New answers tagged

1

As mentioned in the comments, $|\mathbb{N}^m|=|\mathbb{N}|$ for all $m\geq 1$. To get an idea of why this is the case, one can use Cantor-Schroeder-Bernstein, which says if $f:A\rightarrow B$ and $g: B\rightarrow A$ are injections, then there exists a bijection between $A$ and $B$. Informally, it says that if $B$ is no larger than $A$, and $A$ is no larger ...


2

A data point: I once asked a mathematician who did a PhD in algebra how long he would need to be able to understand the then recent FLT proof by Andrew Wiles. He estimated that he would need about 2 years of preparation. :-)


2

Terence Tao taught an advanced graduate course at UCLA called Math 285G: Perelman's proof of the Poincare conjecture. So, maybe the answer is that people tend to learn this stuff at the advanced graduate school level.


5

Regarding math at the graduate level and beyond, these programming debugging techniques can be quite effective: Rubber duck debugging: Find someone/something to whom to explain the bug and your solution. This is also effective for helping undergraduates find bugs in their solutions to word problems. Code review, i.e., presenting your argument to an at ...


1

Nice question! Admittedly, the elusive real numbers that are neither computable nor definable are perhaps of limited use in applications in physics and elsewhere (see links provided in the comments). As you point out, formal aspects such as existence of least upper bound are harder to handle if one works with a subset of the reals. One further extension ...


10

This is a very natural question, even if a little subjective (due to the inherent subjectivity of what does it mean "weird"). But the answer is that the axiom of choice is not the cause of weirdness. The axiom of choice is just a tool with which we can prove there is some uniform "mess" throughout the universe of sets. But it really helps us to rein down ...


0

An amazingly large part of Riemann-Stieltjes integration theory, as given in Apostole's MA, can be developed in the setting of rational number system. Completeness property of real number system is needed, only when we try to prove that every continuous function is integrable. It's the existential proofs of various theorems in Mathematics, that are the ...


2

One approach which might interest you is Automated proof checking, where you write the proof in a form that a computer can check the validity of. It is not easy to write proofs in this form, as humans tend to skip lots of "elementary" steps when writing proofs for other humans to read. A computer-checkable proof will be much longer and more tedious. ...


1

1) Is there any serious problem or inconsistency inside of mathematics if we take 1 as a prime number (other than just re-writing the theorems/lemmas/conjectures/definitions in a different way)? No, no serious problem, only a bunch of minor but unnecessary inconveniences. 2) Should we or should we not take 1 as a prime number? We should not. ...


3

Another method is to apply your method to other inputs that leads to larger errors. The idea is that the larger the error, the more obvious it becomes where the error is coming from. So, instead of 3 T's where the error is just a factor 2, if you consider, say, 100 T's then the error is astronomically larger, making the source of the error much more ...


1

Well one place we could start would be to note that we could define 𝔾 on a larger set; namely any real sequence of coefficients for which the generating series is absolutely convergent or even Abel summable (namely the limit of the series exists as s→1 even if the series diverges when one substitutes s=1). If we are willing to consider quasiprobability ...


1

I think this is the most natural way to define a prime number: An integer $n$ is prime if it has exactly two integer divisors, up to equivalence. If you don't like thinking about equivalence, we can also say: A positive integer $n$ is prime if it has exactly two positive integer divisors. This is analogous to many other concepts in mathematics: ...


0

I have answered an extremely similar question before. I will edit my old answer to better suit this question. I am so old I was taught that definition, that it's enough for a prime number to be divisible only by $1$ and itself. But my children and grandchildren were taught that a prime number must have exactly two distinct divisors among the positive ...


0

Consider the expression of the rotation matrix as angle and axis $$\frac{\delta}{\delta \phi} R(\phi,\mathbf{n}) = \frac{\delta}{\delta \phi} (\cos\phi \mathbf{I} + (1-cos\phi)\mathbf{n}\mathbf{n}^T - sin\phi \mathbf{n}^\times )= -\sin(\phi)\mathbf{I} + \sin\phi \mathbf{n}\mathbf{n}^T - cos\phi \mathbf{n}^\times = - \boldsymbol{x}^\times (\cos\phi \mathbf{I} ...


13

I'm a student currently learning math at the undergraduate level. I mostly learn by self-study, so "debugging" my thinking process during the learning process is crucial. (I will interpret "debug" a little less literally, so more as checking your own work using alternate solutions / shortcuts.) This list is not exhaustive, but off of the top of my head here ...


4

There are two strategies I could use to debug this problem. One is to write the steps out one at a time, with more detail. Let's use this as an example: Claim: There are 24 ways to order 3 Ts and 1 H if the Ts are distinguishable Claim: There are 3 Ts, so if the Ts are not distinguishable, each order is counted 3 times because each one is counted 3 ...


21

A somewhat parallel to debugging would be to hand check the various intermediate steps in your reasoning for an instance of the problem that is small enough to work out everything on paper. In your case "$4$ flips" is indeed such a small number. And it is easy enough to convince yourself that $4$ is the correct answer. So your second computation starts with ...


8

This is a question that can invite a wide variety of answers, but let me give two. The primary way that most mathematicians "debug" their proved theorems is by testing -- much like software is debugged. In your example, you conclude that your reasoning must be wrong because it gives the wrong answer. Theorems are created to be applied: if they are wrong, ...


50

As a comp sci major who dabbles in math, let me throw my perspective into this. Debugging math is a completely different experience from debugging programs. The difference is that while we have programs, they (tend to be) imperative - that is, they list out a series of steps that need to be followed. (I know that this is a gross generalization, and that ...


0

In the Big Ideas direction, I could suggest providing a mathematical formalisation of the intuitive idea that the period of small oscillations of the pendulum is independent of the amplitude, along the lines of this article.


0

It turns out that there indeed exists such a notion of mean, see these slides by Fletcher for example: http://www2.imm.dtu.dk/projects/manifold/Pres/fletcher.pdf To summarize the most important details (and avoid comments about having to do so), we can define the intrinsic mean as the point on the manifold minimizing the sum of the squared distances to the ...


1

In particular the little theorem of Fermat and Wilson's theorem would be valid because $a^{1-1}\equiv 1\pmod 1$ and $(1-1)!\equiv -1\pmod 1$ (since $a\equiv 0\pmod 1$ for all integer $a$). On the other hand $\mathbb Z/(1\cdot \mathbb Z)=\mathbb Z$ is not a field so, according to the corresponding well known theorem, $1$ is not a prime.


0

In my view solving all question of each exercise do not improve mathematics skill. Beacause many question are of same nature and easy as well the best way to do maths is do question levelwise. Firstly very easy which can be done orally. Than easy, difficult, and very difficult. The problem is that maximum student wasting their time in doing easy questions. ...


-3

Long way to explain, it is original method which allows to solve many problems in extremal combinatorics on sequences, text contains some unaccurances, but the idea is (as usual) not difficult. Author


-3

Hard and imaginative problems the whoascum county problem book the green/red book of math problems.


7

On the one hand, there has been a lot of success recently in creating fully formalized and computer-verified proofs of nontrivial theorems, including Hales' theorem, the prime number theorem, the Jordan curve theorem, and Gödel's incompleteness theorems. The sense I get from experts in the field is that the main challenge is time, not theory. It takes a long ...


6

The term "recursive" can indeed be synonymous with "computable". Robert Soare (himself one of the great names of the field) has devoted a number of expository or historical articles to the question of why "computable functions" and "computability" are sometimes called "recursive functions" and "recursion" / "recursive function theory". See, for example, ...


2

According to Wikipedia, Recursion is the process of repeating items in a self-similar way. Recursive functions (in the general sense) are functions that are defined in a recursive way, like recurrence relations are sequences that are defined in a recursive way or recursive data structures are data structures that are defined in a recursive way. In ...


0

There are really far too many examples to be able to give a comprehensive list (at least without spending several years compiling it, and by then it would already be outdated). Some that come to mind, mostly from probability theory: Fourier analysis applied to probability allows one to simplify calculations dramatically using characteristic functions. ...


2

The conversation seems almost equivalent to this hypothetical one: ME: Suppose there are positive integers $m$, $n$ such that $m^2=2n^2$. Then we can generate a list of all prime factors and read off the exponent of $2$. This exponent should be even (because it is in the unique factorization of $m^2$) but also odd (because it is in the unique factorization ...


0

The areas of cyclic polygons is a great place to look for this sort of number. Given the side lengths of a convex cyclic pentagon, you can find the area independent of how the sides are ordered. But as Robbins found, you have to solve a seventh degree equation which in general cannot be expected to be reducible. Succeed, and you still have to take the ...


0

I do not know whether it was expected that this number be transcendental, but it was shown by Firsching in 2014 (so, since this question was asked) that the side length of the largest regular tetrahedon that can be embedded in an icosahedron of unit side length is equal to the root $$1.347442850\!\ldots$$ of the irreducible degree $32$ (even) polynomial ...


-6

The key here is to let it go. If you let it bother you then you yourself risk becoming the "pro-Cantor crank". Brouwer, Poincare and Wittgenstein were all opponents of Cantor's work and many respectable, published, logicians have disputed Cantor's diagonal argument. Since it's a proof by contradiction it relies heavily upon the law of the excluded middle, ...


0

Much of this can be gleaned from the introduction section of John Baez's survey article on the octonions. Hamilton wanted a 3D number system analogous to the complex numbers, in that the length of a number was multiplicative and hence multiplication by unit norm numbers had the effect of rotating space. Hamilton used sum-of-squares identities to eventually ...


1

The prime number theorem is equivalent to the asymptotic estimate $p_n \sim n\ln n$, which is a direct link between logarithms and primes (not just counting primes). So $p_n/n\sim \ln n$.


2

There are heuristic arguments that suggest that the probability that a random integer $n$ is prime is on the order of $\log n$. That's implicit in some of Euler's work. It's why the natural logarithm appears in the prime number theorem. See this question and some of the answers there: "Probability" of a large integer being prime


2

The prime number theorem can also be formulated with the integral logarithm, i.e., as $\pi(x)\sim li(x)$. So we do not necessarily have $x/\log(x)$ directly. Still we can say that the logarithm appears naturally in connection with asymptotic results on primes, e.g., $$ \sum_{p\le x}\frac{1}{p}=\log(\log(x))+c+O(1/x). $$ Perhaps more convincing for you than ...


0

I really like "Math Horizons" (see here). MAA desctibes it as ...a vibrant and accessible forum for practitioners, students, educators, and enthusiasts of mathematics, dedicated to exploring the folklore, characters, and current happenings in mathematical culture. And I have to agree; the articles are quite interesting and most of them can be ...


1

Unfortunately your high-school teacher (and some of the other answers) is wrong. It is false that proving something of the form "$A = B$" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side, not to say necessary. Not necessary Let us first deal with the false ...


2

There are two common errors that these methods are preventing: \begin{align} 1 &< 2 &&\text{True.} \\ 0 \cdot 1 &< 0 \cdot 2 &&\text{Um...} \\ 0 &< 0 &&\text{False.} \end{align} \begin{align} 1 &= 2 &&\text{False.} \\ 0 \cdot 1 &= 0 \cdot 2 &&\text{Um...} \\ ...


4

To prove a statement is true, you must not use what you are trying to prove. The main problem is that you've misunderstood what the second teacher said (and possibly the meaning of the equals sign). What that second teacher is saying is do not take as prior facts what you're trying to prove. As one textbook puts it (Setek and Gallo, Fundamentals of ...


1

I think that the sympler motivation is that in a division algebra we can always solve an equation of the form $AX+B=0$ using the existence of opposite and inverse elements. If the algenbra is also associative and commutative we can solve also more complex equations and, in the case of $\mathbb{C}$ we have the wonderful result that any polinomial equation of ...


3

I think the motivation in pretty clear : the first step was to understand that complex numbers can represent points in the euclidian plane. That was a highly non-trivial and very deep idea. I think everyone will agree that this idea brings a lot to our understanding of the geometry of the plane (and conversely, of course, the geometrical point of view brings ...


5

short answer: equality is symmetric, implication is not (both are however transitive) longer answer: You are right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to conclude that A = C (because "=" is transitive and symmetric) Your teacher is right: if you prove that something true follows from A = B, i.e. A = ...


32

There's no conflict between your high school teacher's advice To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. and your professor's To prove a statement is true, you must not use what you are trying to prove. As in Siddarth Venu's answer, if you prove $a = c$ and $b = c$ ...


3

I agree with @Siddharth Venu. If we have to prove a=b, At times, on proceeding from LHS you may end up in a stage(intermediate step) which you cannot solve any further and you continue to solve RHS to arrive at the same stage i.e, a=c and b=c so you can conclude that a=b these kind of equality problems often comes in trigonometry Many chapters like ...


9

It is enough.. Consider this example: To prove: $a=b$ Proof: $$a=c$$ $$b=c$$ Since $a$ and $b$ are equal to the same thing, $a=b$. That is the exact technique you are using and it sure can be used.


2

They are called algebraists...


0

Yes. Take a look at Flash Anzan competitions in Japan. Basically, you are given 15 numbers, each between 100 and 999. The numbers are shown for a total of 2 seconds. You must then give the sum of the numbers in the next second.


1

Start with a Poisson process $\Pi$ of rate $a+b$ on $(0,\infty)$. Paint each point of the Poisson process either red (with probability $a/(a+b)$) or green (with probability $b/(a+b)$), the colors being i.i.d. and independent of the locations of the points themselves. Let $X:=\min\{t\in\Pi: t$ is painted red$\}$ and $Y:=\min\{t\in\Pi: t$ is painted green$\}$. ...



Top 50 recent answers are included