# BODMAS question

#### mrtwhs

$10 \div 2 \times 5$

Do the division first and you get 25. Do the multiplication first and you get 1.

#### weirddave

$10 \div 2 \times 5$

Do the division first and you get 25. Do the multiplication first and you get 1.
To do the multiplication first, you have to switch them around, so you do 10 x 5 then /2 to get 25.
Another way to do it is convert the divide into a multiply:
$10 \div 2 \times 5$
is the same as
$10 \times 0.5 \times 5$

An unbroken string of multiplies and divides all act on the leading number
for example:
$10 \div 2 \times 5 \div 7 \div 4 \times 49$
I can do these in any order I like, so I'm gonna start with:
$10 \times 49$ to give me 490. Then I'm going to:
$490 \div 7$ to give me 70. Next:
$70 \div 2$ to give me 35. Next:
$35 \times 5$ to give me 175. Next:
$175 \div 4$ to give me 175/4 or 43.75 if you prefer.

The same is true of addition/subtraction
for example, this is an unbroken string of additions and subtractions:
$1 + 1 - 1 - 1$
They all act on the leading number, so I can do these in reverse if I wish and still get 0 as the answer.

#### weirddave

If we replace the the leading 10 in my example, with 1 x 10, we can then start with the $\div 7$
Let's start with that then do the rest of them in reverse order, the new order is:
$1 \div 7 \times 49 \div 4 \times 5 \div 2 \times 10$
The result is still 43.75

#### weirddave

So now, going back to this:
$10 \div 2 \times 5$

Do the division first and you get 25. Do the multiplication first and you get 1.
The problem is that the multiplication isn't really $2 \times 5$, the symbol to the left matters! There are 2 ways to proceed if you really do want to evaluate those two numbers first.

Method 1:
we don't care about the 10 at the start of the problem, let's replace that with N
we now have:
$N \div 2 \times 5$
N is the same a N x 1, so we can say:
$N \times 1\div 2 \times 5$
So the bit we're interested in is
$1\div 2 \times 5$
This is why taking the reciprocal and changing the divide to a multiply works....

Method 2:
Using the same starting point:
$N \div 2 \times 5$
We can change the order to be:
$N \times 5 \div 2$
So the bit we're interested in is
$5 \div 2$

Note that these methods do actually result in doing a divide first.....
Having gone through this, I can see why people are taught to do the 'D' before the 'M', if you can't rearrange correctly you'll get it wrong.

#### skeeter

Math Team
change all division operations to what they really are, multiplication by the reciprocal ...

$10 \div 2 \times 5 \div 7 \div 4 \times 49$

$10 \times \dfrac{1}{2} \times 5 \times \dfrac{1}{7} \times \dfrac{1}{4} \times 49$

... and since multiplication is commutative, order doesn't matter

#### skipjack

Forum Staff
The problem is that the multiplication isn't really $2 \times 5$, the symbol to the left matters!
That's just a convention. Possible interpretations of 10 ÷ 2 × 5 are (10 ÷ 2) × 5 and 10 ÷ (2 × 5). There is a convention that the first of those possibilities is used, but that convention is separate from BODMAS.

#### weirddave

Can you give an example where that convention doesn't give the BODMAS result?

#### skipjack

Forum Staff
Possible interpretations of 10 - 2 + 5 are (10 - 2) + 5 and 10 - (2 + 5). BODMAS doesn't tell you which to use, but the above convention means that the first possibility is used.

#### weirddave

Ah, but as I suggested, the sign to the left matters, so the second 'option' isn't 2 + 5, it's -2 +5, so you can still compute unbroken strings of additions and subtractions in any order you like.
I was thinking more along the lines of:
$-x^2$
But I think this works too... (because it's really $-1 \times x^2$)

#### skipjack

Forum Staff
Your examples aren't matching the BODMAS result. BODMAS doesn't tell you how to resolve these cases, so there isn't a BODMAS result to match. Also, people shouldn't be taught to do the 'D' before the 'M', as a simpler convention suffices.

weirddave