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47

It's purely a matter of how we choose to define the notation. The main reason for it is that it lets us write polynomial expressions (which are extremely common) without parentheses, e.g., $x^3 + 3x^2 y - 41x + 2z$ rather than $(x^3) + (3(x^2)y) - (41x) + (2z)$. However, what really matters is that the notation is clear and unambiguous, so expressions like $...


27

You asked: "How do/would you explain why division by zero does not produce a result." Any such explanation that is not rooted in student understanding would be talking to ourselves, not to students. Therefore both meaning and student understanding are important. Otherwise, what's the point? So I have grounded my response there. Young students (...


26

Have the students tell you that division by zero is a non-sequitur. This is possible at any age where division is understood at all. Teacher: If there are eight cookies and four children, how many cookies does each child get? Student: Uh, two. Teacher: Yes! This is a division problem. $\frac{8}{4} = 2$. Now, if there are 8 cookies shared by only two ...


20

The issue is the implied multiplication in 8÷2(2𝑥). Different calculators actually resolve this differently, so in that sense we would want to say this is ambiguous. If implied multiplication works the same way explicit multiplication does, then we do the division first (left to right) and get 4(2𝑥). If implicit multiplication has a higher priority than ...


15

While there may be legitimate reasons behind the convention In $a \times b $ the $a$ denotes the number of terms and the $b$ denotes the individual terms the larger issue is the mismatch between the teacher's enforcement of that convention and the expressly stated purpose of the formative assessment, which is written at the top of the very same page: ...


15

In third grade we taught division using repeated subtraction. To divide 6 by 2, subtract 2 until you get to 0. 6-2=4, 4-2=2, 2-2=0. It took 3 steps so 6÷2=3. This can also be shown on a number line, where it takes 3 steps of 2 units to go from 6 to 0. Teaching the concept of division this way is just the inverse of what we have done for multiplication. ...


15

If the expression were, say, $48\div 4\times 12$, there would not be much disagreement (multiplication and division are performed from left to right). But an expression such as $48\div 4(12)$ results in disagreement because the parentheses could mean one of two different things: a way of grouping or a way of multiplying. If one interprets the parentheses ...


14

You can use (basically) Newton's method. Take $x_0>0$, and define $$x_{k+1} = \frac{1}{2} (x_k + \frac{n}{x_k}).$$ The seqeunce converges to $\sqrt{n}$. And if one starts with $x_0= n$ one has, as soon as $|x_{k+1} - x_k|< 1$, that $\lfloor x_{k+1} \rfloor = \lfloor\sqrt{n} \rfloor$. But the above is not very feasible for computing by hand, and ...


13

A reason why this form might be preferred is the way one says it: $5 \times 3$ is read out "five times three" so it says take $3$ five times, hence it "is" $3+ 3+ 3 + 3 + 3$. However I doubt there is any real standard. For what it's worth Wikipedia disagrees with itself. On the page on Multiplication it has $a \times b$ as $b + \dots + b$. On the page ...


12

My preferred model of multiplication with integers involves motion. Imagine you are recording a video of a car driving at a certain velocity. If the car is going forward, the velocity is positive; if the car is going backward, the velocity is backward. Now when you play the video, you can choose whether to play it at normal speed, or at 2x, 3x or 4x ...


11

One place in math where this issue actually does come up is in defining ordinal multiplication. From an ordinal perspective, the ordinal $5$ is the order type $a<b<c<d<e$, the ordinal $3$ is the order type $x<y<z$, and $5 \times 3$ is $$a_x < b_x < c_x < d_x < e_x < a_y < b_y < c_y < d_y < e_y < a_z < b_z &...


11

Definition of Division For every real number a and every nonzero real number b, the quotient a$\div$b, or $\dfrac{a}{b}$, is defined by: $$a\div b=a \cdot \frac{1}{b}.$$ Dividing by zero would mean multiplying by the reciprocal of 0. But 0 has no reciprocal (because 0 times any number is 0, not 1.) Therefore, division by 0 has no ...


11

Dividing $1$ disk into $\frac{1}{n}$-ths ($\frac{1}{3}, \ldots$), leads to $\frac{1}{1/n} = n$ pieces ($3,\ldots$): As $\frac{1}{n}$ approaches $0$, the number of pieces $n$ grows without bound. The result upon division by $0$, $1/0$, should be this limit. But there is no limit.


11

One way to see if the student understands the commutative property of addition is to have "fill-in-the-blank" questions such as $$2+3=3+\_\_$$ $$2+3=\_\_+2$$ $$2+\_\_=3+2$$ $$\_\_+3=3+2$$


10

My immediate response is 'wait a few years'. I've spent a fair amount of time with 3 year olds, and most of them are busy learning how to be a person in their own right, how to have a conversation, what the difference is between real and make-believe, and (often) how to tell when they need the toilet. I've read that they can't understand metaphors by that ...


10

Perhaps it is worth pointing out that every programming language defines an operator precedence structure to avoid ambiguities. An example table for C and C++ can be found here. Ambiguities must be avoided in order for the language parser to create the correct compiled (or interpreted) machine code to implement the expression. For example, the expression $4+...


9

There is an algorithm to calculate square roots with paper and pencil (like the long division algorithm). See How to calculate a square root without a calculator.


9

it sounds like the textbook has some poorly written and vague questions. there is no reason why either of the answers would be "better" for the first example. It is simply dependent on how you interpret the words. You would be surprised how many textbooks actually include similar vagueness in some questions and it is concerning to say the least.... keep up ...


9

The problem is due to imprecise specification of the intended result. Here's a more precise way. $\text{Recall that the }{\bf commutative\ law}\ \color{#c00}X+ \color{#0a0}Y = Y + X\ \text{ is true for all reals } X,Y$. $\text{Use the above law to $ $ simplify }\ 2\, +\, \color{#c00}{\pi}\, +\, \color{#0a0}3\ \text{ to the form }\, n + \pi\,\text{ for ...


8

One way this is sometimes done is with factor diamonds. (Google or google image: factor diamond method. Here is a sample result.) In your question, you gave the example of $a \times b = 2$ and $a + b = 3$. One could present this as the following puzzle: I made this image with the goal of including relevant vocab (product and sum) and I purposefully ...


8

Having recently covered this topic in a course for pre-service elementary school teachers, I thought I would write a bit about the somewhat subtle difficulties entailed in tackling this question. I am going to use language that may be at the level of undergraduate majors in mathematics or mathematics education, but I believe that the content can be scaled so ...


8

To build on other answers, you might show how other conventions exist. Use an H.P. calculator for example (postfix), the LISP family of languages (prefix), and the APL language (all right-associative), all of which do not have differing precedence of operators at all, and write expressions in different ways. Given 4 parallel translations of the same ...


8

Here are the two previous pages from those materials (a pre-publication version found with a Google search): And here is the page containing the homework problem in question: The intent now seems pretty clear. Students know that you can join groups by adding them, but in the case where the groups are equal in size (e.g. five bags, each with nine goldfish), ...


8

I am with you on this one. I feel like concatenation (implied multiplication) is of higher precedence than explicit division. For me $8:2x$ means "8 divided by 2 x'es" - $8:(2x)$ not $4x$. Replacing $x$ with $(2+2)$ shouldn't change anything. But the formal answer is that it's undefined. There is no C for concatenation in PEMDAS. For me it should be PECMDAS....


7

Here is my answer to high school underclassmen. If we let $\frac10=n$ for any $n$, we then get $$\frac10 \cdot 0 = n \cdot 0$$ $$1=0$$ This works for any non-zero value divided by zero. Allowing any value here leads to a contradiction. If we let $\frac00=n$ we get $$\frac00 \cdot 0 = n \cdot 0$$ $$0=0$$ This works for any value of $n$, so which value of $...


7

I understand this is not a realistic suggestion, but can you avoid "teaching" "PEMDAS" or "BOMDAS" altogether, and teach your students just the math instead? As pretty much everybody already said, this is not actually a rule -- this is a mnemonic device that's supposed to help students remember the actual rules of the order of operations (in the traditional ...


7

You might explain that BEDMAS is not the whole story when it comes to the order of operations. There is an operation called negation. It reverses the sign on numerical quantifies. It gives the additive inverse of any number, i.e. for all $x \in R$, we have $x + (-x) = 0$. Unfortunately, most textbooks use the same symbol for both subtraction and negation. ...


7

I'm going to answer with something of a polemical frame challenge: FOIL is evil, and probably shouldn't even be taught. Okay... that's a bit extreme. How's this: FOIL is a mnemonic that is, in my opinion, not all that useful, and should not be taught. In my own experience teaching college algebra and precalculus courses, students come to rely on FOIL ...


7

Now, my question - for those who agree with me, what is it about PEMDAS that misses this issue, that the number right before the parentheses multiplies the contents with a higher priority than the division to its left? How do we address that priority? The conventional order-of-operations in textbook math (or any math) simply don't prioritize ...


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