# Tag Info

95

So in mathematics, as far as I know, you can't divide 100% by 3, without having that 0,1..% left.... No! we can in Math and also in real life. This is similar to ask can we divide $1$ into 3 parts? And the answer is again yes. $$1÷3=\frac{1}{3}$$ because adding $\frac{1}{3}$ three times give $1$. Consider 3 sticks of same length. Align these three ...

69

Meanwhile in ancient Greece... For quite a long time, greek (and not only) mathematicians described numbers as lengths of certain line segments. So, when asked "What is $\sqrt{2}$ equal to?" they'd draw a $1\times1$ square (nevermind the unit), draw it's diagonal and say "There it is! This diagonal's length equals exactly $\sqrt{2}$!". So to answer your ...

31

I think you have problems with this because you're thinking in base 10, and 10 (in base 10) doesn't divide evenly by 3. Think in base 3 instead. $100\%$ in base 3 is: $100\% = 10\% + 10\% + 10\%$ Which are trivially demonstrated to be equal parts, with no remainder. Addendum: although I have substituted its meaning analogously, $\%$ is the symbol for ...

23

One way to see this is to distinguish between definition and representation. I think you have began that. Definition: 1/3 is by definition exactly on third of 1. Representaion: "1/3" is a representation that is useful here. However there is no representaion of 1/3 using something like "0.3333333...". Unless you do not have an infinite paper or screen, of ...

18

The fundamental fallacy in your reasoning is that "natural == real". Just because a number never ends doesn't mean it isn't a "real" number with real application. You have three apples. Three is a "natural number"; it is positive and whole, used in counting and other "everyday math". Now, you are converting these three apples to "100%". "Percent" is from ...

13

I object to your teacher's answer (even if we remove physical obstructions, such as indivisibility of some particle or another): after any finite amount of time, Aladdin will have only divided it into pieces of size $(1/2)^n$ for some finite $n$; this is still positive. However, the limit is zero, which is what is meant by $(1/2)^\infty$. As $n$ gets bigger, ...

10

It's been answered above, but I thought I should try for a consise answer. The fact that 100, in base 10, divided by 3 lacks a finite representation does not mean that it's impossible to divide 100 by 3. In fact most real numbers can not be represented by a finite decimal number.

8

Alright - With this sort of topic, we have to bring in a few ideas in order to explain this in full. The first one is: Infinity The concept of infinity is to have an endlessly large number. The fact is that you can't ever write out infinity as an actual number, because you can simply add another digit at the end, and you'll have a number larger than ...

6

This concatenation is A007908 in the OEIS, where you can see that Charles Nicol and John Selfridge ask the same question in R. K. Guy, Unsolved Problems in Number Theory, A3. There are no primes in the first 5000 terms of the sequence, but heuristics suggest that there are infinitely many. In particular, there 'should' be about $$\frac{\log\log n-\log\log ... 5 As a is odd, it is definitely not divisible by 168. But it does not necessarily imply that (168,a)=1 so that 168 will have to divide a^2-1 as 168(=7\cdot8\cdot3) is not prime and 7 and/or 3 can divide odd a. See this . We can follow the following method to establish the proposition: using Fermat's Little Theorem$$a^p-a\equiv0\pmod ...

3

Once you factorize a number as $N=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_n^{a_n}$, $p_i$ prime for every $i$, $a_i>0$ for every $i$ the number of divisors is given by $(a_1+1)(a_2+1)(a_3+1)...(a_n+1)$. It is easy to see why this formula works from a combinatorial point of view, the divisors of $N$ are also of the form $p_1^{b_1}p_2^{b_2}p_3^{b_3}...p_n^{b_n}$, ...

3

The rational roots test does indeed work in $\mathbb{Q}[x]$. If a polynomial has degree less than or equal to $3$, then it is irreducible over the rationals if and only if it has a rational root (Prove why). Wikipedia states that the rational roots test holds for polynomials with integer coefficients. Note that any polynomial in $\mathbb{Q}[x]$ can be ...

3

For a fixed $n$, let $A_r$ be the number of sequences of $n$ terms, whose sum is $r \pmod{11}$. Then, we want the value of $\sum A_r ^2$. As pointed out by ABC, the generating function $f(x) = \left ( 1 + x + x^2 + \ldots + x^9 \right) ^n$ gives us the number of sequences of $n$ terms, with a total sum of $r$. We find the sum of those coefficients with ...

3

What your teacher said was not very mathematically exact. You have to observe the infinite sequence of powers of one half, $$1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots$$ And the proper term is that the sequence tends to zero (it converges to its limit which is $0$). The fact that this goes to zero shouldn't surprise you. But you need an infinite number ...

3

Your result of 457 is correct. The different sets of numbers as table: And as diagram:

3

I don't know any theorems about primary school children. Take any integer $x = 3k-1$ for any integer $k,$ then $x^3 = 27k^3 - 27 k^2 + 9 - 1.$ So $x^3 - 2 = 27k^3 - 27 k^2 + 9 - 3.$ And $$\frac{x^3 - 2}{3} = 9k^3 - 9 k^2 + 3 - 1.$$ So, take $$x=3k-1, \; \; \; y = 9k^3 - 9 k^2 + 3 - 1.$$

3

Even numbers are just numbers that can be cut in half. Or, another way, if you have an even number of objects, you can sort them in pairs of two, or you can split them into two equally sized groups. The odd numbers are those where you can't do that. This is generalizable to other numbers, so $n$ is divisible by $k$ if you can sort $n$ objects into groups of ...

3

Think of indivisible bricks, and houses: $\,\,\,\,$ You can cut the first house into two equal parts. But you're not strong enough to cut the second house into two equal part, since you'd be left with one brick. For divisibility, build houses with bases containing $m$ bricks. Then the total number of full floors is the quotient of the total number of ...

3

I suppose $a,b,c,d >0$ here, as it's certainly not true otherwise (take distinct primes $p,q$ and set $a=p$, $b=q$, $c=-q$, $d=-p$. It's a lot simpler than you're making it. I'll give you a hint... $$ac=bd \Rightarrow a \mid bd \Rightarrow a \mid d\; \text{since}\; (a,b)=1$$ $$ac=bd \Rightarrow d \mid ac \Rightarrow d \mid a\; \text{since}\; (c,d)=1.$$ ...

3

$|a| = a$ if $a\geq 0$. $|a| = -a$ if $a \lt 0$. Now, is it true that if $d\mid a$ then $d \mid a$? And is it true that if $d\mid a$, then $d\mid -a$? If you can justify "yes" for both questions, you are done. Hint: use the definition of $a \mid b$: If $a\mid b$, then $b = ka$ for some integer $k$.

3

Hint $\$ By $\,(m,n)=1\,$ we have $\,(ma\!+\!nb,mn) = (ma\!+\!nb,m)\,(ma\!+\!nb,n),\$ and $\qquad\quad (ma\!+\!nb,m) = (nb,m) = (b,m),\$ and $\ \ \ (ma\!+\!nb,n) = (ma,n) = (a,n)$ Remark $\$ The inference in the first line is the special case $\,(m,n) = 1\$ of this Lemma $\ \ (k,m)(k,n) = (k,mn)\ \$ if $\ \ \color{#c00}{(k,m,n) = 1}$ Proof ...

2

How many primes are there in between $1$ and $10$ ? Four primes: $2,3,5,7$. And what is the greatest power of each such prime, which is also $\leqslant10$ ? $2^3=8$, $3^2=9$, $5^1=5$, and $7^1=7$. Now, just how much is $2^3\cdot3^3\cdot5\cdot7$ ? :-) But if your lower limit isn't $1$, then I'm afraid you'll have to meticulously factor each number, since, for ...

2

Very simply it can be done like this: $gcd(a,b)=d$. Now we ask can: $gcd(\frac{a}{d},\frac{b}{d})=e$ for $e>1$? Well, this implies $e|\frac{a}{d},e|\frac{b}{d} \Rightarrow em=\frac{a}{d},en=\frac{b}{d} \Rightarrow dem=a,den=b \Rightarrow de$ is a common divisor of $a,b$ which is greater than $d$, thus a contradiction as $d$ by definition was supposed as ...

2

The discriminant of $2x^2-(2y+5)x+(19-y)$ as a quadratic in $x$ necessarily needs to be a perfect square. Thus we need values for $y$ s.t. $(2y+5)^2-8(19-y)$ is a perfect square. Now $$(2y+5)^2-8(19-y)=s^2 \iff (2y+7)^2-176 = s^2$$ so we need integer solutions for $$(2y+7+s)(2y+7-s)=2^4 \cdot 11$$ Consider the possible (finite number of) factorisations, ...

2

GCD(277,301): $301 - (277 \times 1) = 24$ $277 - (24 \times 11) = 13$ $24 - (13 \times 1) = 11$ $13 - (11 \times 1) = 2$ $11 - (2 \times 5) = 1$ $2 - (1 \times 2) = 0$ Thus the result is $\mbox{GCD}(277, 301) = 1$. Expressed differently, we have: Divisors of $277: 1, 277~$ (that is, a prime) Divisors of $301: 1, 7, 43, 301$ What is the greatest ...

2

Construct $k$ using $n$ and $m$: Let $p_1, p_2, \dots, p_a$ be the prime factors of $n$ and $m$. Let $x_i$ be the number of times that $p_i$ appears in the prime factorization of $n$. Let $y_i$ be the number of times that $p_i$ appears in the prime factorization of $m$. $$k=\prod\limits_{i=1}^{a}{p_i}^{\max(x_i,y_i)}$$ For example, consider $n=18$ and ...

2

Here is a clean way to describe the situation: factorizations refine in residues. If $a\in A$ and $a=a_1\cdots a_n$ and $\pi:A\to B$ and $B$ is a UFD and $\pi(a)=b_1\cdots b_m$ is its factorization into irreducibles (repetitions allowed) then we can partition the factors $b_i$ in such a way that each multiset of factors is precisely the multiset of ...

2

Claim: $\forall \ p$ such that $p$ is a prime, $$m \nmid {m \choose p}$$ where $m = kp$, $k \in \Bbb Z$ and $k \gt 1$ Proof: $${m \choose p} = {m(m-1)\cdots(m-p+1) \over p(p-1)\cdots1}$$ Observe that since $m$ is a multiple of $p$, none of the other $p-1$ integers are divisible by $p \implies$ p should divide $m \implies m \nmid {m \choose p}$. Also, ...

2

I would also object to your teacher's answer. The advantage of writing $\left(\frac{1}{2}\right)^{\infty}=0$ is that you don't have to answer the question "what do you actually end up with when you cut a carpet in half an infinite amount of times?". Your teacher was probably trying to explain the idea of a limit, which is more like asking the question "If I ...

2

Hint $\$ If $\,d\,$ is a common factor then $\,{\rm mod}\ d\!:\ 2a+1\equiv 0\equiv 11a+5.\,$ Now eliminate $\,a\,$ by cross multiplying $\ 0\equiv \color{#0a0}{11}(2a+1)-\color{#c00}2(11a+5) \equiv \color{#0a0}{11}-\color{#c00}{10}\equiv1\,\Rightarrow\, d\mid 1$. Remark $\$ More conceptually, put the fractions for $\,-a\,$ over the common denominator ...

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