# Tag Info

3

I always say that the most difficult exercise in my undergrad studies was the first question in linear algebra. We were taught about the axioms of a field. Then we had to prove the following thing: For every $x$, $x+0=x$. The catch is that the axioms we were given stated $0+x=x$. So we had to use the axiom of commutativity first, then we could conclude ...

1

I think the answer has three parts Part 1: If you ask a random man at Walmart about what $$\lim_{n\to \infty} \frac{1}{n}$$ is, then you might not get much. If you tell them that it is $0$, then they probably don't think that it is obvious. If you go to a high level research talk, you will hear "it is obvious / it is clear that" a lot. And you might not ...

0

Well those examples mostly use as the definition of the term,for example if you're gonna make a definition for continuity you're gonna start with function $f(x)=x$.Also if you're gonna explain somebody how to prove continuity of a function you'll go with $f(x)=x$.And then again also some things we thought obvious are wrong,like for example that we aren't ...

2

There was just an answer I provided a couple of minutes ago that was wrong. The question was: $$\lim_{n\to\infty}(1+2^{-n})^{2^{n+2}}$$ What I thought: $$\left(1+\frac1{2^{\infty}}\right)^{2^{\infty}}=\left(1+0\right)^\infty=1$$ But the answer was actually $e^4$. Even the computer made a mistake (when $n$ got too high). Sometimes, something that seems ...

4

Because obviously a continuous function must be piecewise monotone. And therefore differentiable at all but at most countably many points. Ampère "proved" this result in 1806 and it was considered a theorem for quite a while. Then Riemann came up with an example of a function than when integrated produces a function (i.e. $x \mapsto \int_a^x f(x)\,dx$) ...

0

The obvious is hard to prove and often wrong.

11

The main procedural reason is to show that your axioms correctly capture what you want them to capture: that is to say they are both "correct" and sufficient. If it turned out that under our axioms $\lim\limits_{n\to\infty}\dfrac{1}{n}\neq0$ then we would probably choose a different definition of $\lim$ (or a different name for it), since it would not be ...

-2

In my opinion, only axioms should be treated as obvious, above all while a theory is being explained to others. I think it is simply immoral for a mathematician who is writing a proof of a proposition in a book not to give every smallest detail in the chain of logical inferences, skipping the task of making the subject perfectly clear through a lazy abuse of ...

23

Because sometimes, things that should be "obvious" turn out to be completely false. Here are some examples: Switching doors in the Monty Hall problem "obviously" should not affect the outcome. Since Gabriel's horn has finite volume, then it "obviously" has finite surface area. "Obviously" we cannot decompose a 3-dimensional sphere into a finite number of ...

8

Things that are obvious to one person are not necessarily obvious to another. Futhermore they dispel (most) skeptics. Just thinking they are true does not mean they are true. For example, before I entered university, I was under the impression that there were twice the number of elements in $\mathbb{Z}$ than in $\mathbb{N}$. I would've called this obvious, ...

0

With leaps of twelve hours each, the sequence is: $$1\,,\,3\,,\,9\,,\,27\,,\,\ldots$$ You have the model of the first elements, and this is clearly a geometric sequence, thus...

0

Mathematics has a pure side that can be very abstract, as you say, but there is also applied mathematics, which, as the name suggests, is more relevant to the "real world" and areas like physics. Analysis and algebra (pure) can be too theoretical and boring for some people. Personally, I find analysis (functional analysis, operator theory, measure theory, ...

1

By using the definition of the derivative and L'Hopital's rule to get \begin{align} f'(1)=\lim_{x\to0}{f(1)-f(1-x)\over x}&\le\ge\lim_{x\to0}{f(1)+x\cos x-f(x+\cos x)\over x}\\ &=1+\lim_{x\to0}{f(1)-f(x+\cos x)\over x}\\ &=1-\lim_{x\to0}(1-\sin x)f'(x+\cos x)\\ &=1-f'(1) \end{align} where, in the first line, the $\le$ sign applies when ...

1

$$\lim_{x\to 0}\frac{f(x+\cos x)-f(1-x)}{x}=\lim_{x\to 0}{\bigg[(1-\sin x)f'(x+\cos x)+f'(1-x)\bigg]}=2f'(1)$$ We also have, for all $x>0\in\mathbb{R}$ that $$\frac{f(x+\cos x)-f(1-x)}{x}\le \cos x$$ and for all $x<0\in\mathbb{R}$ that $$\frac{f(x+\cos x)-f(1-x)}{x}\ge \cos x$$ These imply that $$\lim_{x\to 0^+}\frac{f(x+\cos x)-f(1-x)}{x}\le 1$$ and ...

0

Allow me to join the party guys... This is another proof of the Pythagorean theorem by The 20th US President James A. Garfield. A nice explanation about Garfield's proof of the Pythagorean theorem can be found on Khan Academy.

1

You could try proving theorems you took for granted(ofc don't look for the proof),since most of them have an elementary proof(with this I mean they don't need any higher math stuff to be proven)

4

When you do $m\cdot n$, you start from $0$ and keep adding $n$ as many times as $m$ says: so if $m=0$, you just have $0$ because you have nothing else to do; if $m=1$, you get $0+n$; if $m=2$ you do $0+n+n$. And so on. Exponentiation is exactly the same, but with multiplication replacing addition and $1$ replacing $0$: for doing $n^m$ you start from $1$ and ...

1

1) Learn LaTeX. This can't hurt and it's pretty easy to do when done without time pressure, say like before beginning mathematics studies. 2) Be sure you know as good as you can all the usual high school stuff in mathematics. It can be hard to know exactly what this means, but besides asking people, surfing the internet and etc., you could begin by going in ...

1

You asked for suggestions other than learning Matlab and Latex, I have this suggestion: Self Study Mathematics

4

\begin{align*} n^m &= n\times \ldots \times n\; (\textrm{for}\; m \in \mathbb{Z}_{>0})\\ &\vdots\\ n^2 &= n\times n\\ n^1 &= n\\ n^0 &= 1\\ n^{-1} &= \frac{1}{n}\\ n^{-2} &= \frac{1}{n\times n}\\ &\vdots\end{align*} To go from $n^{m}$ to $n^{m+1}$, you multiply by $n$, and to go from $n^{m}$ to $n^{m-1}$, you divide by ...

11

Exponentiation satisfies the laws of exponents: $a^{b+c} = a^ba^c$ . If we want this law to still be satisfied when we extend to the case $b=0$, we need to have $a^{0+c} = a^0a^c$ , and therefore we need to have $a^0 = 1$.

1

You can think of the rule that $\frac{x}{x}=1$ for any nonzero number $x$. Then, using for example $n^2$, we get: $$1=\frac{n^2}{n^2}=n^{2-2}=n^0$$ Hence it makes sense to have $n^0=1$.

15

HINT: Using Exponent Laws(1) , (2) $$n^{a+b}=n^a\cdot n^b$$ Setting $a=0,$ $$n^b=n^b\cdot n^0$$ Check when $n^b$ can be cancelled

3

There are outlines by Schaum's that have all of those subjects. The outlines are very intuitive and there are problems with given solutions for learning. There are also books called Demystified that offer these subjects with the same format. These books also cover basic algebra, geometry up to topics that include calculus, linear algebra, complex analysis ...

2

Mathematics is obviously contradictory. Predicativist systems disagree with classical systems on the allowable scope of quantification in definitions. Constructivists disagree on the use of excluded middle. Ultrafinitists disagree on induction principles. Not only that, there are mathematical systems that are intentionally contradictory. Paraconsistent ...

0

One thing is when the answer to the first parts leads into the next parts as you said, and I don't have experienced that, in particular. One different thing is an exercise like the integral in your example, where several techniques must be applied in succession in order to solve. Thinking to this second model, comes to my mind the differential equation ...

1

mjqxxxx already answered in the comments: $i$ is indexing the players, so different players may have different payoffs, $u_i$. The potential condition requires that the gain (or loss) from changing the action of one player is the same for the potential function. Maybe the differential case is easy to illustrate: we require that ...

2

Even though you are assuming that unbiased student scores should be normally distributed, you are actually trying to make inferences on a very different object...a sample of 75 actual scores. Of course, these scores will be related to the underlying normal distribution, but that underlying normal distribution only holds exactly when the class size is ...

1

Suppose that $n=2$ and $f(x)=x_1+x_2$, so $f$ is always increasing in both $x_1$ and $x_2$. So the problem then is to find all combinations of $x_1,x_2$ that satisfy the inequality constraints and see which one yields the highest value of $f$. In practice, what you do is that you implicitly cycle through different combinations of constraints in an ...

0

Check out the encyclopaedic "Mathematics for Computer Science" by Lehman, Leighton, and Meyer. It is a set of lecture notes (updated each term or so), which covers lots of ground. I assume they will publish it sometime in the future, some parts are still marked as "work in progress." Best of all, it is available for free as PDF.

0

I've never used Rosen. I've used both the Epp book and the Johnsonbaugh book. Epp is a very nice introductory book. It is geared more towards mathematics than to CS, but it is a nice book if your math is weak. Johnsonbaugh is far more technical, and I think far more relevant to a computer scientist. It is heavy on topics like graph theory, combinatorial ...

2

"Are there contradictions in maths?" It's a great question. Before Russell Paradox It was generally believed that there were not contradictions. Better said, the matter hadn't been spotted yet. After Bertrand Russell the first crisis in maths appeared. And, from my point of view, the issue wasn't solved and it is not solved yet. I know there is the Sets ...

0

I would suggest the book by Rosen. It is easy to read, has lots of examples and has solutions to some of the exercise problems. It is one of the standard prescribed books for discrete mathematics (at least in CS departments)

3

A recent example I found which is credited to Martin Gardner and is similar to some of the others posted here but perhaps with a slightly different reason for being wrong, as the diagonal cut really is straight. I found the image at a blog belonging to Greg Ross. Spoilers

1

$$\log6=\log(1+2+3)=\log 1+\log 2+\log 3$$

1

In mathematics contradiction is a useful tool to prove some theorems. If you want prove some statement, sometimes its easier to disprove that the statement is not true. So you assume its not true and derive a contradiction, which proves that the statement is in fact true. Apart from that, mathematics is build on a few fundamental concepts that don't need a ...

2

Actually in the first development of set theory a very intereseting contradiction was introduced. It is the so called Russel Paradox. I say it is interesting because is one of the very few cases where our intuition on what should be considered a founding truth might fail. Also it is quite interesting the fact that such a paradox is a mathematical translation ...

1

We use cm when we want to measure the length of a one dimensional object. Cm can be used when finding perimeter, or when finding the length of a line. cm$^2$ is used when we want to find the specific area of something. Why cm$^2$? Let's say we want to find the area of a rectangle that is $8$cm long and $7$cm wide. Normally you would just do $8\times 7=56$ ...

2

This has 1 point $$\cdot$$ This has a length of 0 cm: $$\cdot$$ This has an area of 0 cm${}^2$: $$\cdot$$ This has $\infty$ points $$-$$ This has a length of approximately one cm: $$-$$ This has an area of zero cm${}^2$: $$-$$ This has $\infty$ points $$\blacksquare$$ This has a length of $\infty$ cm: $$\blacksquare$$ This ...

1

They measure different things: $\mathrm{cm}$ is a measure of length, and $\mathrm{cm}^2$ is a measure of area (and $\mathrm{cm}^3$ is a measure of volume). If a rectangle has side lengths $a\, \mathrm{cm}$ and $b\, \mathrm{cm}$ then the area of the rectangle is $(a \times b)\, \mathrm{cm}^2$. Likewise, if a cuboid side lengths $a\, \mathrm{cm}$, $b\, ... 7 The most common appearance of contradictions in mathematics is when one inserts their own ideas about a mathematical concept or object that aren't actually true. For example, inserting the idea that$1 + 2 + 3 + 4 + \ldotsis supposed to represent the kind of infinite summation one learns about in introductory calculus. There are a number of other ... 51 Someone told me that math has a lot of contradictions. Correct mathematics does not, as far as we know. However, mathematicians amuse themselves with little "proofs" whose conclusion is absurd. The game is to identify the error. It's important because the "proofs" usually rely on errors that people often make by accident. Finding the error helps ... 8 There is already great general answer by Ittay Weiss, so I will try a different approach. In fact, I will try to explain a bit the infinite sum you stated. As for the infinity, one could write a lot about it (mainly because there are multiple infinities, each with different properties), and I couldn't even hope to fit it here. Unfortunately, I don't know any ... 61 There are no known contradictions in mathematics. That does not mean there aren't, it just says we didn't find any. Considering the fact that thousands of mathematicians are creating new mathematics daily, and not a single one ever encountered any contradiction is quite overwhelming circumstantial evidence that there are no contradictions. However, it is ... 0 What follows is something I posted in the Math Forum discussion group ap-calculus on 13 March 2006 that might be useful. I recently came across a short note on how to convey the idea of a limit to students that I thought readers of this list would find useful. The full text of this short note is given below. James Clyde Bradford, Anecdotes for limits, ... 3 I had an economics professor who thought one result was obvious and spent over 3 hours explaining the "obvious" result to a fellow mathematician. After 3 hours, he realized it might not have been so obvious and found it to be, in fact, false, and published 3 papers from the counter-examples. 1 First of all, it will help if you tell us what mathematical background you have. For example, if you don't know the basics from elementary number theory then it is not such a good idea to try analytic number theory. My advice is the following: If it is not possible to study mathematics and you are self-taught: 1.Post your questions here in mathSE ... 1 This is my favorite. \begin{align}-20 &= -20\\ 16 - 16 - 20 &= 25 - 25 - 20\\ 16 - 36 &= 25 - 45\\ 16 - 36 + \frac{81}{4} &= 25 - 45 + \frac{81}{4}\\ \left(4 - \frac{9}{2}\right)^2 &= \left(5 - \frac{9}{2}\right)^2\\ 4 - \frac{9}{2} &= 5 - \frac{9}{2}\\ 4 &= 5 \end{align} You can generalize it to get anya=b$that you'd like ... 0 The figure below may help explain (apologies for the sub-optimal formatting) 0 I think the question means$4$times as many minutes to$6\$ o' clock from now.

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