Couple of problems i can't think through looking for some help...
Prove there exists a rational number a and an irrational number b such that a^b is irrational.
I was thinking that I could prove they were both irrational, taking square root two^square root two as A and square root two as B and it works. can't put my finger on this one.
Last one disprove statement there is a real number x such that x^6+x^4+1=2x^2.
Thank you for any help
For the second part take a look at the graph of f(x) = x^6 + x^4 -2x^2 + 1
notice that f(x) is always > 0. You can prove this by looking at f'(x) and f''(x). After you find f''(x) find the points where it is 0, then the function is decreasing or increasing in the intervals between the critical points, based on the sign of f'(x) which you can prove from f''(x). This problem is pretty trivial in the way you prove it, but its very ugly.
For the first problem, Tonio had the right idea, and so do you
Consider . If it was rational, what would that imply?
If it was irrational, what would that imply?
For the second one, gmatt's solution will work. However, there is a better way: (answer in spoiler)
However, even if we can establish that is irrational, we would still need to go on to say:We have found such that , where andBut if , then . So now we have the required result.
So, how about it? Why is irrational? And would it be just as easy to prove that is irrational, thus establishing the result much more simply, with ?
Or am I still missing something obvious? If so, please spell it out for me.
I think Grandad is absolutely right. What some posters were showing is that there exist two irrational numbers so that is rational. The original question was about a rational .
In fact, is irrational (see Gelfond–Schneider theorem in Wikipedia), but the proof is in no way easy. Anyway, using Grandad's addition to the argument is one way to give a complete solution.
As for the second problem, Turiski has a nice idea, except that the sum of two squares is not always positive. It is easy to show, though, that at least one of the summands is non-zero.
We've already seen the proof of . In it we considered three numbers, all of which are equal to .
What if we apply the same reasoning to the following three numbers: 2, , and ?
Case 1 is then OK, because you can then go on to re-arrange in the way I indicated in my last reply.
But Case 2...?
I think this works. Just a remark: one should be careful about notation. As far as I know, power operation associates to the right, so .
For example, you wrote earlier:
The original question was this:
So, for my own benefit, and for the benefit of anyone else who has been bewildered by all this, may I amplify (and hopefully clarify) aman_cc's final solution (which is correct). Here it is again, with my additions in red.
We do not know whether is rational or irrational, but in either case we can formulate a solution.
Case 1: is irrational
So this gives the solution (rational), (irrational), (irrational, by assumption)
Case 2: is rational
Note first that
So we now have the solution (rational, by assumption), (irrational), (irrational).