b^k) are multiples of k, and therefore all exponents in the prime factorization of n are multiples of k. It follows the the nth root of n is an integer, contradicting the hypothesis."
When I first read question I thought it was stupid as the fraction you wrote obviously equals 3. However, an interesting question might be to ask exactly when a number of the form (irrational)/(irrational) can be rational? That is, give necessary and sufficient conditions for when that can happen. I'd like to think about that before attempting an answer.
Sorry to burden this discussion, but the Questions page does not function, and this is the only place I know I can send something, so, would some kind person please send this question on:
Why does the site incorrectly say I have an outdated edition of Firefox, when I have the newest available? As per the attached screen shot, if it will be accepted. (The shot is of a version of the question with too many words. I reduced them, but that didn't matter.) Please advise, or please correct the site's misapprehension.