I was reading Principles of Mathematical Analysis - Walter Rudin. In the start (pg-11), it is shown that the equation $p^2 = 2$ is not satisfied by any rational $p$ i.e. $\sqrt{2}$ is irrational.  After this the book says:  "We now examine this situation a little more closely. Let $A$ be the set of all positive rationals $p$ such that $p^2 &lt; 2$ and let $В$ consist of all positive rationals $p$ such that $p^2 &gt; 2$. We shall show that $A$ contains no largest number and $В$ contains no smallest.  More explicitly, for every $p \in A$ we can find a rational $q \in A$ such that $p &lt; q$, and for every $p \in В$ we can find a rational $q \in В$ such that $q &lt; p$.  To do this, we associate with each rational $p &gt; 0$ the number $$ \begin{align} q &amp;= p - \frac{(p^2 - 2)}{p + 2} &amp;(2)\\ &amp;=\frac{2p + 2}{p + 2} &amp;(3)\\ &amp;\Rightarrow q^2 - 2 = \frac{2(p^2 - 2)}{(p + 2)^2} &amp;(4) \end{align} $$  If $p \in A$ then $p^2 — 2 &lt; 0$, (3) shows that $q &gt; p$, and (4) shows that $q^2 &lt; 2$. Thus $q \in A$. If $p \in В$ then $p^2 — 2 &gt; 0$, (3) shows that $0 &lt; q &lt; p$, and (4) shows that $q^2 &gt; 2$. Thus $q \in B$."  I am unable to follow this part, specially how we construct the equations (3) and (4). Can somebody explain?
1461
First, my apologies if this has already been asked/answered.  I wasn't able to find this question via search.  My question comes from Rudin's "Principles of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2.  In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number $q$ that satisfies equations (3) and (4), plus other conditions needed to show that $q$ is the right number for the proof.  As an exercise, I tried to derive his choice of $q$ so that I may learn more about the problem.  If we write equations (3) as $q = p - (p^2 - 2)x$, we can write (4) as   $$ q^2 - 2 = (p^2 - 2)[1 - 2px + (p^2 - 2)x^2]. $$  Here, we need a rational $x &gt; 0$, chosen such that the expression in $[...]$ is positive.  Using the quadratic formula and the sign of $(p^2 - 2)$, it can be shown that we need   $$ x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A, $$  or, for $p \in B$, $x &lt; 1/\left(p + \sqrt{2}\right)$ or $x &gt; 1/\left(p - \sqrt{2}\right)$.  Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting $x = 1/(p + n)$ for $n \geq 2$.  Notice that Rudin chooses $n = 2$ for his answer, but it checks out easily for other $n$.  The Question:  Why does Rudin choose $x = 1/(p + 2)$ specifically?  Is it just to make the expressions work out clearly algebraically?  Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof?  Is there a simpler derivation for the number $q$ that I am missing?
12642
Consider the closed interval $[0,1]$, there is $\frac{2}{3} \in [0,1]$ where $p=2$ and $q=3$. Similarly consider $[2,3]$, one can have $\frac{5}{2} \in [2,3]$ where $p=5$ and $q=2$. Does every interval of the form $[a,b]$, where $a,b \in \mathbb{R}$ contain a rational of this kind. If yes, how can we prove it?
101781
Just out of curiosity I was trying to solve the equation $x^2=2^x$, initially I thought there would be just the two solutions $x=2$ and $x=4$, but wolfram shows that the two equations intersect at not 2 but 3 locations, the third being a negative value of $x$. The third solution isn't obvious like the other two, so I just have a few questions about the negative solution. Is it rational? is it commonly represented with a greek letter? If it is irrational is there a way to approximate it?
25688
Which rational numbers can be written as the sum of two rational squares?  That is, for which rational numbers $a$, are there rational numbers $x$ and $y$ such that $a = x^2 + y^2$.  It is a famous theorem that if an integer can be written as the sum of two rational squares then it can be written as the sum of two integral squares, and then the solution is the famous one by Fermat, but I didn't find anything about the general case.
15972
I have to find the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$. The suggested way of doing it is to prove that $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ first.  I can prove that. It's enough to prove that $\sqrt 2,\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ and that $\sqrt 2 + \sqrt[3] 2\in \mathbb Q[\sqrt 2,\sqrt[3] 2]$. The latter is obvious and $\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ will follow from $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$ It remains to prove $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$  For $\alpha=\sqrt 2 + \sqrt[3] 2,$ we have   $$ \begin{eqnarray} &amp;\alpha^0&amp;=1,\\ &amp;\alpha^1&amp;=\sqrt 2 + \sqrt[3] 2,\\ &amp;\alpha^2&amp;=2+\sqrt[3]4+\sqrt 2\sqrt[3]2\\ &amp;\alpha^3&amp;=2+2\sqrt 2+6\sqrt[3]2+3\sqrt 2\sqrt[3]4\\ &amp;\alpha^4&amp;=4+4\sqrt 2+12\sqrt[3]4+8\sqrt 2\sqrt[3]2\\ &amp;\alpha^5&amp;=32+4\sqrt 2+20\sqrt[3]2+4\sqrt 2\sqrt[3]2 \end{eqnarray} $$  If I can express $\sqrt 2$ as a linear combination of $\{\alpha^0,\alpha^1,\cdots,\alpha^5\}$, I'm done. I don't know how to find out whether $$\{1,\sqrt 2,\sqrt[3]2,\sqrt[3]4,\sqrt 2\sqrt[3]2,\sqrt 2\sqrt[3]4\}$$ are linearly independent over $\mathbb Q$ but I think I don't have to care. I can use Gaussian elimination anyway. I write the matrix  $$\left[ \begin{array}{rrrrrr|r} 1 &amp; 0 &amp; 2 &amp; 2 &amp; 1 &amp; 8 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 2 &amp; 1 &amp; 1 &amp; 1\\ 0 &amp; 1 &amp; 0 &amp; 6 &amp; 0 &amp; 5 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0 &amp; 3 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0 &amp; 2 &amp; 1 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 3 &amp; 0 &amp; 5 &amp; 0 \end{array}\right]$$  and by performing row operations I obtain a row echelon form, which proves that I'll obtain the reduced row echelon form if I want, which will give me the desired coefficients $a_0,\cdots,a_5$ such that   $$\sqrt 2=\sum_{i=0}^5a_i\alpha^i.$$  I don't have to find the coefficients -- I'm just satisfied with their existence.  If this is correct, I have proven that   $$\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2].$$  But I don't know how to proceed further. Of course, I can find the minimal polynomials of $\sqrt 2$ and $\sqrt[3] 2$ over $\mathbb Q$ but I don't know how to find the minimal polynomial of $\sqrt 2$ over $\mathbb Q[\sqrt[3]2]$ or the minimal polynomial of $\sqrt [3]2$ over $\mathbb Q[\sqrt 2].$  What I did is this. I wrote the equation  $$x=\sqrt 2+\sqrt [3]2$$  and by squaring and cubing obtained  $$W(x):=x^6-6x^4-4x^3+12x^2-24x-4=0.$$  I belive this is the required polynomial but I would need to prove that it's irreducible over $\mathbb Q.$ Eisenstein's criterion doesn't work, which means I'm lost.  Could you please help me with this? I would like to know three things.  1) Is my proof of $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ correct?  2) How can I use this fact? I believe it's supposed to make the solution easier to find.  3) How can I tell whether $W(x)$ is reducible over $\mathbb Q$ or not?  EDIT: 1) has been answered by André Nicolas in a comment. All of this answers 3) but I would like to ask whether we could know that the polynomial is irreducible without knowing that $\sqrt 2+\sqrt[3]2$ is its zero. Also, it would be to know whether  4) there is a theorem that would give me $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ without the need of performing the Gaussian elimination?  For these numbers it was possible to do, but a bit higher dimension and it would turn out impossible without a computer.
76126
I want to prove that for $p,q,r$ different primes, $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is irrational.   Is the following proof correct?  If $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is rational, then  $(\sqrt{p}+\sqrt{q}+\sqrt{r})^2$ is rational, thus $p+q+r+2\sqrt{pq}+2\sqrt{pr}+2\sqrt{qr}$ is rational, therefore $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational.  If $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ is rational, then $(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})^2$ is rational, therefore $pq+qr+pr+\sqrt{p^2qr}+\sqrt{pq^2r}+\sqrt{pqr^2}$ is rational, therefore $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$ is rational.  Now suppose $p&lt;q&lt;r$. If $p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}$  and $\sqrt{pq}+\sqrt{pr}+\sqrt{qr}$ are rational, then $$p\sqrt{qr}+q\sqrt{pr}+r\sqrt{pq}-p(\sqrt{pq}+\sqrt{pr}+\sqrt{qr})$$  is rational, therefore $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational.   If $(q-p)\sqrt{pr}+(r-p)\sqrt{pq}$ is rational, then $((q-p)\sqrt{pr}+(r-p)\sqrt{pq})^2$ is rational, thus $(q-p)^2pr+2(q-p)(r-p)\sqrt{p^2qr}+(r-p)^2pq$ is rational, thus $\sqrt{qr}$ is rational.  But $q,r$ are distinct primes, thus $qr$ can't be a square. Thus $\sqrt{qr}$ is irrational. Contradiction.   Also, is there an easier proof?
123218
I am having a problem understanding this proof.   Suppose that $\sqrt{2}$ was rational.   Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.   We have  $s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.  Because $s\sqrt{2} = 2t$ and $s$ are both integers,  $s\sqrt{2} - s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$  must also be an integer.   Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} &gt; 1$.   It is less than $s$, because $\sqrt{2} &lt; 2$ so that $\sqrt{2}-1 &lt; 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.  My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?   If someone can explain this proof to me it would be great! thank you
107423
Why does the rational root theorem only work when the polynomial has integer coefficients?  the rational root theorem (or rational root test, rational zero theorem, rational zero test or $p/q$ theorem) states a constraint on rational solutions of a polynomial equation $$ a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{0}=0$$ with integer coefficients ${\displaystyle a_{i}\in \mathbb {Z}}$. The theorem states that each rational solution $\,x = p⁄q\,$ written in lowest terms (so  $p$ and $q$ are relatively prime), satisfies:  $p$ is an integer factor of the constant term $a_0$, and $q$ is an   integer factor of the leading coefficient $a_n$.
13378
More specifically, I have to find a $q$ in $A$ such that $q$ is larger than any $p$ in $A$.  The only thing I can think of is using (${2 - {p^2}}$) somehow, or ${(p+x)^2}&lt;2$, but other than that I don't know how to proceed. This exercise is taken from the very first chapter of Walter Rudin's Principles of Mathematical Analysis. Quoting the exercise:     To do this, we associate with each rational $p &gt; 0$ the number:     $$q = p - \frac{{{p^2} - 2}}{{p + 2}} = \frac{{2p + 2}}{{p + 2}}$$   But he didn't explain how he came up with that number.   The only thing I've thought of is the following procedure:  $$\eqalign{   &amp; {p^2} + (2 - {p^2}) = 2 \Rightarrow {p^2} + (\frac{{2 - {p^2}}}{2}) &lt; 2,{\text{ and since }}2 - {p^2}{\text{ is always positive,}}  \cr    &amp; {p^2} &lt; \left[ {{p^2} + (\frac{{2 - {p^2}}}{2})} \right] &lt; 2  \cr    &amp; {\text{Therefore, let }}q = \sqrt {{p^2} + (\frac{{2 - {p^2}}}{2})} ,{\text{ then:}}  \cr    &amp; p &lt; q &lt; \sqrt 2  \cr} $$  However, my result doesn't necessarily belong to ${\Bbb Q}$. I wonder how did he obtain the value of $q$ in the book.  EDIT: Feel free to point out if there are more appropriate tags for this question.
49584
I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$  Squaring both sides, $$p^2/q^2 = 2\tag{a}$$  or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.  Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.  For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.  Now, from $(b)$, $q^2$ can be even or odd.  Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.  Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2&gt;1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)  Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.  Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.  Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.  PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing  2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as  1/ $2^(n-1)$ in which n has to be existent.  therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1  Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments?
16666
Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.
73502
Suppose $i_1$ and $i_2$ are distinct irrational numbers with $i_1 &lt; i_2$.  Is it necessarily the case that there is a rational number $r$ in the interval $[i_1, i_2]$?  How would you construct such a rational number?  [I posted this only so that the useful answers at  could be merged here before that question was deleted.]
29865
The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable? e.g. Suppose $\sqrt{4}$ is rational, $$\begin{align}        \sqrt{4} &amp;=p/q        \qquad\text{where pand q are coprimes} \\         4 &amp;=p^2/q^2\\       4q^2&amp;=p^2       \tag{1} \\          4&amp;\mid p^2\\          4&amp;\mid p\\      \text {let }p&amp;=4m  \qquad\text{for some natural no. m}  \\        p^2&amp;=16m^2\\       4q^2&amp;=16m^2   \qquad\text{(from (1) )}\\        q^2&amp;=4m^2\\          4&amp; \mid q^2\\          4&amp;\mid q \end{align} $$ but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence  $\sqrt{4}$ is not rational. But we know that it is a rational. Why?
102109
Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.   Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.   Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2&lt;n^7-n^3=\text{r.h.s}.$ This means that either $p&gt;n^2$ or $p&lt;n^2$ must hold.  Assuming $p&gt;n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p&gt;n^2)$.  Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.  Finally, we get $(5,2)$ as the solution to the given equation.      But how do I conclude this is the only solution possible? Also, why does'nt $p&lt;n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem.
91536
I think of the number zero as a whole number.  It can certainly be a ratio = $\frac{0}{x}, x \neq 0.$  Therefore it is rational.   But any ratio equaling zero involves zero, or is irrational, e.g.$\frac{x}{\infty}, x \neq 0$ is not a ratio of integers.  Can a rational number that is rational only when the number itself is involved still be rational?  I realize that it "should" be rational but it doesn't seem to fit into the same category with other rational numbers. (I'm glad mse has a soft question category for questions like this...).
121472
A popular proof of the irrationality of $\sqrt2$ is to first assume that the number is rational. This means that $\sqrt2=a/b$ where $a$ and $b$ are integers. Another assumption is that $a$ and $b$ are coprime. It turns out that this leads to a contradiction. And the conclusion is that $\sqrt2$ is not rational (because the assumption of rationality led to a contradiction). But what about the second assumption? Logically, the conclusion could also be that $a$ and $b$ are not coprime..  How can this be resolved?  Edit:  Thank you all for your comments. It cleared up my mind.  Three aspects of this kind of proof are: mathematical, logical, and didactical.  Mathematically, it is the case that any rational number can be written as a fraction with coprime numerator and denominator. But this point is often described in a way that is either literally a logical assumption, or in a way that is natural to interpret as a hidden logical assumption. In this case, the proof will have a flawed logic, because any of the two assumptions could be the cause of the contradiction. Didactically, this can cause confusion in the mind of the student, who might not be so accepting of the proofs conclusion. I used to be that student.  So, how to resolve it? Many of you provided explanations on the mathematical aspect, which are all correct. But my main point is not of not understanding the proof, but of the use of flawed logic which leads to not understanding the proof. I am sure that a lot of teachers think or say something along the lines of WLOG, but for many students, this makes the proof weaker, because it is not so convincing. When we present this proof to our students, we need to have this in mind.  Thank you.
98578
Suppose $\sqrt 2$ were rational.  Then we would have integers $a$ and $b$ with $\sqrt 2 = \frac ab$ and $a$ and $b$ relatively prime.  Since $\gcd(a,b)=1$, we have $\gcd(a^2, b^2)=1$, and the fraction $\frac{a^2}{b^2}$ is also in lowest terms.  Squaring both sides, $2 = \frac 21 = \frac{a^2}{b^2}$.  Lowest terms representations of rational numbers are unique, so we have $a^2 = 2$ and $b^2=1$.  But there is no such integer $a$, and therefore we have a contradiction and $\sqrt 2$ is irrational.  I am not interested in the pedagogical value of this purported proof; I am only interested in whether the logic is sound.