Here is a brief overview of what we covered so far for our arithmetic lesson last week. If you would like to access our handout, please access the following document:
PEMDAS. Recall that PEMDAS is just an acronym that helps you remember in which order operations should be performed. Note here that after we perform the “P” and the “E”, we should make sure that we perform ALL multiplication and division from left to right. Similarly, we would want to perform addition and subtraction in the same level, from left to right.
- P: Parentheses. We evaluate a certain expression in the parentheses using “EMDAS”, and then proceed evaluating the stuff outside of the parentheses.
- E: Exponents. Evaluate any exponents.
- M/D: Multiplication/Division. Perform multiplication and division from left to right.
- A/S: Addition and subtraction. Perform addition and subtraction from left to right.
A common mistake that many students make is performing all multiplications before divisions, since logically, “M” comes before “D”. However, the order of operations requires us to perform multiplication and division at the same level, from left to right, doing whichever one comes first.
Let’s take this following example:
Section 1.1 Problem 1 (handout): What is \( 1-2+0 \div 3+3 \times 7 \)?
Let’s evaluate this using our handy acronym PEMDAS:
- P: Parentheses. We don’t see any parentheses in this expression, so we can skip this step.
- E: Exponents. We don’t see any exponents, so let’s move on.
- M/D: Multiplication/Division. Be careful! We want to evaluate all multiplications and divisions from left to right, considering whichever one comes first!
Read the expression from left to right. We first encounter a division sign between the \( 0 \) and the \( 3 \), which means that we should evaluate \( 0 \div 3 \) first. That’s just zero, so the expression is now
\( 1-2 + 0 + 3 \times 7 \)
We next see a multiplication symbol between the \( 3 \) and the \( 7 \), which tells us to multiply three by seven. That’s a simple fact from our times table: \( 3 \times 7 = 21 \). Our expression is now
\( 1-2 + 0 + 21 \)
Seeing no more multiplying or dividing that need to be done, we can finally finish off with our last step:
- A/S: Addition/Subtraction. Same rule with multiplication and division, except this time we are considering subtractions and additions. By this point, you can simply evaluate the expression from left to right.
Write down the expression again:
\( 1-2+0+21 \)
\( 1-2 = -1 \). So we want
\( -1 + 0 + 21 \)
\( -1 + 0 = -1 \), so our final answer is just
\( -1 + 21 = 20. \)
I know following these steps may sound tedious, but after a little bit of practice, this process will become second-nature, and you won’t even need to write PEMDAS before every problem!
Daniel introduced three important properties of arithmetic that can speed up calculations. What you need to understand is how to apply them to problems and not the names of the properties themselves.
- Associative Property of Addition/Multiplication. For any quantities \( a,b,c \),
\[ (a \times b) \times c = a \times (b \times c) \]
\[ (a + b) + c = a + (b + c) \]
In other words, if you have a series of only multiplications or only additions, you can group them however you want.
2. Commutative Property of Addition/Multiplication. For any quantities \( a,b,c \),
\[ a \times b = b \times a \]
\[ a +b=b+a \]
If you have a series of only multiplications or only additions, you can reorder them however you want.
Remember that the associative and commutatve properties don’t work for division and subtraction! Why is that?
3. Distributive Property (Daniel’s favorite!): For any quantities \( a,b,c \):
\[ a \times (b+c) = a\times b + a \times c \]
This fact is very useful especially if you see a common number being multiplied. In algebra, we call that a common factor.
Let’s try these in action!
Section 1.1 Problem 9: Find \( 2 \cdot 3 \cdot 25 \cdot 5 \cdot 4 \cdot 10. \) (\( \cdot \) just means multiplication).
You could multiply these numbers out from left to right. And that wouldn’t be too bad.
But is there a better way? Can you make some observations?
For one, this is problem consists of just multiplications. Thus, we can re-order or regroup the numbers however we wish. Typically, multiplying by \( 10 \) is very convenient. For integers, you just add a zero after the number. For instance, \( 7 \times 10 = 70 \) and even \( 2342343298 \times 10 = 23423432980 \).
Let’s try regrouping or rearranging the numbers so that numbers multiply to a power or multiple of 10.
We notice that \( 25 \times 4 = 100 \) and a \( 10 \) is present, so we can write
\( 2 \cdot 3 \cdot 25 \cdot 5 \cdot 4 \cdot 10 = (2 \times 3 \times 5) \times (25 \times 4) \times 10 \)
\( = (2 \times 3 \times 5) \times 100 \times 10 \)
We can evaluate the product of the last two numbers to get
\( = (2 \times 3 \times 5) \times 1000 \)
Yay! We can evaluate the expression in the parentheses to get the answer of \( 30000 \).
Notice that in this problem, we could break PEMDAS a little bit because our problem was solely consisting of multiplications. We can re-order and rearrange the numbers however we want. Be flexible; sometimes re-writing an expression is necessary if you have something like \( 2 \times 5 \times 5 \times 5 \times 2 \times 2 \times 2 \times 2 \). Since \( 5 \times 2 = 10 \), we can pair up 2’s and 5’s to form products of ten, and eventually all you have left is a simpler multiplication problem with some added zeroes.
Let’s close with my favorite: what is \( 117 \times 63 + 117 \times 37 \)?
Hold up bro, don’t multiply the two products and add them up.
Notice that in both products, the number 117 is involved. Thus we can factor the 117 out using the distributive property:
\( 117 \times 63 + 117 \times 37 = 117 ( 63 + 37) = 117 ( 100 ) =11700. \)
It becomes as easy as adding two zeroes. With more practice, you’ll even be able to identify that \( 63 + 37=100 \), so having them sum up in some way will help solve the problem.
Stay curious, and keep learning math. See you tomorrow!