# John

Senior Members

417

1. ## How to start a formal proof in symbolic logic?

No. You've got the assumptions listed above, and you're assuming K is true. Starting with the assumption that K is true, you have to reason your way through the other assumptions you have to arrive at the conclusion that H is also true. This will prove that K implies H. So far, we've seen that since K is true, ~K must be false. And since ~K is false, I must also be false. Now given that I is false, what can we conclude, given the various other statements we assume to be true?
2. ## How to start a formal proof in symbolic logic?

Almost. The biconditional is true by assumption. Since $\neg K$ is false, the fact that $I \iff \neg K$ tells us $I$ is false.
3. ## How to start a formal proof in symbolic logic?

Yes, and given that $I \iff \neg K$, what does this tell us?
4. ## How to start a formal proof in symbolic logic?

Well, since $K \wedge \neg K$ is a contradiction, and $K$ is true, what do we know about $\neg K$?
5. ## How to start a formal proof in symbolic logic?

Hm. Well, the way a proof works in general is we start with some assumptions, and using new conclusions we can logically deduce from those assumptions (and further conclusions we can logically deduce from those conclusions, and so on), we work towards some end goal. So for this proof, our assumption is that $K$ is true. We also have that $I \iff \neg K$. Since $K$ is true, what do we know about $\neg K$? And what does this tell us, given our biconditional?
6. ## How to start a formal proof in symbolic logic?

I'm not sure what that program is, but I'm assuming $\neg F \implies (G \vee H)$, $(F \vee G) \implies (I \wedge J)$, and $I \iff \neg K$ are propositions you can take to be true, and thus your goal is to use these to show that $K \implies H$. Is that correct? Is the problem that you don't see how to work from K to H, or is it that you're not sure how to express the proof using this program?
7. ## Question about square roots!

In the case of 16, $-\sqrt{16} = -4$ and $\sqrt{16} = 4$. The negative radical sign doesn't denote a positive value unless we take the radical sign to specifically refer to the negative root, which is something we probably agree would be unusual. As for the rest, I agree with you. If we take the radical to denote either root (thanks for the correction of my sloppy wording), then the $\pm$ is redundant. But usually we do prepend $\pm$ in contexts where either root is a solution, which lends further credence to my claim that the radical itself is usually taken to be the principal square root. I absolutely agree. My point is that, without an explicit declaration by the mathematician in question, the radical sign is normally taken to denote the positive root. I have stated that $\sqrt{x}$ denotes the principal square root of x in every post I've made in this thread (though a couple of times it was somewhat indirect). I've irritated even myself with the repetition, so surely you've noticed.
8. ## Question about square roots!

I've explained and shown in several sources that the radical sign is used to denote specifically the principal square root. I've also explained that notation isn't sacred, and one can easily use the radical to denote both roots, but this isn't the common usage. This is sort of the definition of "convention" in this context. In the context of the square root function, 16 would be the input and the radical sign would be the function. In general, $\sqrt{16}$ represents the principal square root, as I've stated many times. I reject the last one. Edit: Also, why are we even still discussing this? We agree on the concepts involved. The only disagreement is on the default meaning of a single symbol. It's silly to get this caught up in notation.
9. ## Question about square roots!

The passage you quoted doesn't seem to mention the radical sign at all, but simply explains that even roots may be positive or negative, which is not in dispute. I'm talking about the definition of the symbol itself, which is the principal square root. As an example of ubiquitous use, check out the quadratic formula. Further examples are given by the Wikipedia articles on the square root and nth root, and the results of plugging something like x = sqrt(16) into Wolfram Alpha (which also has a page of information about the square root here). While these may not be century-old books, they are the results of the collaborative efforts of hundreds or thousands of individuals, which at least suggests the definition is common. These sources are also more up-to-date than a book written in 1880. In any case, I think all points that can be made about this have been made. I appreciate you forcing me to reexamine my thoughts and statements on this, even if I have ended up reaching the same conclusions I had before.
10. ## Question about square roots!

I am following the (usual) definition. $x^2 = 16$ has the solutions $x = \pm \sqrt{16} = \pm 4$. $x = \sqrt{16}$ has the solution $x = 4$. When Amaton says, in his post, that $\sqrt{16} = 4$, he is correct. There is no issue here. I think we may be arguing about different things.
11. ## Question about square roots!

That's perfectly true, but $\sqrt{x}$ normally refers to the positive root specifically. There is no conflict here.
12. ## Question about square roots!

I'm stating the (usual) definition of notation. You might as well ask me to logically develop why we normally use "+" to denote addition. If you did, then I'd say the same thing: It's a matter of convention, and a Google search should provide sufficient evidence to show this. Notation isn't sacred. If I like, I can define $\sqrt{x}$ to refer to $e^{x}$ or $x \choose {x-6}$ or anything else. It'd be odd, and there's no good reason--or at least, no reason that outweighs the confusion it'd likely cause--to do so, but I could. However, the usual definition of $\sqrt{x}$ is the principal square root of x. This doesn't require rigorous development. It's simply the way notation has developed over the years. This definition also doesn't conflict with the quoted passage, unless you're referring to the fact that I translated $\sqrt{x}$ as "the square root of x." But if that bothers you, then ignore it. The rest of my post remains true.
13. ## Question about square roots!

We're just talking about the definition of the radical sign. By default, it indicates the principal (i.e. positive) square root of the radicand. In situations where both square roots of the radicand are desired, we prepend $\pm$ to specify this. So if someone asks for $\sqrt{16}$, i.e. "the square root of 16," the answer is 4. If someone asks for solutions to the equation $x^2 = 16$, i.e. "what numbers are square roots of 16," the answers are 4 and -4, and we can express these as $\pm\sqrt{16}$. If you want to use $\sqrt{16}$ to mean both 4 and -4, then that's fine, but it may cause some confusion unless this usage is specified (though often the intended meaning will probably be clear from the context).
14. ## Question about square roots!

The difference between an equation and a function is irrelevant. The notation $\sqrt{x}$ refers to the principal square root of x. It's not strange. It's not controversial. I don't know why this is such a sticking point.
15. ## Question about square roots!

It's not my convention. It's the convention used by mathematicians in general. While the fact that it's a convention means there's probably no absolute authority to make the definition, a Google search for definitions of the square root or the radical sign should convince you. And I'm honestly not sure what your point is. All I've said is that $\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|$. It's simply a matter of notation. Questions about the roots of various polynomial equations are somewhat unrelated. But for the particular example of $\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)}$, so long as $1 - 2x + 5{x^2} - 4{x^3} + 4{x^4} > 0$, then yes, $\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)} > 0$. Using Wolfram Alpha, it seems the two square roots of $1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}$ (which, as it turns out, is positive for all real x) are $\pm(2x^2 - x + 1)$, so $\sqrt {\left( {1 - 2x + 5{x^2} - 4{x^3} + 4{x^4}} \right)} = |2x^2 - x + 1|$, which is positive. Edit: Just saw your edit. The notation indicates the principal square root, so $\sqrt{(-4)^{2}} = |{-4}| = 4$.
16. ## Question about square roots!

Er, no, why? Finding the roots of a general polynomial equation isn't the same as taking the square root of a real number. And as stated before, if you look at $x^{2} - 16 = 0$, then there are two solutions, $\sqrt{16} = 4$ and $-\sqrt{16} = -4$. But the notation, $\sqrt{16}$, refers to the principal square root of 16, i.e. 4. Disagree if you like, but that is the convention.
17. ## Question about square roots!

There are two square roots of 16, but $\sqrt{16} = 4$ unless otherwise specified. In general, $\forall x \in \mathbb{R}, \sqrt{x^{2}} = |x|$. If we have the equation $x^{2} - 16 = 0$, then $x = \pm \sqrt{16} = \pm 4$. It's not a huge deal, regardless, but Amaton's post is correct.
18. ## Question about square roots!

The notation $\sqrt{x}$, for $x\in\mathbb{R}_{\ge 0}$, refers to the principal square root (which is positive) unless otherwise specified. Amaton's post is correct (except that notation is somewhat flexible--it just depends on how much you want to annoy your readers ).
19. ## Question

Although it's not a book, you might like Wikipedia's Introduction to quantum mechanics. This is an article intended for readers who haven't necessarily had the physics and mathematics education necessarily for a rigorous introduction to the subject. Section 8 specifically discusses quantum entanglement. I admit I haven't read it yet, but if it's anything like the similar introductions to special and general relativity (which I have read), then it's a pretty good read. There are a few formulas, but these articles are primarily conceptual in nature.
20. ## Two Questions Regarding a Corollary about Uniqueness

I'm assuming these are questions you're asking for yourself, and not questions you have to answer for an assignment. Mods forgive me if I'm incorrect. I'm assuming N is supposed to be the natural numbers. We have some function f that has the Cartesian product of N and X (i.e. all the ordered pairs (n, x) for n in N and x in X) as its domain and X as its codomain. The corollary states that if we take any x in X, then there exists a unique sequence starting with x (this is what x(1) = x means) and defined by the recurrence relation x(n + 1) = f(n, x(n)), or alternatively x(n) = f(n - 1, x(n - 1)). Thus, letting x(1) = x, we have that x(2) = f(1, x(1)), x(3) = f(2, x(2)), etc. The precise form of this sequence will depend on the definition of f. All we know is that each term in the sequence must be an element of X (since X is the codomain of f). As an example, let X be the rational numbers, and let f be defined such that f(n, x) = nx. Then, taking x = 3/5, we have that x(1) = 3/5, x(2) = f(1, x(1)) = (1)(3/5) = 3/5, x(3) = f(2, x(2)) = (2)(3/5) = 6/5, x(4) = f(3, x(3)) = (3)(6/5) = 18/5, x(5) = f(4, x(4)) = (4)(18/5) = 72/5, and so on.
21. ## Mod feedback

Replacement of an unfamiliar word with a perhaps more common synonym doesn't help? This is news to me. To go with the cancer analogy, the situation is more akin to removing moles or polyps. The mods don't close every new thread in an effort to prevent problems, but they do close threads that seem likely to cause problems. My statement is the result of hanging around these forums, talking to mods, and knowing roughly how their process works. I did see your thread, before it was locked, and I'm not sure whether I would have locked it, though I might have at least suggested a change in title. However, I'm not a mod, and even if I were, if several others had agreed with locking the thread, the outcome would have been the same. As for extrapolation, if not knowing is grounds for rendering extrapolation invalid, then any extrapolation is invalid. Any extrapolation comes with a certain degree of uncertainty. It's the nature of inductive reasoning. In my experience, moderator actions are almost always justified, and those few that aren't are quickly reversed--it's that whole multiple moderators thing in action again. I can't be certain that all future moderator actions will be justified, but based on my experience, I think it's safe to say any given new action is. It's difficult to set aside one's ego in the face of criticism, especially when it's public. You seem to have a pretty good track record here, and no one thinks you're an idiot just because a couple of your posts have resulted in moderator response. Whatever you think about the age, qualifications, or whatever else of a moderator, the fact remains that the moderators are given the authority to make decisions regarding possible violations of the forum rules. Also, you have the ability to justify yourself, through the proper channels.

23. ## High school physics or chemistry?

If you're planning to go to college and do a STEM degree (or at least, take calculus), then I recommend holding off on physics until you've done calculus and can take calculus-based physics courses. Otherwise, as mentioned previously, it's a matter of personal preference and deciding which instructor is more competent. Either class has the potential to be pretty interesting and useful.
24. ## Vectors

This is what I was alluding to earlier. There's actually an even easier formula to use.
25. ## Math Team Questions That I Have; Already Did The Questions, Just Asking For Another Way To Solve

For the first, you could let the first digit be x and the second be y, and use the equation 10x + y + 9 = 10y + x. Simplifying, you end up with x = y - 1, i.e. the second digit is one more than the first, which is the pattern you found. I'm not sure your second example (increasing by 6) is possible, since using the same reasoning (and assuming you're still talking about two-digit numbers), we end up with x = y - 2/3. For the second problem, what you're looking for is the set of solutions to the equation 83 = nx + 2, or 81 = nx, i.e. x divides 81. The factors of 81 are 1, 3, 9, 27, and 81. Clearly 1 isn't a solution. For 275 and a remainder of 3, we have that x must divide 272. The factors of 272 are 1, 2, 4, 8, 16, 17, 34, 68, 136, and 272. A quick way to find these factors is to find the prime factorization of 272, which is 2^4 * 17, and realize that since each of the prime factors divides 272, any product consisting of the prime factors also divides 272. For the last, since the number is divisible by 24, it must be divisible by both 8 and 3. The divisibility rule for 8 says a number is divisible by 8 if its last three digits are divisible by 8. So you have 65B, and you must find B. An immediate thing that pops into my mind is 640 is divisible by 8, so if we add two 8's to that, we arrive at 656 being divisible by 8. Therefore, B is 6. Now, the divisibility rule for 3 says a number is divisible by 3 if the sum of its digits is divisible by 3. At this point, we have 8A656, so we need to find A such that 8 + 6 + 5 + 6 + A = 25 + A is divisible by 3. We want the smallest number, so we can start with A = 0 and work our way up. Obviously 0 doesn't work, and letting A = 1 doesn't work either (the sum will be 26). But when A = 2, we have 8 + 2 + 6 + 5 + 6 = 27, which most certainly divides 3. Thus, 82656 is the smallest number of the form 8A65B that divides 24.
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