Here is a hint. Starting in similar way as you did, assume that , where is some rational number.
Now consider ...
I know how to prove that and are both themselves irrational, but how would you go about proving that
is irrational?
I've tried making the assumption that where and are both natural numbers, but it hasn't gotten me anywhere yet. I've manipulated the expression and made some observations when I see certain expressions in the form of or , therefore being either even or odd, but I can't seem to get to a proper proof. I have though, after making the rational assumption, run into multiple "contradictions", things that just can't be equal. But I'm not certain if these are the type of contradictions that count towards proving the irrationality, or if I made a simple mistake to arrive at the contradictions.
I'm also trying to prove the irrationality of , its next to the other question in my textbook, and I also have had the same experience with this number as well. thanks in advance for any help.
I found this question rather interesting, as I have never actually thought about what operations
we can apply on certain classes of irrational numbers (as in the class of all irrational
square roots of rational numbers), and still get a rational number.
Although I support melese's method, it will not always succeed in telling when is irrational.
For example when (although it perfectly solves the original problem and I suggest
that you use that method in your excercise, since that's what was probably meant to be used).
Here's a nice way (in my opinion) to prove that if
with , then is irrational.
Suppose the contrary. Then . Therefore
But then , a contradiction.
Or in general: has minimal polynomial (wich can be found by looking at powers of )You can also note that is a root of the polynomial . If has any rational roots, then they must be of the form . But we can see that none of these are roots of . Therefore, is irrational
If has any rational roots, it must be a divisor of . Now, we need to check that any divisor can not correspond with . Thus showing is not an integer is also enough.
Still, Unbeatable0's method is cutest
I've just spent some time trying to find the proof, with little success, after noting for some rational , and then considering as you said. I'm sure the proof is right under my nose, and I'm just not seeing it, do you have another hint? This is how far I got (granted, I could be "moving" in the wrong direction with these steps):
therefore
that leads to this equation:
this led me to:
But I feel like this is to complicated to be anywhere in the right direction, I did the isolation of the and squared both sides in an attempt to get rid of irrationals so that I could use natural number factors to prove a common divisor between and .
A series of subproofs by contradiction should suffice.
You got
Since by assumption r in Q, then r^2 in Q. Then 8 + 2*sqrt(12) is rational if and only if sqrt(12) is rational. (Prove by contradiction.) Then you can prove sqrt(12) is irrational in the manner that sqrt(2) is typically shown to be irrational. (Again, contradiction.) Alternatively, rewrite sqrt(12) as 2*sqrt(3) and prove sqrt(3) is irrational.
EDIT: melese's post #9 simplifies this by removing the first contradiction step.
Um, I think melese's method works for
where is rational.
, where is rational
After that, it's just a matter of proving that is irrational
Also, be careful when you say that , because if and could be equal to and respectively, where , respectively ... then .