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21

First off, your definition of rational numbers is correct. So $3/2$ is a rational number, but in the definition it does not say that $p/q$ should be an integer, only that both of $p$ and $q$ should. Rational numbers can have decimals and even an infinite decimals, BUT any rational number's decimals will have a repeating pattern at some point whether it be ...


16

Put $x^2-ax=b$ and $x^3-ax=c$ with $b,c\in \mathbb{Q}$. As $x\in \mathbb{R}$ is a real solution of $y^2-ay-b=0$, we have $D=a^2+4b\geq 0$. Now: $$x^3=xx^2=x(ax+b)=ax^2+bx=a^2x+ab+bx=(a^2+b)x+ab$$ This is also $ax+c$. If we suppose that $x\not \in \mathbb{Q}$, we get that $a=a^2+b$. Hence $D=a^2+4b=4a-3a^2<0$ as $\displaystyle a>\frac{4}{3}$, a ...


15

3 out of the 4 options on that exam can be written in $\frac{p}{q}$ form, where p and q are integers. Note that integers can not have decimals. For example: a) $\sqrt{25} = 5 = \frac{5}{1}$ b) $\sqrt{45} = \frac{?}{?}$ c) $\sqrt{\frac{256}{225}} = \frac{\sqrt{256}}{\sqrt{225}} = \frac{16}{15}$ d) $\frac{3}{4}$ So a, c, and d can be written in ...


6

Let $x^2-ax=c$ and $x^3-ax=d$. Let $P(y)$ be the minimal polynomial of $x$ over $Q$. Then $P(Y)| Y^2-aY-c$ and $P(y) | Y^3-aY-d$. In particular the degree of $P$ is at most two. Case 1 $\deg P=1$, it follows immediately that $x \in \mathbb Q$. Case 2 $\deg P=2$, then $P(Y)=Y^2-aY-c$ and hence $$ Y^2-aY-c| Y^3-aY-d \,.$$ This shows that $$Y^3-aY-d = ...


4

Whether a number is a rational number does not depend on how it is expressed, but on whether it is possible to write it as $\frac pq$ where $p,q$ are integers and $q\neq 0$. So any terminating decimal, for example $r=abc.defg$ is a rational number, because $10000r=abcdefg$ and $r=\frac {abcdefg}{10000}$ Also any decimal which eventually repeats represents ...


3

As shown in this answer, if the solution of $$ x^2-45=0\tag{1} $$ is rational, then it is an integer. $x=6$ is too small and $x=7$ is too big, so there is no integer that satisfies $(1)$. Thus, there is no rational number that satisfies $(1)$. That is, $\sqrt{45}$ is irrational.


3

$$\sqrt{45} = \sqrt{9\cdot 5} =3\sqrt{5}, $$ but $\sqrt{5}$ isn't rational. Short proof: if $\sqrt{5}=\dfrac{a}{b}$, $a,b\in\mathbb{N}$, and $GCD(a,b)=1$, then $$ 5 = \dfrac{a^2}{b^2}, $$ $$ a^2 = 5b^2, $$ so $~~5|a^2$ $~~\Rightarrow~~$ $5|a$ $~~\Rightarrow~~$ $5|b^2$ $~~\Rightarrow~~$ $5|b$ $~~\Rightarrow~~$ $GCD(a,b)=5$. Contradiction.


2

A rational has to be able to be shown as a fraction with integers on the numerator and denominator. Some decimals can be written as fractions (examples below) $3=\frac{3}{1}$ $1.5=\frac{3}{2}$ $1.66666...=\frac{5}{3}$ The problem arises in the fact that $\sqrt {45} =3 \sqrt{5}$ and $3$ is rational, but $\sqrt{5}$ can't be written in any fraction with ...


2

You are correct, a rational number is one that can be expressed in the form $m/n$, where $m,n$ are integers (and $n\ne0$ - also we usually assume they have no common factors). So 1.5 is a rational number because it can be expressed as 3/2. Indeed, any finite decimal can be expressed in the form $m/n$ (think about what 0.123 means). Some infinite decimals ...


2

The rigorous mathematical definition of rational numbers is that they are equivalence classes of ordered pairs of integers $(m,n)$ with $n\neq 0$ for the equivalence relation $$(m,n)\equiv (o,p)\iff mp = no.$$ I don't really understand how you first wrote the rigorous mathematical definition of $\mathbb Q$, then asked is there a rigorous mathematical ...


2

I differ slightly from the other comments and answers in that I don't think the real-world meaning you gave is different from the formal meaning. Maybe we can break it into steps. What I hope is that each step can be deduced from the next. Two-thirds of a pizza is what you get when you divide a pizza into three equal pieces, then take two of them. ...


2

For any three integers $m,n,b$ such that $$m \ge 1,\quad n, b > 1\quad\text{ and }\quad\gcd(m,n) = \gcd(m,b) = \gcd(n,b) = 1$$ Let $\displaystyle\;\tau\left(\frac{m}{n}, b\right)\;$ be the "period" of the expansion of $\displaystyle\;\frac{m}{n}\;$ in base $b$. Let $\lambda = \lambda(n)$ be the value of Carmichael function for $n$, we have ...


2

We will show more: that if $x \ne 0, 1$ is algebraic (possibly non-real), then $\sin(\log(x))$ is transcendental (regardless of which value of the logarithm we take). To do this, we apply the Gelfond-Schneider Theorem, which states that if $\alpha, \beta$ are algebraic numbers with $\alpha \ne 0,1$ and $\beta \notin \mathbb{Q}$, then any value of ...


2

Suppose the diagonalised number $d = 0.536\ldots$ is rational. Then $d + 0.333\ldots = d + \frac13$ is also a rational number in $(0,1)$. But $d + 0.333\ldots$ differs from $d$ in every decimal place, so it can't occur in your list $-$ a contradiction.


2

Hint: suppose $f\colon {\bf Q}\to {\bf Z}$ is continuous, and $k$ is in the range of $f$. What can you say about $f^{-1}[\{k\}]$?


1

I would avoid using the symbol $\sqrt{2}$ and stick to rational numbers only. I would also be more careful about proving the inequalities; how do you know that the manipulations that you did are reversible? Here's my version. Suppose, towards a contradiction, that $F$ contains some smallest element, say $p$. Now consider the element: $$ q = \frac{2p + 2}{p ...


1

"Then why people saw something good in this decimal expansion representation of rational numbers and constantly use it?" One reason is that it's easy to add decimal numbers, but to add two fractions you have to find a common denominator. Consider calculating $${1\over5}+{1\over7}+{1\over8}+{1\over9}+{1\over11}+{1\over13}$$ by finding a common denominator, ...


1

Now, $\sqrt{45} \approx 6.708$. Can someone explain why this not rational? Is it about decimal points? Your answer to this is below: $\sqrt{45}$ is not rational because it cannot be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers. This also means it can’t be expressed in decimal form with terminating or repeating digits. ...



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