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28

Richard Feynman made some interesting comments about this in chapter 14 of volume 1 of the Feynman Lectures on Physics: In learning any subject of a technical nature where mathematics plays a role, one is confronted with the task of understanding and storing away in the memory a huge body of facts and ideas, held together by certain relationships ...


9

I'd like to throw some words around, as an undergraduate myself. I don't really have an answer per se, but I do have some thoughts on the question based on my experiences and understanding so far. I think it's kind of a question you ask yourself, and answer yourself. There is not one exact standard. I expect that in mathematics, no matter how far you go, ...


9

This follows quite easily from the fact that digit sum arithmetic is equivalent to arithmetic modulo 9. To see this we express a given integer $n$ as $n = k_1 + 10 k_2 + ... + 10^j k_j$ where the $k_i$s are integers (in fact the digits of $n$). Then $n = k_1 + k_2 + ... + k_j \, (mod \, 9)$. Then the question is simply which numbers are square modulo 9 ...


8

When people say $\mathbb{R}^2$, it is commonly implicit all its "canonical" structure (as a vector space, topological space, metric space etc). But you must know what this canonical structure is! Imagine you enter a room where everyone knows Bob. You also know Bob. But everyone also knows Bob's dog, which everyone simply assumes everybody else knows: it is ...


8

My UK undergraduate degree was assessed solely on "final examinations" at the end of 3 years (and again for the 4th master's year). So for my degree the principle was that for a particular couple of weeks one summer, it was necessary to be able to prove at the drop of a hat all the theorems on the syllabus in all topics examined, and to be able to complete ...


7

The notation $\Bbb R^2$ already refers to the vector space you're referring to, so there's no need to define the vector space structure every time you use the notation. But we do need to define everything, even common notation, at least once in our lives so that terms even have meaning. Whether or not you need to rewrite the definition again in another ...


7

We can prove this using the well known identity $$\cos (a-b) = \cos a\cos b + \sin a\sin b$$ Here $a = \dfrac{\pi}{2}, b = x$ so now we have \begin{align*} \cos\left(\frac{\pi}{2}-x\right) &= \cos\frac{\pi}{2}\cos x + \sin\frac{\pi}{2}\sin x \\ &=0\cdot\cos x + 1\cdot \sin x \\ &= \sin x \end{align*}


6

If I toss a coin an infinite amount of times, can I be sure to get an infinite amount of heads? According to the Borel-Cantelli lemma, since each coin toss is an event of probability $\frac12$ and a sum of $\frac12$ diverges, the probability of $\limsup_{n\to\infty}\{\text{heads at $n$-th flip}\}$ is 1. But the $\limsup$ is precisely the event of ...


6

So most of us encounter $\mathbb R^2$ as the cartesian plane in which we create axes, sketch functions, explore geometry, encounter vectors etc. It is a familiar object. Then we meet, in algebra, the abstract concept of a vector space. We understand the idea - one thing we want to do is to take what we've learned in $\mathbb R^2$ and other examples which ...


5

To use language effectively, you have to consider your audience... In the context of an introductory upper-division course on linear algebra, it's good to be explicit about the vector space operations. After all, they do have to be defined, and there's value in emphasizing that point. In a professional context, it would be insane. The reader knows what you ...


5

We have that $F_n>F_{n-1}$ then $$F_{n+1}=F_n+F_{n-1}>2F_{n-1}>2\cdot2^{(n-1)/2}=2^{(n+1)/2}$$


5

No cases are really required: Lemma: $\;\lvert x\rvert=\lvert- x\rvert$. Obvious from your definition. Hence we can suppose $x,y\ge 0$, in which case there is nothing to prove, really. Variant: Prove first that $\;\lvert x\rvert=\sqrt{x^2}$. Then use $\sqrt{ab}=\sqrt a\cdot\sqrt b\;$ for any $a,b\ge 0$.


5

Given an injection $f : S \to \mathbb{N}$, where $S$ is infinite, you can define a function $h : \mathbb{N} \to S$ by, for each $n \in \mathbb{N}$, defining $h(n)$ to be the element $s \in S$ for which $f(s)$ is the $n^{\text{th}}$ least element of $f(S) \subseteq \mathbb{N}$. This is possible because $S$ is infinite, so such an element $s \in S$ always ...


5

The Principle of Mathematical Induction says that for all "properties" $P$, $$\left(P(0)\land\forall k\in \mathbb N\left(P(k) \implies P(k+1)\right)\right)\implies \forall n\in \mathbb N(P(n)).$$ So you're basically asking how to write the $\forall k\in \mathbb N\left(P(k)\implies P(k+1)\right)$ bit. It's a universal statement. It's common to start those ...


4

Mathematics is a lot about outsmarting your definitions. Namely, "show where it says that you can't do that". This is why so many definitions in mathematics have "trivial" examples, sometimes to the point where they can be a cumbersome addition to each proof when first learning the subject. If the definition of a vector space is such that nowhere it is ...


4

1. You just bound in by making the denominator smaller and numerator bigger, this is a common trick in these proofs. $$\frac{11}{2(2n+7)}<\frac{11}{2(2n)}=\frac{11}{4n}<\frac{12}{4n}=\frac3n$$ These proofs have TWO STEPS. Basically, you want to do the "prep work", then the proof. Let me write out this proof in full. The above is the PREP WORK and not ...


4

Consider the number $30$. It is divisible by $6$ and $10$ and not by $60$. We have found a counterexample and therefore the statement is not true. The statement $ab$ divides $n$ if and only if $a$ divides $n$ and $b$ divides $n$ is only true if $a$ and $b$ are relatively prime (share no prime factors). Otherwise we can take the least common multiple of ...


4

There is never too much rigor in math. ;) A vector space is a set of vectors over a set of scalars, with two operations: vector addition and scalar multiplication. If you don't have operations, you don't have a vector space. And you will discuss different operations on $\Bbb{R}^2$ other than the standard addition and multiplication, so yes it is necessary ...


4

This is Mark Bennet's answer, with a different emphasis. I am also operating under the assumption, like many others, that you saw this toward the beginning of a set of linear algebra, abstract algebra, or analysis notes. I agree with you that the check that $\Bbb R^2$ is a vector space is generally boring, obvious, and tedious, but I posit that, in this ...


4

Consider the problem systematically: Starting with $1$, the sequence cannot contain the number $2$ until there are $2$ adjacent $1$s. Similarly, the sequence cannot contain $3$ until there are either $3$ adjacent $1$s or $3$ adjacent $2$s. Since the former will occur first, this occurs when there are $3$ adjacent $1$s. Continuing this trend, it is not ...


4

Geometrically: given a right triangle with angles $A$, $B$, and $C=\pi/2$ and sides $a$, $b$, and $c$ opposite the respective angles, you have $$\sin(A)=\frac{a}{c}$$ and $$\cos(B)=\frac{a}{c}$$ And we know that $$B=\pi-A-\pi/2$$ so the result is immediate. Using Euler's formula: We know that $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, so that ...


4

Hint: 1) The difference of two continuous functions is continuous; 2) Define $h(x) = g(x) - f(x)$. Then $h(a) > 0$ and $h(b) < 0$; 3) Apply the Intermediate Value Theorem to $h$.


4

This is the base $9$ equivalent of all the squares ending in $0,1,4,5,6,9$ base $10$ because the "digital root" you are calculating is also the remainder on dividing by $9$. This works in turn because any power of $10$ leaves remainder $1$ when it is divided by $9$ ($10^r=\dots 999+1$ where there are $r$ nines). It is a nice thing to notice and explore. So ...


4

Hint: a four-digit palindrome must look like $ABBA$ where $A$ and $B$ are digits, and $A$ is nonzero. The number of ways can you choose $A$ and $B$ is the number of four-digit palindromes.


3

Your proof seems a bit confused, all you need is that $f - g$ is positive at one end point and negative at the other, and thus must be $0$ somewhere in between because $f - g$ is continuous (Why?). I don't see how you get, for example, $c = a = b$


3

Your wording is completely confusing. Your statement: Suppose that the sequence of prime numbers is finite, so the sequence $R_n$ is finite is wrong. Also, you then say that $R_{n+1}$ is not a prime number, but you did not determine what the value of $n$ is. Furthermore, you then proved that $R_{n+1}$ is divided by $1$ and itself, and concluded that ...


3

If $Ax = 0$ only for $x = 0$ exactly means that the matrix $A$ considered as a linear mapping is injective hence bijective (since it is $n \times n$). Then trivially also $A^k$ is bijective, in particular injective.


3

Your proof is valid, but I think you're making the situation more complicated than it really is. Your claim has two parts: a hypothesis, and a conclusion. The hypothesis is "for all $x$, $f(x) = g(x)$." The conclusion is "for all $x$, $f(x + c) = g(x + c)$." Both the hypothesis and conclusion discuss, separately, a claim about "all $x$", and the fact that ...


3

HINT: Note that $$\left|a_n-\frac{1}{3}\right|=\left|\frac{2n+5}{6n-3}-\frac{1}{3}\cdot\frac{2n-1}{2n-1}\right|=\left|\frac{2n+5-(2n-1)}{6n-3}\right|=\left|\frac{6}{6n-3}\right|=\left|\frac{2}{2n-1}\right|$$ Taking $n_0\ge\max\left(\frac{1}{\epsilon}+1,2\right)$ we have the inequality, since $$n\ge ...


3

Part (1) Essentially whoever wrote the proof wants to get an $n$ in the denominator (to be explained shortly). Note that $\frac{11}{2(2n+7)}=\frac{11}{4n+14}<\frac{11}{4n}<\frac{3}{n}$, so that we got a nice bound with $n$ in the denominator. Part (2) Note that we have $|a_{n} - \frac{3}{2}| < \frac{3}{n}$ but really what we want is to show that ...



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