乘正数

Multiplication with Non-Negative Integers

One of the most important multiplication facts is that when you multiply two positive numbers together, the result will always be positive. When you multiply any number by $0,$ the product will be $0.$ In other words, we'll be learning how to multiply whole numbers together!

Multiplication Review

We already learned that multiplication with one-digit numbers is the same as adding a number to itself several times. E.g. $2 \times 5 = 2 + 2 + 2 + 2 + 2$.

The same rule applies for two-digit numbers, but that can be a bit tedious to work out. E.g. $12 \times 13 = 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12 + 12+12+12+12$.

Exhausting! A better way is to re-use what we know about multiplying one-digit numbers to our advantage.

The Steps

Let's say we wanted to find $12 \times 13.$ We know the product will be positive, because $12$ and $13$ are positive.

The first thing to do is line up the numbers vertically, like this:

$$\begin{align} &12\\ \times \phantom{0} &13\\ \hline \end{align}$$

Then multiply the top number by the ones digit of the bottom number:

$$\begin{align} \boxed{12}&\\ \times \phantom{0} 1\boxed{3}&\\ \hline \boxed{36}& \end{align}$$

Next, drop down a $0$ underneath the ones digits like this:

$$\begin{align} &12\\ \times \phantom{0} &13\\ \hline &36\\ &\phantom{1}0 \end{align}$$

Now, multiply the top number by the tens digit of the bottom number.

$$\begin{align} \boxed{12}&\\ \times \phantom{0}\boxed{1}3&\\ \hline 3\phantom{|}6&\\ \phantom{|}\boxed{12}0& \end{align}$$

Lastly, just add the two products you just found!

$$\begin{align} 12&\\ \times \phantom{0} 13&\\ \hline \boxed{36}&\\ + \phantom{0} \boxed{120}&\\ \hline \boxed{156}& \end{align}$$

So, $12 \times 13=156.$ Here's a graphic with another example.

Let's do one more example!

We are trying to find $$\begin{align} &45\\ \times \phantom{0} &86\\ \hline \end{align}$$

Step One

First, we find the product of the top number and the ones digit of the bottom number. $$\begin{align} \boxed{45}&\\ \times \phantom{0} 8\boxed{6}&\\ \hline \boxed{270}& \end{align}$$

Step Two

Drop down a $0$ under the ones digits. $$\begin{align} 45&\\ \times \phantom{0} 86&\\ \hline 270&\\ \phantom{1}0& \end{align}$$

Then find the product of the top number and the tens digit of the bottom number. $$\begin{align} \boxed{45}&\\ \times \phantom{0}\boxed{8}6&\\ \hline 27\phantom{|}0&\\ \phantom{|}\boxed{360}0& \end{align}$$

Step Three

Now add the products you just found! $$\begin{align} 45&\ \times \phantom{0} 86&\ \hline \boxed{270}&\

• \phantom{0} \boxed{3600}&\ \hline \boxed{3870}& \end{align}$$

So, $45 \times 86=3870.$

(Video) Multiplying Integers

by mathman1024

Here’s a video by mathman1024 that covers multiplying positives or zero, multiplying a negative by a positive or zero, and multiplying a negative by a negative.

$\large{\text{Summary}}$

This video goes a little bit into variable substitution in order to give general forms of each case.

There are many ways we denote multiplication. A familiar one might be the $\times$. $$a\times b$$ The other ways to represent multiplication are with a raised dot, a star, or parentheses. $$\begin{align} &a\cdot b\\ &a(b)\\ &(a)(b)\\ &a*b \end{align}$$

Multiplication with two factors with different signs (one's negative and the other's positive) results in a negative product.

Multiplication with two factors with the same sign (both are positive or both are negative) will result in a positive product.

To see this let's multiply the number $5$ by different numbers: $$\begin{align} 3\cdot5&=15\\ 2\cdot5&=10\\ 1\cdot5&=5\\ 0\cdot5&=0\\ -1\cdot5&=-5\\ -2\cdot5&=-10 \end{align}$$ Now let's do the same for the number $-5$: $$\begin{align} 3\cdot-5&=-15\\ 2\cdot-5&=-10\\ 1\cdot-5&=-5\\ 0\cdot-5&=0\\ -1\cdot-5&=5\\ -2\cdot-5&=10 \end{align}$$ Notice also that when we decrease the first factor by $1$ each time, the product is changing by the same amount, $5$. This is not a coincidence. In effect we are writing out the multiples of $5$.

Let's try some questions to check our understanding.

Our product will be positive. Now let's find the product.

(Videos) Multi-Digit Multiplication

by Math Antics

Here's some videos by mathantics showing you how to do multiplication with multi-digit numbers. The first video shows you how to multiply a multi-digit number by a single digit. In the second video, he shows you how to multiply a multi-digit number by a multi-digit number.

$\large{\text{Summary}}$

With multiplication of multi-digit numbers, we break the problem down into single digit multiplication. Because of this, it's important to have your times table memorized.

The video starts by jumping in with an example. $$137\times5$$ Start by rewriting the problem in stack form. $$\begin{align} &137\\ \times\quad&\phantom{13}5\\ \hline \end{align}$$ Next we take the $5$ and multiply it by the ones place of the top number, $7$. We write this below the line. $$\begin{align} &13\color{#29A603}{7}\\ \times\quad&\phantom{13}\color{#29A603}{5}\\ \hline &\phantom{1}35 \end{align}$$ We then carry the $3$ in the tens place up top. $$\begin{align} &\phantom{1}\,\tiny{3}\\ &137\\ \times\quad&\phantom{13}5\\ \hline &\phantom{13}5 \end{align}$$ Next, we multiply $5$ with the digit in the tens place, $3$, and add the $3$ we carried up: $5\times3+3=18$. $$\begin{align} &\phantom{1}\,\tiny{3}\\ &1\color{#29A603}{3}7\\ \times\quad&\phantom{13}\color{#29A603}{5}\\ \hline &185 \end{align}$$ Carry the $1$ in the hundreds place up top. $$\begin{align} &\,\tiny{1}\,\tiny{3}\\ &137\\ \times\quad&\phantom{13}5\\ \hline &\phantom{1}85 \end{align}$$ Finally, multiply $5$ by the digit in the hundreds place $1$, and add the $1$ we carried up: $5\times1+1=6$. $$\begin{align} &\,\tiny{1}\,\tiny{3}\\ &\color{#29A603}{1}37\\ \times\quad&\phantom{13}\color{#29A603}{5}\\ \hline &685 \end{align}$$

When multiplying two multi-digit numbers, we take a similar approach with an extra step.

Let's see what we do by looking at $324\times46$. The first step is to write out the multiplication vertically. $$\begin{align} &324\\ \times\quad&\phantom{3}46\\ \hline \end{align}$$

We start with the ones place digit of the second number, $6$. We want to multiply this through the first number, $324$, just like we did in the last example. $$\begin{align} &\phantom{1}324\\ \times\quad&\phantom{13}4\color{#29A603}{6}\\ \hline &1944 \end{align}$$

Here's where we add a new step. Whenever we move a place value in the second number, we put the multiplication on the next line down. We also need to put a $0$ or an $X$ in the ones place since, in our multiplication number, we're focusing on the tens place. Let's start by setting this up. $$\begin{align} &\phantom{1}324\\ \times\quad&\phantom{13}\color{#29A603}{4}6\\ \hline &1944\\ &\phantom{194}0 \end{align}$$

We multiply our top number by $4$ now. Make sure to place this product next to the $0$ in the bottom row. $$\begin{align} &\phantom{11}324\\ \times\quad&\phantom{113}\color{#29A603}{4}6\\ \hline &\phantom{1}1944\\ &12960 \end{align}$$

Finally, we add these two products together. $$\begin{align} &\phantom{1}1944\\ +\quad&12960\\ \hline &14904 \end{align}$$