BODMAS question

Feb 2010
732
157
$10 \div 2 \times 5$

Do the division first and you get 25. Do the multiplication first and you get 1.
 
Apr 2014
1,010
347
UK
$10 \div 2 \times 5$

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.
 
Apr 2014
1,010
347
UK
Additionally, we can change the leading (starting) number too.
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 :)
 
Apr 2014
1,010
347
UK
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
Jul 2011
3,292
1,776
Texas
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

order_mult.png
 

skipjack

Forum Staff
Dec 2006
21,393
2,412
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.
 
Apr 2014
1,010
347
UK
Can you give an example where that convention doesn't give the BODMAS result?
 

skipjack

Forum Staff
Dec 2006
21,393
2,412
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.
 
Apr 2014
1,010
347
UK
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
Dec 2006
21,393
2,412
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.
 
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