4
$\begingroup$

I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.

This was my attempt:

Here's how this question works. To motivate what I'll be doing, consider \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.} \end{equation*} This is because when 5 is divided by 3, 3 goes into 5 once (hence the $1$ term) and there is a remainder of $2$ (hence the $\dfrac{2}{3}$ term). Note the following: every division problem can be decomposed into an integer (the $1$ in this case) plus a fraction, with the denominator being what you divide by (the $3$ in this case).

So, when $n$ is divided by 14, the remainder is 10. This can be written as \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14} \end{equation*} where $a$ is an integer.

We want to find the remainder when $n$ is divided by 7, which I'll call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,} \end{equation*} where $b$ is an integer.

Here's the key point to notice: notice that \begin{equation*} \dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.} \end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$.

Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) = 2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) = 2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) + \dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a + 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.

To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?

$\endgroup$
  • 6
    $\begingroup$ $$ n = 14 w + 10 = 7 (2w) + 10 = 7 (2w+1) +3 $$ $\endgroup$ – Will Jagy Aug 31 '14 at 20:14
  • $\begingroup$ When $n$ is divided by $10$ the remainder is $7$ (or $8$). What is the remainder when $n$ is divided by $5$? Undoubtedly the person has intuition plenty to deal with thst, and it might be a helpful beginning. $\endgroup$ – André Nicolas Aug 31 '14 at 20:20
8
$\begingroup$

I wouldn't use fractions, instead use the usual division algorithm, note that every $7$ numbers, there is a multiple of $7$, ever $14$ a multiple of $14$, et cetera to motivate writing a number as

$$n=14q+r$$

with $0\le r < 14$ each time. Then say every number is also of the form

$$n=7q'+r'$$

with $0\le r'< 7$ and emphasize that clearly $r$ is unique. This is, of course, because you just count how many up you have to go from the nearest multiple of $7$, if you are $4$ more, then you are clearly not $3$ more.

If you like visuals you can demonstrate to the student with a simple list

$$\underbrace{\color{red}{0}}_{7\cdot 0},1,2,3,4,5,6,\underbrace{\color{red}{7}}_{7\cdot 1}, 8, 9, 10, 11, 12, 13, \underbrace{\color{red}{14}}_{7\cdot 2},\ldots$$

If the student knows enough about well-ordering, you can make this rigorous rather than simply intuitive since you can look at natural numbers of the form

$$\{n-7k: k\in\Bbb Z\}$$

and just define $r$ to be the minimal element of this set.

From either approach, you can write

$$14q+10=7(2q+1)+3$$

so that $q'=2q+1$ and $r'=3$.


Addendum: If you want to emphasize how things are evenly space for the other remainders, you can make the same list with different highlighting, here I'll do $14$ and highlight the related $7$ information

$$\underbrace{\color{red}{0}}_{14\cdot 0},1,2,\underbrace{\color{orange}{3}}_{7\cdot 1+3},4,5,6,7,8,9,\underbrace{\color{blue}{10}}_{14\cdot 0+10=7\cdot 1+3},11,12,13,\underbrace{\color{red}{14}}_{14\cdot 1},$$ $$15,16,\underbrace{\color{orange}{17}}_{7\cdot 2+3},18,19, 20,21,22,23,\underbrace{\color{blue}{24}}_{14\cdot 1+10=7\cdot 3+3},25,26,27,\underbrace{\color{red}{28}}_{14\cdot 2},$$ $$29,30,\underbrace{\color{orange}{31}}_{7\cdot 4+3},32,33,34,35,36,37,\underbrace{\color{blue}{38}}_{14\cdot 2+10=7\cdot 5+3},\ldots$$

This illustrates exactly how the $7q'+3$ numbers are distributed, and it's easy to see how they overlap with the $14q+10$ numbers every other one.

$\endgroup$
  • $\begingroup$ This approach makes good use of "algebra" in explanation. $\endgroup$ – hardmath Aug 31 '14 at 20:22
3
$\begingroup$

To motivate this to students who are just beginning to learn about remainders, it often helps to use real-world examples, e.g. if changing $\,n\,$ pennies (1 cent coins) into dimes (10 cent coins) leaves 8 cents, then changing these $\,n\,$ pennies into nickels (5 cent coins) leaves 3 cents, because we can first change them to dimes, then change each dime to 2 nickels, then change the remaining 8 pennies to 1 nickel and 3 pennies. Algebraically $$\ n = 10q + 8\, =\, 5(2q) + 5 + 3\, =\, 5(2q+1) + 3$$

In modular language, $\ {\rm mod}\,\ \color{}5\!:\ \begin{array}{r}\color{#c00}{10\equiv 0}\\ \color{#0a0}{8\equiv 3}\end{array}\,\ \Rightarrow\ \begin{array}{r}\color{#c00}{10}q+\color{#0a0}8\\ \,\equiv\ \color{#c00}0\,+\,\color{#0a0}3\end{array}\ $ by $ $ Congruence Sum, Product Rules.

Most students can easily visualize the money-changing. The goal is to help them translate that intuition into rigorous algebra, or, more powerfully, congruence algebra - as above.

$\endgroup$
2
$\begingroup$

Colloquially (as I find sometimes works with my daughters at high school)

When we knock out $14$s the remainder is $10$. We knock out $7$s - well the $14$s all go because $14 = 2\times 7$ and we're left with $10=7+3$.

$\endgroup$
  • $\begingroup$ Of course use this to motivate general algebraic ideas ... that's the joy of teaching (today's lesson was exponential functions from scratch) $\endgroup$ – Mark Bennet Aug 31 '14 at 20:26
1
$\begingroup$

$n=14k+10=7\cdot2\cdot k+7+3=(2k+1)7+3$, the remainder is $3$

$\endgroup$
0
$\begingroup$

We have $n=14a+10$ for some integer $a$. If we take this equation modulo $7$ we have that $n \equiv 10 \mod 7 \equiv 3 \mod 7$ (This because $14 \equiv 0 \mod 7$ and $10 \equiv 3 \mod 7$). You should look up modular arithmetic where you will see how to deal with this sort of questions. Modular arithmetic is very important in much branches of mathematics mostly in structures that have a finite number of elements in them (finite groups, rings, fields,...).

$\endgroup$
0
$\begingroup$

It might help to go back to the basic definition of multiplication as repeated addition:

$$2 \cdot 5 = 2 + 2 + 2 + 2 + 2 = 10$$

and that division is the opposite of multiplication:

$$\frac{10}{2} \mapsto 10 \underbrace{{}- 2 - 2 - 2 - 2 - 2}_{\text{5 times}} = 0.$$ So $\frac{10}{2}=5$, remainder $0$

Also, for $\frac{11}{3} = \rightarrow 11 \underbrace{{}- 3 - 3 - 3}_{\text{3 times}} = 2$

So $\frac{11}{3}=3$, remainder $2=3+\frac{2}{3}$

If $\frac{n}{14}=a+\frac{10}{14}$, that means $n\underbrace{{}-14-14-14-\cdots}_{a\text{ times}}=10$

Here's the simple/intuitive part:

Since $2(7)=14$, division by 7 will also have a "remainder" of 10
But 10 > 7, so we can subtract out one more 7: $10 - 7 = 3$.
Therefore, the remainder ($r$) is 3: $r=3$

(Also note, $b=2a+1$. Because $14=2(7)$, we double $a$.
Because we subtracted one more 7, we add 1.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.