Google is not being helpful here. Since the algebraic numbers contain the rationals, they must be dense, but what if you remove the rationals? Are you still left with a dense set?
Yes. If a and b are irrational algebraic numbesr, then (a+b)/2 is also irrational, and also an algebriac number (unless I’m missing something major).
You’re thinking of closure under addition, and that doesn’t hold here (take a = 1 + sqrt(3) and b = 1 - sqrt(3)). What I wonder is, given any two real numbers r and s with r < s, is there an irrational algebraic number q such that r < q < s?
For any given irrational algebraic number x, all nonzero rational multiples rx are also irrational algebraic numbers.
Oops. You’re right. Omphaloskeptic has provided an actual proof.
Is the difference between an irrational and a rational ever rational?
I don’t think so. (But I’m asking so you can correct me if I’m wrong.)
Also, is an irrational divided by a rational ever rational?
Again, I don’t think so. (But again, not a mathematician. Correct me!)
Is there always a rational between any two irrational algebraics?
That’s conjecture number three. I suspect it’s true. (Sure would be nice if I had arguments for these conjectures, wouldn’t it?)
If I’m right (that any irrational divided by a rational is irrational, that any irrational minus a rational is irrational, and there is always a rational between any two irrational algebraics), then the irrational algebraics are dense.
For (and all of the following assumes the above) between any two irrational algebraics P and Q there is a rational R, and the interval between R and (say) Q is irrational and algebraic, (since its length is an algebraic minus an algebraic) and so the midpoint of that interval is also to be found at an irrational algebraic number (since it’s an algebraic divided by an algebraic).
I am sure one of the three assumptions–probably the third–has an obvious counterexample, though.
ETA: The first two assumptions are trivially true. If you could divide an irrational by a rational and get a rational, then you could multiply a rational by a rational and get an irrational. But you can’t. Similarly, if you could subtract a rational from an irrational and get a rational, then you could add a rational to a rational and get an irrational. But you can’t. But what about my third assumption?
I think I can prove the third assumption by proving that between any pair of numbers P and Q there is a rational number. For take the difference between P and Q. Whatever that difference is, there is some rational number smaller than that difference, for the difference must be greater than zero (we’re assuming P is not equal to Q of course!), and this means that there is some number k large enough such that 1/k is smaller than Q minus P. But some multiple of 1/k (call it m/k) must lie between P and Q. (There must be a multiple of 1/k that is the last one before P. The next multiple can not be greater than Q since 1/k is less than Q - P.) And m/k is rational, since it’s a multiple of 1/k and 1/k is rational.
What am I missing? How does his reminder that rational multiples of irrational numbes are also irrational show that the irrational algebraics are dense?
If r is irrational, then for any rational s, r+s is irrational (else the difference between the rational numberfs r+s and s would be irrational). And the sum of two algebraic numbers is algebraic (not entirely evident, but true). So the irrational aglebraics are certainly dense.
That’s the fact that the rationals are dense, which is known (and the proof is basically like yours).
Omphaloskeptic is hinting at the following proof: let r and s be two reals with r < s, and let x be an irrational algebraic number. Then there is some rational q with r/x < q < s/x. This implies that r < qx < s, and qx is an irrational algebraic. So the answer is yes.
Got it.
But just to be sure, does my proof (in post 6) work too? 
Assume x - r/s = t/u
x = t/u + r/s
x = (st + ru)/us
therefore x e Q
Assume x/(r/s) = t/u
x = (rt)/(su)
therefore e Q
General proof (very informal): Between any two irrationals there is at least one rational.
Let i and j be irrational with j>i.
In looking at each place value, there exists a digit in j larger than i. Note that since i and j are assumed irrational, there is no problem with the whole 1.000000… = 0.999999…
Let k = the number made up of the same digit of i & j until they get to that first different digit. Choose the digit from j and terminate the decimal.
i<k<j and k e Q