Base 10 only, and no funny stuff.
Archive for the ‘Mathematics’ Category
Consider the product 1! x 2! x 3! x ... x 99! x 100! — a very big number, but that doesn’t faze us mathematicians (since you won’t need to multiply out in any event).
The puzzle: Can you, by omitting exactly one of the factorials from the product, produce a perfect square? (For instance, omitting 3! would make the product be 1! x 2! x 4! x 5! x ... x 99! x 100!.)
no..it is not part of even numbers for sure they are 2, 4, 6, 8,...... [EV: For sure!]
it feels like an even number just as I feel that one is an odd number. I suppose it has to do with the maleness and femaleness of numbers.
“For mathematicians the answer is easy: zero is an even number.” Bollocks again, BBC, An even number is a number that is wholly divisible by two. A child knows this. You cannot wholly divide 0 by 2. Idiotic. Sensible people would point out that in the case of a zero you would step to the second number, in which case the two numbers are going to be divisible by ten and hence two. Perhaps this is where the dipshittery has sneaked in?
If zero is a number, then it must be a prime number too. I do not agree on thet.
If you take a number x and add/multiply/divide/substract 0, you will always end up with the same number [EV: !], but if you replace the zero with anything but zero you will always end up with a different number. So we can then cleary see that zero represents absence and not quantity. So isnt zero more the opposite of a number?
No zero is not even since it doesn’t gives any value when divided by it
Just to make things clear, zero is even, under the standard mathematical definition of evenness; if divided by two, it yields no remainder (0/2 = 0 with remainder 0), and that’s all it takes. And zero is a number, though it often refers to the absence of things, just as the word “nothing” is a word, though it often refers to the absence of things. (Likewise, negative integers are numbers, and can be even or odd, even though they don’t directly correspond to, say, the number of beans on your plate or the number of sheep in your flock.)
To be sure, zero is neither positive nor negative, and one is neither prime nor composite, two things that might confuse some people by analogy. But zero is even.
UPDATE: When I go to the individual page for this post, I see an ad with the headline, “Fix Defiant ODD Children” — I assume it’s this ODD, not the one in the title of this post.
A few days ago, over the breakfast table, my 9-year-old tells me:
Dad, I told a boy at school that any number to the power zero is one. He said that was idiotic.
Indeed, I had told Ben earlier that any (nonzero) number to the power zero is one. I think I had even tried to explain the reason to him, but I wasn’t surprised that the reason hadn’t stuck. So I felt I had to explain it again. And then yesterday, in the math puzzle thread, that boy-at-school’s more polite uncle (OK, I’m making up the family relationship) commented:
So I’m a lawyer and therefore math dumb, but I don’t understand how you raise a number to a square root power, or to any non-whole number power for that matter. So 2^2 is the same as 2 x 2. 2^3 is the same as 2 x 2 x 2. What is 2^1.4?
Here’s the thing: As Leopold Kronecker supposedly observed in the 19th century, God created the positive integers (and maybe zero), and all else is the work of man. Less metaphysically, when we talk about 2+3, or 2×3, or 6/2, or 3-2, we’re talking about things that obviously correspond to real world phenomena: combining piles of things, or splitting piles of things. When we exponentiate integers, we can also easily conceptualize this as corresponding to an obvious operation on integers: 3^4 is four threes multiplied together. We might throw in the rational numbers to this real-world math as well.
But, seriously, doesn’t it seem a bit idiotic to talk with a straight face about multiplying zero threes together? And even beyond that, negative five — just what is that? Does anyone actually have negative five of anything? (I don’t mean owing five of something, which it turns out can be conveniently represented as having negative five, I mean actually having negative five of something.)
Here’s how I explained things to my 9-year-old: Let’s look at 1234.56. What does the 1 stand for? “Thousands.” And 10 to the power three is ... “One thousand.” How about the 2? “Hundreds.” And 10 to the power two is ... “One hundred!” (Always good to have some confidence-building questions there.) The 3? “Tens.” And 10 the power one is ... “Ten.”
Now at this point, we get to the 4, in the ones place and it sure looks very convenient just to go from 10^3 to 10^2 to 10^1 to 10^0 — the thousands to hundreds to tens to ones. It’s useful to just say that 10^0 = 1. And then since the 5 after the decimal is tenths, you’d want to say that 10^-1 = 0.1, and then that 10^-2 = 0.01, and so on.
There’s another way of thinking about this, I told my son: We know that a^b x a^c = a^(b+c), for instance 2^2 x 2^3 = 2^5 (4 x 8 = 32); always good to see that with familiar, observable integers. If that’s so, then if c = 0, a^b x a^0 = a^(b+0), which means that a^0 = a^b/a^b = 1. And you can do the same with negative powers, too; adapting the formula I gave at the start of the paragraph to negative numbers, we see that it fits well to say that a^(-b) equals 1/a^b.
How about my commenter? Well, let’s say this: We know that (a^b)^c = a^(b*c). Thus, for instance, (2^2)^3 = 4^3 = 64 = 2^6 = 2^(2*3). We can even easily show that this is true, in a way that makes it intuitively clear for the sorts of numbers for which exponentiation is intuitively clear. If that’s so, then what’s 9^(1/2)? Well, (9^1/2)^2 = 9^(1/2*2) = 9, so (9^1/2) = square root of 9. Raising something to the power 1/2 is taking the square root. Raising it to the power 1/10 is taking the tenth root. (1024^0.1 = the tenth root of 1024 = 2.) Raising a number to the power 1.4 is multiplying it by the square of its fifth root (i.e., raising it to the power 1+2/5), though it’s not clear that conceptualizing it this way is that helpful.
So once we get past some basic operations involving the positive integers, zero, and the positive rational numbers, which correspond pretty directly to the real world, most other things — negative numbers, raising something to the power 0, raising it to a negative power, raising it to a fractional power, and so on — have properties that mathematicians have agreed on because they’re useful.
We make some mathematical statements because they are “true” in the sense of corresponding to the observable world; the commutativity of addition of positive integers, which is to say (x + y = y + x). But other mathematical statements we make because that creates a more coherent way of dealing with certain sets of problems (though this coherence will often also help us deal with the observable world).
It’s not so much idiotic, I’d tell my son’s classmate, as postmodern. In fact, young man, it all depends on what “is” is. (Well, maybe I wouldn’t tell him exactly that, because then I might have to explain the reference, and that might give me a bad reputation around school.)
If by “a number to the power zero is one” you mean something like what we normally mean by “two plus three is five,” which is to say “there’s a real-world phenomenon that clearly corresponds to raising something to the power zero, and the result of that is one,” then that’s not really so. If you mean “when you multiple zero copies of a number by itself, you get one,” then that’s not really so, either. But if you mean “it really is very convenient to treat a number to the power zero as being one, and because of that pretty much all mathematicians — and those students of theirs who paid attention — define the operation in a way that the result is one,” then a number to the power zero indeed “is” one.
Idiotic? Postmodern? Sensible? Maybe all three.
Yesterday, I posed this math puzzle: Treating “^” as meaning exponentiation, and treating the exponentiation chain as going on infinitely, solve for x:
x^(x^(x^(x^(x^(x^...))))) = 2
Many people in the comments got the right answer — the square root of 2 — some people were unsure, so I thought I’d briefly blog a follow-up.
Here’s a way to figure it out: Let’s focus on the material in the outermost set of parentheses, which we set in bold below:
x^(x^(x^(x^(x^(x^...))))) = 2
That material is the same as the entire left-hand side of the equation, right? It’s x^(x^(x^(x^(x^...)))). But we were told as part of the problem what the left-hand side equals — it equals the right-hand side, which is 2. So substituting that in, we get
x^2 = 2
x must therefore be the square root of 2. No need for logarithms or anything more exotic. If you doubt the result, and want to see it illustrated, build an Excel spreadsheet with block A1 set to =sqrt(2) and A2 set to =sqrt(2)^A1, and then copy A2 down into another fifty-odd cells in column A. (The way Excel works, that will set each cell to sqrt(2) to the power of the previous cell, so A5, for instance, becomes =sqrt(2)^A4.) You’ll see the cells getting closer and closer to 2. Naturally, that’s not a mathematical solution — I began with the solution, once we assume that the question is correct and the series converges on 2 — but it might be a helpful illustration. (Note that proving that the series converges to 2 when x=sqrt(2), rather than solving the problem for x when we know the series converges to 2, is somewhat harder.)
The same technique might be familiar to some of you if you’ve ever studied chain fractions, such as (thanks to Wikipedia for the figure):
If you get rid of the 1+ on the left, and just call what’s left x, you see that
x = 1 / (2+x)
That means x^2 + 2x = 1, which means x^2 + 2x + 1 = (x+1)^2 = 2, or x = sqrt(2)-1. Bring back the 1+ on the left, and we see the chain fraction adds up to sqrt(2). Solving the chain exponentiation problem ends up using much the same sort of recursive solution.
UPDATE: I originally said that -sqrt(2) was also a solution, but a comment leads me to be uncertain of that, for reasons related to the zaniness of non-integer powers of negative numbers, so let me set that aside for now.
Pretty easy, but I found it amusing. Treating “^” as meaning exponentiation, and treating the exponentiation chain as going on infinitely, solve for x:
x^(x^(x^(x^(x^(x^...))))) = 2
I’m on a family trip, and will be blogging little if at all this week; but I had a chance to look at a New York Times op-ed titled “Is Algebra Necessary?” and thought it was worth passing along to see what our readers thought of it.
My own quick reaction to the op-ed is negative — though I’m not certain of this, I suspect that algebraic problem-solving teaches useful mental habits that both open up possible future careers and also help train people’s general problem-solving abilities — but I don’t have time to say more about it. So instead of substance, I thought I’d note this sentence:
(How many college graduates remember what Fermat’s dilemma was all about?)
I remember both Fermat’s last theorem and his little theorem, but not Fermat’s dilemma — and neither does Google Books, which reports one hit for “Fermat’s dilemma,” referring to a problem in a book on math teaching in which a hypothetical math teacher named Mr. Fermat faces a dilemma.
Am I missing some thing that really is called “Fermat’s dilemma”? Or is it an erroneous reference to the seemingly very obscure Fermat’s Lemma (7 Google Books hits)? Or is it perhaps some deep joke on the author’s part that I’m missing?
UPDATE: Thread-winner from Orin Kerr: “Fermat’s Dilemma is whether to admit that you don’t know the proof for a theorem or just to pretend you know the proof but you don’t have space in the margin to explain it.”
Say that the only point-scoring events in a football game are field goals (3 points) and touchdowns with one-point conversions (7 points). Some point totals cannot be scored in such a game — for instance, 1, 2, and 4. What is the highest integer point total that cannot be scored using just 3-pointers and 7-pointers?
Now say that we exclude field goals, but allow touchdowns with missed-conversions, so the only point-scoring events are 6 points and 7 points. What is the highest point total that cannot be scored using just 6-pointers and 7-pointers?
And now let’s generalize. Say that there are two point-scoring events, one which yields a points and one which yields b points. If a and b have a common divisor, then of course there are an infinite number of positive integer point totals that can’t be scored; for instance, if all you have is 4-pointers and 6-pointers, then all the scores will be even, and any odd score will be unachievable. So let’s assume a and b are relatively prime, which is to say that they don’t have any common divisors. What is the highest point total that cannot be scored using just a-pointers and b-pointers?
UPDATE: Thanks to commenter Nick, I now know this is the Frobenius coin problem.
I was shocked at the number of people who took the view that 0 was neither even nor odd. (I was even more shocked by those who thought 0 was odd, and by those who thought 0 was both even and odd, but there were comparatively few of those.) Under every definition of even that I’ve ever seen, and under every one that to my knowledge has any mathematical utility, an integer is even if it is divisible by 2 with no remainder. 0 is divisible by 2 with no remainder (0/2=0). Therefore, 0 is even. End of story, though if you want much more of the story, read this monster comments thread.
But what shocked me even more was a link to McGraw-Hill’s Catholic High School Entrance Exams p. 213 (2d ed. 2009), which asserts (twice) that “The number zero (0) is an integer but is neither even nor odd.” As I said, this departs from all that I’ve ever seen of actual mathematical definitions. And the material in the book is actually inconsistent with that very definition; for instance, later in the page, it says that
(even integer) +/- (even integer) = even integer
(odd integer) +/- (odd integer) = even integer
But that of course is wrong if 0 isn’t even, and right only if 0 is even. (Consider 2-2 and 3-3.) And of course these equations, and many others, are part of the reason that having 0 be even is such a useful definition, one that mathematicians have settled on.
In any case, this assertion in the book can’t be doing its readers any good. I tried to find an e-mail address to which I could complain, but I couldn’t. If any of you can let me know whom I can contact on this, I’d much appreciate it.
I had discussed this before, but I thought it would be good to do a quick survey on this. Please note that this is not a trick question. Also please give what you seriously believe to be the correct answer; and please vote even if you think the answer is obvious.