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32

Here is a proof which does not require Galois theory. Write $$f(z)=z^n-x^n\quad\hbox{and}\quad g(z)=(z+1)^n-(x+1)^n\ .$$ It is clear that these polynomials have rational coefficients and that $x$ is a root of each; therefore each is a multiple of the minimal polynomial of $x$, and every algebraic conjugate of $x$ is a root of both $f$ and $g$. However, if ...


28

I will prove the following fact: Let $x$ be a real number such that $x^n$ and $(x+1)^n$ are rational. Then $x$ is rational. Without loss of generality, I assume that $x \neq 0$ and $n>1$. Let $F = \mathbb Q(x, \zeta)$ be the subfield of $\mathbb C$ generated by $x$ and a primitive $n$-th root of unity $\zeta$. It is not difficult to see that ...


0

It seems the following. We can prove that $x$ is rational provided $n$ is a power of an odd prime $p$. Denote $x^n=r$ and $(x+1)^n=s$ with $r$, $s$ rationals. If both $r$ and $s$ are $p$-th powers of rational numbers, then we descent from $n$ to its $p$-th root. So without loss of generality we can assume that one of the numbers $r$ and $s$ (for instance, ...


2

To add another proof to yours (which is perfectly correct), but avoiding relying on unique factorization, I would do it as follows: First, prove that $\gcd(ab,c)$ divides $\gcd(a,c)\cdot\gcd(b,c)$. This can be done noting that, by the Euclidean algorithm, we can find integers $n_1,n_2,m_1,m_2$ such that: $$\gcd(a,c)=n_1a+n_2c$$ $$\gcd(b,c)=m_1b+m_2c.$$ ...


3

As I say in the question, I think the answer is yes - the result is automatically in least terms. If $a/b$ is in least terms, $a$ and $b$ don't share any prime factors. If they shared a prime factor $p$, you could reduce the fraction to $a'/b'$ where $a'=a/p$ and $b'=b/p$. When you calculate $x'=x^n$ for any positive integer $n$, $x'$ has the exact same ...


1

As was already said, solution is the limit of infinite sequence of field extensions of rational numbers with square roots of all primes. But I’d point out there is more explicit description. The smallest ring (and ${\mathbb Q}$-algebra) which contains all square roots of positive rational numbers is, obviously, an infinite(countable)-dimensional vector ...


5

Others have already pointed out that your proof was wrong. A different way to see that your proof is wrong is as follows: if I would replace 2 in your proof everywhere by 10, I would get the result that log(10) is also irrational. Any proof you have for the irrationality of log(2) should not work for log(10). (If you take 10 as the base of your logarithm.)


3

The proof for irrationality of $\log 2$ can be done in few lines using elementary divisibility properties or natural numbers. Assume $a, b \in \mathbb{N}$, $\text{gcd}(a,b) = 1$, and $a < b$. Thus: $\log 2 = \dfrac{a}{b} \to 2 = 10^{\frac{a}{b}} \to 2^b = 10^a = 2^a\cdot 5^a \to 2^{b-a} = 5^a$. We see a contradiction here because the $LHS$ is even while ...


17

A proof can be carried out after modifying your calculation a bit. $$2=10^{\frac{a}{b}}\implies \color{blue}{2^b=10^a}\implies2^{b-a}=5^a$$ Which is a contradiction when both $a$ and $b$ are non-zero integers. Check the colored step carefully and you will understand in which step you have made a mistake.


18

As has been pointed out in comments and in another answer, $10^{a/b}\neq 10^{a}10^{\frac{1}b}$. This is a rather subtle error, however there's a notable warning flag that could alert you to it: Your proof does not use the hypothesis that $a$ and $b$ are integers. This is a serious issue, because it means you've proved the (false) statement that $\log(2)$ ...


2

The equality $2=10^a10^{\frac{1}{b}}$ is not true, because $10^a10^{\frac{1}{b}}=10^{a+\frac{1}{b}}.$


3

Be careful, $10^{\frac{a}{b}}$ equals $(10^{a})^{\frac{1}{b}}$ not $10^{a}10^{\frac{1}{b}}$


0

The question of why we need real numbers is a good question. The basic answer is that real numbers are vital to the theoretical foundation of analysis (calculus) - "completeness property", is the key. It is true that people of an engineering or physics background who rely on calculus techniques do not require the theory of real numbers, since they can just ...


2

No, there is not always a unique closest natural number, as mentioned in comments. In more detail: If $q$ is a rational number of the form $q= a/2$ with $a$ an odd positive integer, then the two natural number $(a-1)/2$ and $(a+1)/2$ are both at distance $1/2$ from $q$. Every natural number other than these two is either less than $(a-1)/2$ or greater than ...


2

The integral of any function over a set with measure $0$ is equal to $0$.


0

The exact condition is that $S$ intersects $\{\pm\sqrt{r}: 0 \le r \in {\mathbb Q}\}$, where $\mathbb Q$ is the rationals. It's a dense set, but countable, and therefore very "easy to miss" by anything that doesn't have to contain an open interval.


1

That's not the definition of irrational numbers. A rational real number is a number that can be written as the ratio of two integers, $\displaystyle \frac{p}{q}$. An irrational real number is a real number that is not rational. It turns out that this will imply an irrational number has a decimal expansion which is not a repeated sequence of digits. ...


3

Question 1. Yes, you do need to define your relations. In particular, your three axioms are consistent with a further axiom "$\mathrm{frac}(a,b) = \mathrm{frac}(c,d)$ for all $a,b,c,d$: i.e, that all fractions are equal to each other. None of your axioms say anything about when fractions are not equal. So you need something else. The suggested axiom $$ ...


1

To prove 2), begin with the following facts: If the interval $(a,b]$ contains some $q_k$ for $1\le k\le n$, then $f(b)-f(a)\ge 2^{-n}$. If the interval $(a,b]$ does not contain $q_k$ for $1\le k\le n$, then $f(b)-f(a)\le 2^{-n}$ The proof of 1. amounts to observing that $f(b)-f(a)$ contains a term equal to $2^{-n}$. The proof of 2 is ...


0

Hypothesis: $\sqrt[n]{\sqrt{2}+1} + \sqrt[n]{\sqrt{2}-1} \in \mathbb{Q}$ First transform given value. Note that $(\sqrt{2}+1)(\sqrt{2}-1) = 1$. $$\sqrt[n]{\sqrt{2}+1} + \sqrt[n]{\sqrt{2}-1} = \sqrt[n]{\sqrt{2}+1} + \frac{\sqrt[n]{\sqrt{2}-1} \cdot \sqrt[n]{\sqrt{2}+1}}{\sqrt[n]{\sqrt{2}+1}} = \sqrt[n]{\sqrt{2}+1} + \frac{1}{\sqrt[n]{\sqrt{2}+1}}$$ Let $x ...


1

I might want to use a property of integer arithmetic $a \lt b \iff a+1 \le b$, as in: $\qquad\;\;\lceil x/y \rceil = n $ $ \iff n-1 \lt x/y \leq n $ $ \iff (n-1)y \lt x \leq ny $ $\iff (n-1)y \le x-1 \lt ny$ $\iff n-1 \le (x-1)/y \lt n $ $ \iff \lfloor(x-1)/y\rfloor =n$ so $\lceil x/y \rceil = \lfloor(x-1)/y\rfloor$.


1

starting from line 2, the inequality can be rewritten as \begin{equation*} n-1+1/y \leq x/y < n+1/y \end{equation*} since x,y are integers, the left inequality is equivalent to \begin{equation*} n-1 < x/y \end{equation*} and the right inequality is equivalent to \begin{equation*} x/y \leq n \end{equation*}


2

If $y$ divides $x$ then: $\lceil\frac{x}{y}\rceil=\frac{x}{y}$ $\lfloor\frac{x-1}{y}\rfloor=\frac{x}{y}-1$ Hence $\lceil\frac{x}{y}\rceil=\lfloor\frac{x-1}{y}\rfloor+1$ If $y$ does not divide $x$ then: $\lceil\frac{x}{y}\rceil=\lfloor\frac{x}{y}\rfloor+1$ $\lfloor\frac{x-1}{y}\rfloor=\lfloor\frac{x}{y}\rfloor$ Hence ...


1

Even if you wanted to preserve only addition (without the constraints on multiplication and order) it wouldn't work: since $\forall q \in \mathbb Q, f(\frac{q}{2}+\frac{q}{2}) = 2 f(\frac{q}{2}) = f(q)$, and thus $f(\frac{q}{2}) = \frac{f(q)}{2}$, you could build an infinite series $\left( \frac{f(q)}{2^n} \right)_n$ with values in $\mathbb N$, which is ...


0

HINT: If $f\colon\Bbb{N\to Q}$ would be such map, what could you say about $f(0)$ and $f(1)$? (Double hint, it suffices to consider the order.)


1

You have to be careful with these kind of things if your base is not a non-negative real number. For example, $$1=1^{1/2}=[(-1)^2]^{1/2}=-1.$$ The reason for this can be found when you look at the "true" definition of $x\mapsto a^x$ when $a\in\mathbb{C}\setminus[0,\infty)$. We define this as: $$a^x:=e^{x\log a}$$ which of course requires some kind of ...


0

I think you can solve it with a Farey fraction algorithm. Let's assume we start with $0\lt a\lt b\lt 1$. Any time you have $${p\over q}\lt a\lt b\lt {r\over s}$$ with $p/q$ and $r/s$ in reduced form, you can check if $$a\le{p+r\over q+s}\le b$$ If so, you're done. If not, replace the appropriate upper or lower bound and continue. By starting with ...


1

We may assume $0 < a < b$. (If $a \le 0 \le b$, then $0/1$ is the solution; and if $a < b < 0$, then solve for $[-b,-a]$.) Let the continued fractions of $a$ and $b$ be $[a_0;a_1,a_2,...]$ and $[b_0;b_1,b_2,...]$. Let $n$ be the first position in which they differ: that is, $a_i = b_i$ for $i < n$, and $a_n \ne b_n$. Let $c = \min(a_n,b_n)$. ...


0

If $p$ and $q$ are integers such that $a \leq \frac pq \leq b,$ then $qa \leq p \leq qb.$ In fact, since $p$ is an integer, $\lceil qa \rceil \leq p \leq \lfloor qb \rfloor.$ For any integer $q,$ then, if $\lceil qa \rceil > \lfloor qb \rfloor$ then it is impossible to write $\lceil qa \rceil \leq p \leq \lfloor qb \rfloor$ for any integer $p$ and ...


0

To add to André's hint: Set $a=b=1$, so that $1^{p}+1^{q} = c^{r} \implies 2=c^r$ What does this say about the rationality of $\sqrt[n]{2}$?


3

Hint: Suppose that $\sqrt{1}+\sqrt{2}$ is of the shape $c^r$, where $c$ is a positive rational and $r=\frac{m}{n}$ where $m$ and $n$ are integers, with $n\gt 0$. Show by taking the $n$-th power of both sides that this implies that $\sqrt{2}$ is rational.


1

Proof guideline: Show that for all $ x \in \mathbb{Q}$ any line $L_x = \{(x,y) , \forall y \in \mathbb{R}\}$ is path-connected Show that for all $ y \in \mathbb{Q}$ any line $L'_y = \{(x,y) , \forall x \in \mathbb{R}\}$ is path-connected From (1) and (2) follows that $L_0$ and $L'_0$ are path-connected Show that if $X$ and $Y$ are path connected and $X ...


0

Path connected means that for arbitrary pairs of points in $X$, $(x_1, y_1)$ and $(x_2, y_2)$, you can find a continuous map $f : [0,1] \rightarrow X$ such that $f(0) = (x_1, y_1)$ and $f(1) = (x_2, y_2)$. One way to do this is to construct a path $(x_1, y_1)$ to $(x_1,0)$ to $(0, 0)$ to $(x_2, 0)$ to $(x_2, y_2)$. That is, every member of $X$ is path ...


2

Indeed, a "nearest rational" does not exist. If $x$ is rational, you can find an obvious path from $(x,y)$ to $(x,0)$ and then on to $(0,0)$. Similarly, if $y$ is rational, you can find a path to $(0,0)$ via $(0,y)$. Thus any $(x,y)\in X$ is pathconnected to $(0,0)$.



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