# Tag Info

35

When you move from $\exists s\in S$, to specifying "Let $s$ be an element of $S$" you are using what is known as existential instantiation. This is an inference rule of the underlying logic, stating that if there are objects satisfying some property, we can add a new symbol to the language with the statement that this symbol satisfies our property. So you ...

18

Three concepts should always be a part of your mathematical problem-solving process. Documentation. Write out each step carefully, using consistent and precise notation. Don't skip steps and don't be sloppy. Each step should be understandable and justifiable, as if you were explaining to a reader what you are doing. Double-checking your computations. ...

12

This is a misunderstanding about the essence of axiomatic set theory. In axiomatic set theory, you don't assume that a set exists because you can think of it; in a sense, the entire point of axiomatic set theory is not to do that, to decouple the notion of existence of sets from such pre-existing (pun intended) notions. When you say "Define $f(S)=s$", you'...

10

I recommend reading John Littlewood's Miscellany (more recent editions ed Bela Bollobas, another great mathematician but who is still alive). Littlewood was one of the great mathematicians of the last century. He noted that mathematicians tend to be good on different timescales. Few are good on all timescales. To become a "great" mathematician, you have to ...

9

Good question! This is really subtle stuff; its impossible to give a proper answer without picking an actual formal system for first-order logic and trying to formalize your argument inside. After a lot of struggle, you'll notice that it can't be done. An informal explanation, however, is that you're implicitly using the "logicians" axiom of choice: AC....

7

This is a confusing matter, mainly because the kind of reasoning you use in your proof is usually taken to be valid. However, in order to formalize that reasoning in axiomatic set theory, we need to reduce it to particular symbolic formulas in a formal logic system. And it turns out that the rules of symbolic logic and set theory that are sufficient to ...

7

An important thing is to first find a resolution strategy. Your intuition should tell you how the computation will proceed. In this case, I noticed that the goal is to eliminate the variables $M$ and $N$, and you can do that by completing the square in the first two equations. So a possible strategy is to explicit $M$ and $N$ and plug them in the third ...

6

Strictly speaking "will this notation I like ever catch on?" Is inappropriate for this site. But we can still weigh the pros and cons of the suggestion and mention any variants we know. Cons: Introduces yet another symbol to memorize weak improvement over the obvious alternative Rarely do you talk enough about irrationals collectively to warrant special ...

5

I believe a counter-example (check) is $$f(x)=\begin{cases} x & \text{if }x<0;\\ x+1 & \text{if }x\geq0. \end{cases}$$ You can take your interval to be $[-1,1]$, but anything including zero suffices. As for your soft question, my thought process was as follows: I proved it for continuous functions, then I noticed that it seemed "too good to ...

5

I think it is much clearer to stipulate that $U$ is an open set. Expressing it as an element of the topology does not increase the clarity of what you are saying.

4

In a basic sense, algebra is about sets that have certain kinds of structures and about the functions that preserve those structures—namely, morphisms. Morphisms reveal the structure of spaces. For example, if you know about the properties of one space $X$, the morphisms into another space $Y$ may tell you about $Y$. As for why isomorphisms might not ...

4

A few suggestions: Look at the places where you made mistakes. Is there a pattern to what kinds of mistakes you're making? "Careless mistakes" is a bit broad, and perhaps a bit unfair, because you may be being overly harsh on yourself. It could be that there is a gap or two in one of the earlier math courses that you took, and it's only coming to light ...

4

If $f(x) \leq f(y)$ dividing both sides by $f(x) f(y)$ we get $$\frac{1}{f(x)} \geq \frac{1}{f(y)} \mbox{ whenever } f(x) f(y) > 0 \\ \frac{1}{f(x)} \leq \frac{1}{f(y)} \mbox{ whenever } f(x) f(y) < 0$$ This tells you that $\frac{1}{f(x)}$ is monotonic exactly when $f(x) f(y)$ doesn't change sign (and it is easy to see that this happens exactly when ...

3

There is a closed form. $\mbox{diag}(x)=I_n\circ (xu)$ where $\circ$ is the Hadamard product, $I_n$ the identity matrix and $u=[1,\cdots,1]$.

2

This is perhaps not exactly what you had in mind (it is about real-world applications of sets of measure zero in general, as opposed to the rational numbers in particular), but I will float it out here anyway. Considering the power iteration algorithm for finding eigenvalues: https://en.wikipedia.org/wiki/Power_iteration. If the matrix $A$ is singular (has ...

2

I think this is probably not possible in general. For a simple example, the matrix $$\left(\begin{array}{cc}1 &1 \\ 0 & 1\end{array}\right)$$ has conjugates $$\left(\begin{array}{cc}1 &x \\ 0 & 1\end{array}\right)$$ for $x \neq 0$. So you can make its Gershgorin radii arbitrarily small, but not zero...

2

Take comfort in the fact that real mathematics is not done under timed conditions like the examinations. When I was an undergraduate, I too found the introductory calculus and linear algebra courses one of the hardest, simply because I could not do computations as fast as other people. But mathematics is ultimately about theorems and proofs, not computation (...

1

One technique is to do the problem in two very different ways. If the methods are different enough, it's unlikely that you'll repeat the same mistake both ways, so comparing the answers gives you a way to check. And if there's a difference, you can often use what you learned from one method to validate your intermediate results from the other and find out ...

1

Obviously the definition of two edges being adjacent would be different from the definition of two vertices being adjacent. The only example you've given is that two adjacent edges are also said to be incident. Incident and adjacent are almost synonymous and I don't think you've really pointed out enough examples of overlapping or seemingly random terms. ...

1

The Frobenius inner product of matrices $A, B \in \mathbb R^{m \times n}$ is defined by $$\langle A, B\rangle := \mbox{tr} (A^T B) = \mbox{tr} (B A^T) = \mbox{tr} (A B^T) = \mbox{tr} (B^T A)$$ For example, the standard basis "vectors" for $\mathbb R^{2 \times 2}$ are $$M_{11} := \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} \qquad \quad M_{12} := \... 1 I think it might raise eyebrows for those who are new to set theory and topology, but that is precisely why I like this notation. One better get used to certain sets having elements which are sets containing further elements themselves. That said, if you want to “translate” Let U be an open set that contains x a better solution in this case ... 1 One of the best classics is Calculus Made Easy by Silvanus Phillips Thompson. It is a book on infinitesimal calculus originally published in 1910. 1 Let \delta_{ijk} be the tensor such that$$\delta_{ijk}=\begin{cases}1&\text{if}\;i=j=k\\ 0&\text{otherwise}\\ \end{cases}$$(in the particular basis you're working in). Then$$\mathrm{diag}(x)_{ij}=\sum_k\delta_{ijk}x_k or in the summation convention just $\delta_{ijk}x_k$.

1

Disbelieving the theorem is a good way to start, I find. Every time a statement is made, try to break it. Ask "what if $x=-1$?", or whatever seems likely to make an equation or an inference break down. After all, the whole point of a proof is that the thing wasn't obvious to start with, which is why it needed proving. So put the proof to the proof. Once ...

1

The trick to memorizing/remembering proofs of theorems is to understand them. If you understand what a theorem is about and you remember the strategy for proving it, it is often not hard to fill in the details on the go. Some theorems do require some "tricks" that can be useful to remember. I think the worst thing you can do is to just mindlessly memorize ...

1

Gödel/Scott themselves use both Ax. 2 and Ax. 3. The only difference is in Axiom 1 where you omit the box operator inside the scope of the allquantifier. Here is why I think this is a troublesome assumption. It might well be that it is accidentally the case that all individuals in our World that have a certain positive property A also have a Property B. For ...

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