Touché. I'll think about where my proof went wrong.
EDIT: Since 18n2=m2, m2≥18. In fact, I showed that n≠1 when k is non-square, so m≥72. I get your point though.
EDIT2: Then what I should say is that since 18|m2, gcd(m,18)≠1. Here is a revised proof.
Let √k be a natural number where k is not...
mod() basically returns the remainder after a division. For example, 18 mod 3 returns 0, but 19 mod 3 returns 1. The arguments would be something like mod(#,base). Mods have some pretty cool uses in programming and mathematics :)
It looks like people are making this too complicated (sorry if I am wrong :/). I usually attack these problems using parity. First, I suggest checking x and y parities. In base 2, x and y can only have 2 parities, so this should not be difficult. You will get a result that yields parity of...
EDIT: This proof is flawed (thanks Norwegian!). Hopefully this works better :)
There is a generic proof for showing that the square root of a non-square integer is irrational. Let √k be your number where k is not square. Assume it is rational. That is, let \sqrt k=\frac{m}{n} where m,n\in Z...
Since that is a valid representation of eu, you can see how it leads to representing an imaginary power of e directly to a function of cosine and sine. I think this is how Euler showed the identity, too.
Basically, it works like this. Assume n=p1c1p2c2p3c3... Now, divide by all the primes with odd power. For example, if c3 is odd, divide by p3. now you have a product of primes to the first power multiplied by the product of primes to even power. the second product, then, is a perfect square, so...
Glad you asked :P
Also, I wish I had replied yesterday. Micromass is completely correct. We needed to define how exponentiation works when it is extended to complex powers. Until that point, we cannot simply assume how it works. In my post, all I had to say was that complex numbers are closed...
And for those that didn't catch the simpler nature of the problem, as dextercioby said, you can use basic laws of exponents:
eixe-ix=eix-ix=e0=1.
Alternatively:
eixe-ix=eix/eix=1.
:)
You know that all numbers can be expressed as a product of primes and 1. Then, if you write your number n as n=p1c1p2c2p3c3... (a general prime factoring), what happens when you square this number? (Hint: The "p" values are not the important part.)
Once you have found this, what can you say...
Hi gonzalol, the TI-NSpire will handle complex numbers (like a+bi), so I am sure you will be able to calculate phasors.
However, if you have never used a TI-NSpire and your test is in a few days, it might be difficult for you. The TI-NSpire has different controls and a different layout than...
Sorry, I am still blaming my lack of sleep for me not being clear. What I mean is something more along these lines...
If I have a set that has a cardinality of (m!)a as m approaches infinity and where a is some natural number >1, is the cardinality countable or uncountable? If the set is...
How "big" is this number?
I have had absolutely no sleep for a while, so my math brain has been failing me. Last night, I was working on a problem and I believe that one of my connections is wrong. Is this number countably infinite or uncountably infinite (as m approaches infinity)...
Ah, right, sorry. It is not exactly an identity, but yes, you are right. It is because I was still thinking (a+b) as opposed to (sqrt(a)+sqrt(b)). The identitiy there is a-b, which in thios case happens to be 1 (which is extremely useful).
Thanks for the catch, I'll edit my previous post so...
I know, that was what I was saying. I was responding to him, not you, sorry :/
Also, I read that as a different syntax. I thought he was saying "congruent to exactly one 0,2,or 4mod 7" meaning that it was one of the values in the list {0,2,4}
EDIT: This is one of those cases where you could...
This post contains incorrect info (after the quote, of course)
To avoid misinforming the person asking the question, I would like to note that a conjugate is not an inverse. A conjugate of (a+b) is (a-b) whereas the inverse of (a+b) is (a-b)/(a2-b2). Applying this:
If √3 + √2 = n/m, then...
@bonfire09:
Depending on the situation, you may want to show that the cases repeat. For example, showing these 3 computations should be satisfactory (though if this is for an introductory class, it might be prudent to be more rigorous than this):
0^2 mod 7 = 0
1^2 mod 7 = 1 = 6^2 mod 7
2^2 mod...
Wow, it seems a lot of people are having issues. Some more information that might be helpful is:
What kind of apps or programs do you have installed? (some games and utilities might be buggy or intended for older or newer hardware versions).
Also, when it says Installation in progress, it can...
For future reference, for anybody else that has this problem, here are two ways to fix this:
In the Mode settings, change Exact/Approx to Approximate.
Alternatively, if you add a decimal point instead, like 950.cos(75), the decimal value will be returned as opposed to the exact value with...
The problem seems to be that you have stored some value to x. What you may need to do is clear all those variables (there is an option from the homescreen under F1, I believe). Basically, you may have done something like this:
37→X
So then factor(x^2-5x=6) would do something like this in...
This feels like homework. However, I will give a proof just to make sure I still can:
For purposes of contradiction, assume \sqrt{2}+\sqrt{3} is rational. That is, there are some integers n and m (m≠0) such that \sqrt{2}+\sqrt{3}=\frac{n}{m} and gcd(n,m)=1 (in other words, n and m are...
Hmm, this looks neat, so you are going to add all the odd terms in f(x)? So you need, as a function, \displaystyle \sum_{x=a}^n f(x)\frac{1-(-1)^{f(x)}}{2}. That looks like a really cool and useful function! I have literally filled notebooks finding sums like that, so I have a huge collection...
That is a much more interesting problem (to me). For certain types of irrational numbers, there is indeed a pattern (you can check out Pell's equation and Continued Fractions to find ways to very closely approximate square roots).
However, here is how you would find such integers for pi. We...
I think they mean that it won't be outdated too soon. For example, most cell phones and computers are outdated within a year, so they are not future proof. However, calculators take much longer to become outdated and since the TI-NSpire CX CAS is the top of the line right now, it is the most...
It depends on which algorithm is used, as mentioned before. Basically, the TI-84+ (and the other z80 models) use 9-byte Floating point numbers to do math. A number in the form of data will look like this, for example:
00 80 31 41 59 26 53 58 98
the first 00 tells several things...
To define a function, you can enter the Y= menu. In that case, Y= translates to f(x)=. You can also change from the Mode menu to parametric, polar, and sequential graphing modes. Alternatively, if you want 3D graphing or graphing of differential equations, Graph3D is a wonderful app that...
You could write your own parser. I have am not familiar with your model, but I assume it has a similar syntax to the TI-89 or 92? If so, then writing a parser should not be too difficult. The only difficulty I can see is in how tedious the process will be. I've never made the attempt myself, but...
It seems the issue is that the calculator is parsing the characters as the name of a variable which only allows lowercase letters. If you place a quotation symbol around the text, there will be no modifications to your input :)
Hmm, making a program to parse set theoretical operations could be a rather daunting task. It would be useful, but also keep in mind that this series of graphing calculators has some notable limitations including the small quantity of RAM (and memory in general). Still, to take on the task, you...