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16

Check for yourself by trying some numbers! For example, if $a= b =1$, then $1/(a+b) = 1/2$, while $1/a + 1/b = 2$. Since $1/2 \neq 2$, we have that $1/(a+b) \neq 1/a + 1/b$ in this case. So clearly $a/(b+c) \neq a/b + a/c$ in general. Why, on the other hand, does $(a+b)/c = a/c + b/c$? The answer is that this really is just using the usual distributive ...


10

Since you know algebra, here is a proof that may satisfy you. Consider: $$\frac{1}{a+b} \stackrel{?}{=} \frac{1}{a} + \frac{1}{b}$$ If they are equal, then multiplying them by the same thing should keep them equal. Similarly, if they are unequal, then multiplying them by the same thing (except 0) should keep them unequal. Now, multiply both sides by $ab$ ...


10

Define $f\colon \mathbb{Q}\to \mathbb{Q}$ by $$f(x) = \begin{cases} 1 &, x \leqslant 0 \\ 1 &, x > \pi^{-1} \\ k &, \pi^{-1} - \pi^{-k} < x < \pi^{-1} - \pi^{-k-1},\end{cases}$$ where $k\in \mathbb{N}\setminus \{0\}$. Since the jumps of $f$ are at the irrational points $\pi^{-1} - \pi^{-k},\, k \geqslant 2$ and at $\pi^{-1}$, $f$ is ...


9

Perhaps a simpler example than the one given by Daniel, $$f(x)=\frac1{x^2-2}$$ It is not hard to show that this function is continuous on every rational number (simply note that it is continuous on $\Bbb R\setminus\{\pm\sqrt2\}$ and so its restriction is continuous). But it is not bounded on the interval $[1,2]$.


9

Your argument is fine and squares are not a problem. You are not pairing up the divisors, you are just reordering them. Let's walk through your calculation with $n=9$, where the sum of divisors is $1+3+9=13$. We then are asking what the value of $\frac 11 + \frac 13 + \frac 19=S$ is. We multiply by $9$ and get $9S=9+3+1=13, S=\frac{13}9$. The term $3$ ...


8

Why would it be equal? Think of $1/3$ or $1/4$ as a single third or a single fourth. (I would use $1/c$ here, but then there isn't a convenient word to go with it.) We have $a/3+b/3=(a+b)/3$, since $a$ thirds plus $b$ thirds is $a+b$ thirds. However, we don't have $1/3+1/4=1/7$; a seventh is smaller than a third or a fourth.


6

If you regard $\dfrac82=4$ as true because $8 = 2+2+2+2$, i.e. you can write $8$ as the sum of four $2$s, then you can regard $\dfrac{-2}{-1}=2$ as being equally true because $-2 = (-1)+(-1)$.


5

What happens is that what you were taught (or found out) does not apply to negative numbers. If you multiply a negative number $a$ with a number $b > 1$, the result is not larger than $a$, but smaller. Essentially, multiplying $a$ by $b > 1$ "amplifies" $a$: if $a$ is negative, then the result is "more negative" than $a$ (i.e., smaller). Now, $a/b$ can ...


4

it is not correct for eg: $$ 2\in \mathbb{Q},\frac{1}{2}\in \mathbb{Q}\;\;\;but :\sqrt{2}=2^{\frac{1}{2}}\notin \mathbb{Q}$$


4

Yes, there's a basic rule that you're missing, and it's a pretty useful one. When you have $\frac{N^a}{N^b}$, you can simplify to $N^{a-b}$. In your case, $a = \frac{2}{3}$ and $b = 1$, so the final exponent is $\frac{2}{3} - 1 = -\frac{1}{3}$. If you're wondering why the rule is true, it's probably easiest to see when $a$ and $b$ are integers. If you ...


4

You may compute the continued fraction for $\frac{\pi}{C}$ where $C$ is your constant: $$ \frac{\pi}{C}=[4;2,1,1,21,1,35,\ldots] $$ and by expanding $[4,2,1,1]=\frac{22}{5}$ we get: $$ C = 0.714286 \approx \frac{5\pi}{22}.$$ A better approximation is given by expanding $[4,2,1,1,21,1]=\frac{497}{113}$, so $$ C = 0.714286 \approx \frac{113\pi}{497}.$$


3

Write $\frac{1}{(3n-2)(3n+1)}=\frac{1}{3}(\frac{1}{3n-2}-\frac{1}{3n+1})$ before summing.


3

Let $p$ and $q$ be positive integers (whole numbers). In the fraction $\frac{p}{q}$, the numbers $p$ and $q$ play distinct roles: $q$ measures the "size" of (or denominates) $\frac{1}{q}$. Intuitively, $\frac{1}{q}$ is the quantity obtained by dividing one unit into $q$ equal portions. For example, $\frac{1}{2}$ represents the result of dividing $1$ into ...


3

For $B\not=0,C\not=0$, we have $$A\div\frac BC=\frac{A}{\frac BC}=\frac{A}{\frac BC}\times 1=\frac{A}{\frac BC}\times\frac{C}{C}=\frac{A\times C}{\frac{B}{C}\times C}=\frac{AC}{B}=A\times \frac{C}{B}.$$


3

As Thomas Andrews says, remember that multiplication and division have the same operator precedence. That is, the following is true: $$\frac{A\cdot B}{C\cdot D} = (A\cdot B)/(C\cdot D) = A\cdot (B/C)/D = A\cdot (B/(C\cdot D)) = A\cdot (B/D)/C...$$ So what I do when I do division involving negative numbers, is I always factor out the minus sign as a ...


2

We have $$\frac{a+b}{c} = \frac{a}{c} + \frac{a}{c}$$ But on the other hand $$\frac{a}{b+c} \not= \frac{a}{b} + \frac{a}{c}$$ Why is that? Consider that $x/y$ can also be written as $x \cdot (y^{-1})$. As you probably know, $$(a+b)^2 \not= a^2 + b^2$$ And indeed, in general, $$(a+b)^n \not= a^n + b^n$$ It just doesn't work like that. Given this ...


2

The two "small" fractions are $\frac {11}3$ and $\frac 3{11}$ Two methods of proceeding come to mind - the first is to see whether the factors $3$ and $11$ cancel (there are easy tests to show that both do), and see what happens. The second would be to use the Euclidean Algorithm to see whether there are any common factors of the "large" fraction which can ...


2

The difference is the topology you put on $\mathbb{Q}$. In these two cases you're generating an order topology, one based on the standard ordering, the other is an order inherited by the specific bijection you have with $\mathbb{N}$. For the last question, definitely yes. You take the inverse of the bijection you were using before. Whereas you were ...


2

It's almost OK. You lack some rigour: you should write $\,\cos\dfrac\theta2=\pm\sqrt{\dfrac{1+\cos\theta}2}$, and explain why you choose +. What is wrong is fractions: $$\sqrt{\dfrac{\cfrac{32}{25}}{2}}=\sqrt{\dfrac{16}{25}}=\dfrac 45.$$


2

Note that: arccos(7/25) lies in either 1st quadrant or fourth quadrant. (cos is +ve), so 2*arccos(7/25) lies in 2nd quadrant and 3rd quadrant, where cos is negative. So the answer should be -4/5.(by dividing the numerator and denominator by $sqrt(2)$) Other than this, the answer is correct.


1

HINT: use this formula: $$\cos(\theta/2)=\pm\sqrt{\frac{\cos(\theta)+1}{2}}$$


1

You think about the rationals in two different ways. One the one hand, you look at it as a dense linear order without endpoints and on the other hand as another linear order without endpoints, which is order isomorphic to the integers. You can do this with any countable set $X$ (and in many different ways): Pick bijections $q: X \rightarrow \mathbb Q$, $i: X ...


1

There is nothing incomplete in the proof. The proof shows the following statement: If there exist such integers $a,b$ that $r=\frac ab$ and there exist such integers $c,d$ that $s=\frac cd$, then there also exist such integers $p, q$ that $r+s = \frac pq$. Remember that this is all you need. A number $x$ is rational if and only if there exist some ...


1

Generally the sum of the reciprocals of the divisors of $n$ is equal to $\frac{\sigma(n)}{n}$ where $\sigma$ is the sum of divisors function. This quantity is sometimes referred to as the abundancy ratio or abundancy index of $n$. It can be used to tell whether $n$ is abundant, deficient, or perfect.


1

Your answer still correct if $n$ is a perfect square since it only has one positive root, so for the term $\frac{n}{d_{root}} = d_{root}$ you get exactly what you need.


1

Using a simple calculator, the big fraction is $$r=0.27272727\cdots$$ Then $100r-r=27.27272727\cdots-0.27272727\cdots=27$ and $$r=\frac3{11}.$$ Indeed $\dfrac{116690151}3=\dfrac{427863887}{11}=38896717$. The rest is trivial.


1

The "canonical" interval to try is $(0,1)$. Based on that, go to the Online Encyclopedia of Integer Sequences and look up the sequence of numerators (which is the concatenation, for all $k\ge2$, of the $\phi(k)$ integers less than $k$ which are relatively prime to $k$: $1, 1, 2, 1, 3, 1, 2, 3, 4, 1, 5,\ldots$) Your sequence appears as sequence A038566; ...


1

The proportion of fractions in lowest terms is $6/\pi^2$. Using JHance's comment, there will be around $$\frac{12}{\pi^2}(M-1)(N-1)$$ different slopes. Here is why that proportion of fractions in lowest terms is $6/\pi^2$: $3/4$ of the time, $m$ and $n$ are not both multiples of $2$. $8/9$ of the time, they are not both multiples of $3$. $24/25$ of the ...


1

Hint: $$ \tau=\dfrac{2f(aN-bN^{\frac{2}{3}})}{N} =\dfrac{2faN}{N}-\dfrac {2fbN^{\frac{2}{3}}}{N} = 2fa -2fbN^{(\frac{2}{3}-1)}= $$ $$ =2fa -2fbN^{(-\frac{1}{3})} $$


1

By the way, a presentation of $(\mathbb{Q},+)$ is $$\langle x_1,x_2,\ldots \mid x_n^n=x_{n-1}, \ n \geq 2 \rangle.$$ To understand why, it is possible to consider the morphism induced by $x_n \mapsto \frac{1}{n!}$. For more information, see Johnson's book, Presentations of groups, Chapter 5.7.



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