How would I begin a proof to prove that [tex] \forall k \in \mathbb{F},\;\left( { - 1} \right)^k \notin \mathbb{R} [/tex] (F denotes the set of all irrationals)
Have you tried anything yet? If not the first thing that jumps to my mind is a proof by contradiction.
[tex](-1)^k=e^{k{\pi}i}[/tex] so [tex](-1)^k\in\mathbb{R}\rightarrow sin(k\pi)=0[/tex] so show sin(k pi) has no zeros when k is not rational
Good idea, but a little flawed. [tex](-1)^k\in\mathbb{R}\rightarrow sin(k\pi)=0[/tex] is not true. [itex]k = \frac{1}{3}[/itex] is a counterexample. I would amend the proof slightly, like so : [tex](-1)^k=e^{(2n + 1)k{\pi}i}[/tex] where [itex]n\in\mathbb{Z}[/itex] so [tex](-1)^k\in\mathbb{R}\rightarrow \sin{[(2n + 1)k\pi]}=0[/tex] Then observe that [tex]\sin{(m\pi)} = 0 \rightarrow m\in\mathbb{Z}[/tex] Obviously [tex]k\in\mathbb{F} \rightarrow (2n + 1)k \notin\mathbb{Z}[/tex] and we're done.
No it is not. [tex](-1)^{\frac{1}{3}}=\frac{1+i\sqrt{3}}{2}[/tex] Unless you ignore all the conventions of math, in which case you might assert. [tex]1^{\frac{1}{4}}=-i[/tex] The question poster might have stated the definition of exponential being used, but it is reasonable to assume the standard definition is intended.
Exponents/roots can be multivalued. You're assuming the principal value. The question did not stipulate only principal values should be considered. I interpreted it as "If k is an irrational, prove that no possible value of (-1)^k can be real". I believe this is what the question is asking for. (-1) does have a real cube root (-1). It's just not the principal value. That's what I meant.
Our approches are equivelent. It is implicit in your work. I reduced the problem to show sin(k pi) is not 0 for all irrational k you prefered show sin((2n+1)k pi) is not zero for all irrational k and integral n clearly they are equivalent I can let my k=your (2n+1)k and you can let your (2n+1)k=my k. In other words my assumption of principle value was not only reasonable and in accordence with standard convention, but also without loss of generality. If someone wants their (-1)^pi to be my (-1)^(5pi) it is not a problem as I have that case. The proof can be thought of in terms of element chasing, and the principle values just keep track of the elements.
The approaches are equivalent. That's because I simply amended your proof to come up with what I feel is a more complete one. I never claimed otherwise. Your proof is a good one. I do not disagree with this part : "show sin(k pi) is not 0 for all irrational k" I disagree with this one : [tex](-1)^k\in\mathbb{R}\rightarrow sin(k\pi)=0[/tex] That only applies when the principal value of (-1)^k is being considered. The question never stipulated that, so I don't see why it should be assumed. If we take all quadrants, then there are counterexamples to your assertion, I've already stated that k = 1/3 is an obvious one, because one cube root of (-1) is real (-1) whereas sin(1/3pi) is nonzero. You speak of mathematical convention. But let's be fair here, the question is pretty dumb if we're only going to be considering principal or first quadrant values for that expression. Because if that were the case, the principal value of (-1)^k is not real for all rational non-integral values of k. I think the question is taking a broader view than this, otherwise it seems stupid.
This is far more simple than you two seem to have made it out. As I say you start with a proof by contradiction and assume it is true: [tex](-1)^k = \alpha \quad \text{for} \quad k \in \left. \mathbb{R} \, \right\backslash \left\{ \mathbb{Q} \right\} \quad \text{and} \quad \alpha \in \mathbb{R}[/tex] Then: [tex]\left| (-1)^k \right| = |\alpha|[/tex] [tex]\left| (-1) \right|^k = |\alpha|[/tex] [tex]1^k = |\alpha|[/tex] [tex]\alpha = \pm 1[/tex] Trying 1: [tex](-1)^k = 1 \quad \text{then} \quad k = 2c_1 \quad \text{for some} \quad c_1 \in \mathbb{Z}[/tex] Now trying -1: [tex](-1)^k = 1 \quad \text{then} \quad k = 2c_2 + 1 \quad \text{for some} \quad c_2 \in \mathbb{Z}[/tex] So the solution set for k is [itex]\mathbb{Z}[/itex] and as [itex]\mathbb{Z} \subset \mathbb{Q}[/itex] and k is defined as [itex]k \in \left. \mathbb{R} \, \right\backslash \left\{ \mathbb{Q} \right\}[/itex] we have a contradiction!
HMm, however, [itex] k [/itex] does not necessarily need to be an integer; it is not required that [itex] k \in \mathbb{Z} [/itex] so that [itex] (-1)^k \in \mathbb{R} [/itex] For [itex] k = 2{/}3 [/itex], [tex] (-1) ^ k = (-1) ^ {2/3} [/tex], which has a real solution equivalent to 1.
I think you will find you are wrong [tex](-1)^{\frac{2}{3}} = \frac{1}{2} + \frac{\sqrt{3}}{2}i \notin \mathbb{R}[/tex]
Why can't we just say What is [itex](-1)^{2/3}[/itex]? Well it is the number x such that [itex](-1)^2 = x^3[/itex]? Heck, even my calculator agrees. Btw it's [tex](-1)^{\frac{2}{3}} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \notin \mathbb{R}[/tex] I think.
Why so? [tex] \left( { - 1} \right)^{2/3} = \sqrt[3]{{\left( { - 1} \right)^2 }} = 1 [/tex] and even [tex]\left( { - 1} \right)^{1/3} = \sqrt[3]{{ - 1}} = - 1 [/tex] There exists a real cube root of -1. It is possible for k=1/3, and 1/3 is rational but not an integer. Sure, you can have complex solutions for [itex] (-1) ^ {2/3} [/itex]. But you can have real ones as well , and you see, **The point of the original statement is basically: "If k is irrational, then no real solutions at all will exist for (-1)^k "
No, you seem to have a misunderstandings of how powers work for negative numbers. You can equally say: [tex](-1)^{\frac{1}{2}} = (-1)^\frac{2}{4} = \left( (-1)^2 \right)^{\frac{1}{4}} = 1^\frac{1}{4} = 1[/tex] However [itex]1^2 \neq -1[/itex]. Do you see what went wrong here?
Not quite-->you see, in my example, [tex] (-1)^{1/3} = -1 [/tex] Because [tex] (-1) ^ 3 = -1 [/tex] *Then again, would it be incorrect to say that [tex] (-1)^k \in \mathbb{R} \; \text{if} \; \exists c \in \mathbb{R}\;{\text{such that }}c^{\frac{1}{k}} = - 1 \; {?} [/tex] What exactly went wrong here?
Try this: [tex]\left( a^p \right)^{\frac{1}{q}} = \left( a^\frac{1}{q} \right)^p \quad \text{for} \quad p \in \mathbb{Z} \quad \text{and} \quad q \in \left. \mathbb{Z} \right\backslash \left\{ 0 \right\}[/tex] What domain for a does this hold true? I urge you to try this question, a lot of important basic maths concepts are held within
Also in which you mean [itex] p \ne 1 [/itex], because where [tex] p = 1 [/tex], [tex] \left( a^p \right)^{\frac{1}{q}} = \left( a^\frac{1}{q} \right)^p \Rightarrow [/tex] [tex] a ^ {1/q} = a ^ {1/q} [/tex] Again, why is it incorrect to say: [tex] \left( { - 1} \right)^{\frac{1}{3}} = - 1\;{\text{because }}\left( { - 1} \right)^3 = - 1 [/tex] I even searched for this problem in Dr. Math, at http://mathforum.org/library/drmath/view/55604.html. It says: ***Then again, I would have to look up what exactly a ***principle value is. I'll try MathWorld() or google() Until then....
Actually I meant: [tex]\left( a^p \right)^{\frac{1}{q}} = \left( a^\frac{1}{q} \right)^p \quad \forall p \in \mathbb{Z} \quad \text{and} \quad \forall q \in \left. \mathbb{Z} \right\backslash \left\{ 0 \right\}[/tex] It's really worth considering and understaing why negative powers raised to rationals is defined on the princaple value. If this wasn't the case then there would be a lot of problems. Seriously try and prove where this is valid for a.