This is a bit naive question bu help me out here. We know that in Arithmetic,  we follow the BODMAS rule where first preference is given to Brackets then Exponent then Division and so on. My question is how did we arrive at the conclusion to give the following preference. Was it conventional or has some theoretical twist in it?
123805
After reading through a few other questions I was just asking myself: How was the Order of Operations defined, and why is it this specific order and not a different one?  Most of us know things like multiplication/division before addition/subtraction, parentheses first, etc - but what's the actual reason behind it? I'm probably biased by following those rules since childhood, so I can't really think of any other way.  2 + 2 x 2 = 6 and not 8  But if the order would be changed, let's say to "addition/subtraction before multiplication/division" would that order still work if we assume that mathematics would build up on it? Or is there some strange mathematical problem if we would be using a different order?  Of course, parentheses have a grouping function so they should always come first - I'm mainly talking about the exponents, multiplication/division, addition/subtraction (and maybe other operations left out).
21599
If I know for two numbers a and b that    $${a &lt; b }$$  Then is it correct to say that    $$ a \leq b $$  I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information.
64085
After browsing through related threads, I am now of the understanding (please correct me if I am wrong) that the rule of thumb is to use the \bigl,\bigr pair with brackets, parentheses, etc. for operators like sums, products, and integrals, and \left,\right in all other cases.  I only recently found out about \middle , which has solved what had been a vexing problem for me: delimiters for function arguments (slashes, bars) not scaling properly. Leafing through my cheatsheet has not been of much help, so I wish to ask: is there an equivalent of \middle for \bigl and \bigr, or would the use of \middle along with that pair not be gauche?
53663
My question is told in a few words: Why do you write $E[X]$ in square brackets instead of something like $E(X)$? Probably it is not a "function". How would you call it then? This question also applies for $Var[X]$.
48437
So, I know that we first need to have an initial point. The answers I have say it's $(-1, 0)$ which makes sense because it satisfies the equation. But for example $(1, 0)$ satisfies it too. Why did we choose $(-1, 0)$? In general, how do we go about choosing the initial point?
78026
In my math text book the proof for the Sum to Product rule starts with a substitution:  let A = (u+v)/2, let B = (u-v)/2  ...and then uses this substitution to prove that A+B = u  and  A-B = v  ...then substitutes (u+v)/2 and (u-v)/2 in the Product-to-Sum equation to demonstrate that the Sum-to-Product rule is true.  As it appears to me from what is presented in the text book, the substitutions  (u+v)/2 and (u-v)/2 were pulled out of thin air.  How was this determined to work? I doubt it was just a lot of guess and check until something worked.   Is there a point in math where one just needs to accept an equation for what it is and not worry about why or how it is?  I'm getting the feeling that understanding some of the how's and why's right now requires an understanding of math concepts that I haven't learned yet.  Any insight would be appreciated. Thanks!
98951
There are so many available bases. Why is the strange number $e$ preferred over all else?  Of course one could integrate $\frac{1}x$ and see this. But is there more to the story?
61448
I'm having trouble remembering PEMDAS (in regards to precedence in mathematical equations), ie:   Parentheses Exponentiation Multiplication &amp; Division Addition &amp; Subtraction   I understand what all of the above mean, but I am having trouble keeping this in my head. Can you recommend any tricks or tips you use to remember this.  Thanks
22539
$\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector?  Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?   So why is cross product defined in the way that it is? I am mainly interested in the right hand rule defintion too as it is out of reach?
37306
$ab = ba$  This is, inherently, true. Some texts drop it like an axiom without any justification. But I'm a bit curious where it stems from or basically why/how it works. If anyone could enlighten me a bit further, I'd be most grateful. Peano axioms? From set theory? Help.  What bugs me with this definition is it's real world "application":  Let's say I have a power output of 5 W ( $kg\cdot m^2 \cdot s^{-3}$) and I want 30% of that power output (to have some units and natural context).  Naturally, 30% is 30 1/100 (% is the unit). For simplicity, let's express that as:  $30$%$ = 30/100 = 3/10 = 3d$ where $d = 1/10$, a simpler unit (because %W would look... Wrong?).  So 30% of power is then $5 W \cdot 3d$, and we can force it to "make sense" if we associate d with W or basically scale the unit of power by d (which is intuitively understood as being 10 times smaller then the output of 1 W) $dW$. And we want to scale these 5 units of $dW$ three times.  $5 dW \cdot 3$  But the commutative property says it is the same as scaling 3 units of $dW$ 5 times or:  $3 dW \cdot 5$  And the result is the same. This is the bit that hurts my head, the fact it is the same. I try to interpret the same as, for example 50 J of work, it's either applying 50 N over 1 m of distance or 1 N over 50 m of distance.    Down here is additional stuff I think I have (you don't have to read it if you know how to answer immediately). I don't claim it's correct, if anyone parses this information, feel free to correct me.  $m\sum\limits_{i=1}^{n}{1} = n\sum\limits_{i=1}^{m}{1}$  This is how I've been trying to break it down, as it is repeated addition. It is repeated $n$ times and multiplied by $m$, and reverse on the right side. There is $x = m - n$, so when I multiply $n$ $m$ times, each repeated addition of $n$ lacks exactly $x$ to $m$.  $mn = nm$  $x = m-n$  $m(m-x) = (m-x)m$  $m^2 - mx = m^2 - mx$  $ true $  And yes, sadly, I realise that I am using the very property I'm trying to prove ($xm$ shifted as $mx$ on the right-hand side.)  And it even calls in the distributive property of multiplication over addition which I derive (for myself, informally) from the very nature of multiplication and the way we "process" numbers:  For example, number $55$ times $2$  $55 = 5 \cdot 10^1 + 5 \cdot 10^0 = 5 \cdot 10 + 5 \cdot 1$  $(5 \cdot 10 + 5 \cdot 1) \cdot 2$  This is the basis of our positional notation which appends digits of varying orders of magnitude or units together in a way it makes "sense". Each one is b times bigger than the one to the right. Basic stuff. It is why I expressed 10 and 1 explicitly, those are what I consider units in this case.  Now, from the definition of multiplication, which is at its heart just repeated addition, it is truly the same if you add together $2$ $50$ times and then again $5$ times or "all at once" (figuratively, we are always doing the former mentally) $2$ $55$ times.  That added together gives 110, which is true.  To me, the distributive property (in case of integer multiplication) is a repercussion of the very definition of multiplication which is repeated addition (evading strict concepts). I am not certain whether this is a good way to look at it.
8308
There is a huge debate on the internet on the value of $48\div2(9+3)$. I believe the answer $2$ as I believe it is part of the bracket operation in BEDMAS.  yields the same answer. I also believe that if $48\div2\times(9+3)$ was asked it would be $288$ which Mathway agrees with as well. However,  says it is $288$ either way. A friend of mine (who is better at math) told me that there is no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didn't explicitly give a reason) What is the answer and why?
21018
In my textbook on elementary number theory from a class last year, as well as elsewhere through my academic experience and even posts here, I often see the greatest common divisor notated as $(a,b)$ (where this represents the greatest common divisor of $a,b$).  My question is, why is such an unusual notation used? I say it's unusual (if not a bit absurd), for a couple of reasons:   It is unintuitive. It does not grant the reader an immediate knowledge of what it means. In particular, I can easily see it being misinterpreted as an ordered pair of points, and not as a function (which it basically is and I touch on in a second) or representing a single value. Good notation would be immediately understood, no? One should not struggle with parsing through the context or meaning. A clearer notation is often used, and just three letters away. You could certainly think of $(a,b)$ as an ordered pair, if we consider it a function. I often see the alternate $gcd(a,b)$: a function, which takes in two inputs ($a$ and $b$), and outputs their greatest common divisor (or factor, if you prefer). Merely adding the three letters makes the notation clearer by clarifying it is  a function, and using a relatively well-accepted acronym. (Or, at least, I often hear it abbreviated "g.c.d." in school.)   I've mulled this over for a while and I can't think of much without borrowing inspiration from other "unusual" notations. For example:   Common understanding in context: consider the $\log(x)$ function. Depending on context, without a base stated as seen here, it can be presumed to mean whatever is most "appropriate." It could be base $e$ (common in mathematics), base $10$ (common in engineering), or base $2$ (common in computer science). The ambiguity is essentially resolved by context in other words - personally, not a fan of that (I like the explicit stating of bases for full clarity), but if such a motivation holds elsewhere, it might stand to reason that $(a,b)$ could just be taken as a shorthand and "understood" in the context of number theory to simply be the greatest common divisor. That it is in fact used despite $gcd(a,b)$ being clearer and somewhat common also does seem to suggest this playing a role. Perhaps it just started that way and it kept being used. For example,  (based on a ) regarding how logarithms, roots, and exponentiation can be unintuitive from a notation perspective, and proposes an alternative, easier to understand notation. I haven't been in middle school in nearly a decade, but guessing from discussions online, that obviously hasn't really picked up too much. In that sense, perhaps it's like that for $(a,b)$: like we still notate exponentiation, etc., the same way today despite how counterintuitive it is (at least for students), perhaps it just "stuck" and proliferated?   I wasn't able to easily find any reason why online, and it bugs me a bit, so I ask:     Why do we so unintuitively notate the greatest common divisor of two numbers $a,b$ as $(a,b)$?   I of course recognize that this is somewhat moot in light of how easily it's made clear - as established above, all I have to do is append the front of the parenthetical by $gcd$ or $gcf$ to make it way better. And of course, people do that. But I'm more curious in where this ambiguous former notation arose and, in light of a better notation, why it is still used today.
7294
I've always heard this reasoning, and it makes obvious sense, but how do you actually show it for some arbitrary product? Would it be something like this?  $$(a(b(cd)))e=((ab)(cd))e=(((ab)c)d)e=abcde?$$  Do you just say that the grouping of the parentheses now corresponds to just multiplying straight through? Thanks.
60186
What is the logic/thinking process behind deriving an expression for even and odd functions in terms of $f(x)$ and $f(-x)$?  I've been pondering about it for a few hours now, and I'm still not sure how one proceeds from the properties of even and odd functions to derive: $$\begin{align*} E(x) &amp;= \frac{f(x) + f(-x)}{2}\\ O(x) &amp;= \frac{f(x) - f(-x)}{2} \end{align*}$$ What is the logic and thought process from using the respective even and odd properties, $$\begin{align*} f(-x) &amp;= f(x)\\    f(-x) &amp;= -f(x) \end{align*}$$   to derive $E(x)$ and $O(x)$?    The best I get to is: For even: $f(x)-f(-x)=0$ and for odd: $f(x)+f(-x)=0$  Given the definition of $E(x)$ and $O(x)$, it makes a lot of sense (hindsight usually is) but starting from just the properties. Wow, I feel I'm missing something crucial.
54806
In Peano arithmetic (first order), we first define natural numbers using a successor function and Peano axioms, then we define addition (and multiplication), and then, we define inequality, as:  $a\leq b\leftrightarrow\exists c\left(a+c=b\right)$   Is there any way to define inequality first, directly from the successor function and Peano axioms? (I mean, if I don't need addition for my purpose, why define it?).
88389
In my textbook (S. Abbott. Understanding Analysis 1 ed. pp 47 Theorem 2.3.3.iv), the author proves $b_n \to b$ implies $\frac{1}{b_n} \to \frac{1}{b}$ the following way:   $$\left|\frac{1}{b_n}-\frac{1}{b}\right|=\frac{|b-b_n|}{|b||b_n|}$$  therefore, since we can make the numerator as small as we like by definition, we just need to find a "worst-case estimate" for $\frac{1}{|b||b_n|}$ by considering $|b_n| \geq \lambda &gt; 0$. He then argues that we need to look far into the sequence so that  the terms are closer to $b$ than to $0$. So, considering $\epsilon_0$=$|b|/2$, we can use $|b_n-b|&lt;|b|/2$ for all $n \geq N_1$. Thus, $|b_n| \geq |b|/2$. Then we choose $N_2$ s.t. $n \geq N_2$ implies $|b_n-b|&lt;\frac{\epsilon |b|^{2}}{2}$. Finally, if $N$ = max {$N_1, N_2$}, then $n \geq N$ implies  $$\frac{|b-b_n|}{|b||b_n|}&lt;\frac{\epsilon |b|^{2}}{2} \frac{1}{|b|\frac{|b|}{2}}=\epsilon.$$  What confuses me is why we need to pick elements closer to $b$ than to 0. What makes picking $\epsilon_0$=$|b|/3$ and $|b_n-b|&lt;\frac{\epsilon |b|^{2}}{3}$ wrong for the proof? Thanks.
115477
Where $(50)_5 = 50! / 45! $ and $50\choose 5$ $= 50!/(5!45!)$   I've narrowed it down to these two possibilities (though even these may be wrong), but I think I'm confusing myself as to which one I use. Can anyone suggest a good way of thinking of this problem that will make it apparent whether I need use use the falling factorial or the binary coefficient?   Edit: thank you for all the answers! Indeed, I meant $50!/45$ as the falling factorial. Also, the question doesn't specify whether it was meant to be ordered or unordered, but I think I understand my confusion now!