Order of operation logic, why should multiplication division done first?

Question

I understand how order of operations works but i didn't get it's logic why should multiplication division done first. Please give me a logic to solve an expression without mnemonics like PEMDAS BODMAS,etc.

Answer 1

Think of it this way(this might help or not) but it is one way that helps me understand:

Say I have the following expression:

$$ 5 + 4*5$$

Remember that $4*5$ is the value $5$ added together $4$ times; $5+5+5+5$. This is also restating that multiplication is repeated addition.

Now if I expanded it, I get the same operator, so no problem if I go from right to left:

Start here: $5+5+5+5+5$

Picture this on a real number line, beginning from $0$, you move $25$ steps to the right.

Now if we trace our steps back, remembering that we expanded, so we are going to condense the $4$ "$5$s" back into multiplication. But wait, in doing so where did we start?

Picture the number line again. We ignored the first $5$ and moved straight to the second, so we started $5$ units ahead in the first place. Then by condensing, we moved $20$ spots, because we already added those $4*5$s, hence we needed to finish whatever we started in the first place before moving on to the next operation. So once we moved $20$ spots, the equation becomes:

$$ 5 + 20$$

So what am I trying to say here?

When "multiplying", we have already had performed repeated addition(that was our starting point), but we stopped halfway by leaving it condensed which resulted in multiplication(or as we call it), so we need to finish the operation by removing all possible operators that come about and get a numerical value before proceeding on to the next operation.

What about the others?

Subtraction is addition in the other direction(yes numbers can be treated as vectors) and division is just repeated subtraction until the $0$ or a quotient and remainder as small as possible is obtained.

Answer 2

A priori, an expression such as $$ 1+2\cdot 3$$ is ambiguous, and one could get rid of the ambiguity by always using parentheses to express the intended order of operation, such as $$ (1+2)\cdot 3=3\cdot 3=9\qquad \text{vs.}\qquad 1+(2\cdot 3)=1+6=7.$$ Now the idea is to be lazy and agree that one can simply drop the parentheses in one of the two cases. Choosing either of the options would have been feasible, but the convention chosen is that dropping parentheses is allowed for the product.

Similarly, for $5-3-2$, could decide to agree that its disambiguited form should be $5-(3-2)=4-1=4$ instead of $(5-3)-2=2-2=0$. Convention has chosen the second path, perhaps guided by the fact that the ultimate effect of a term is determined only by its immediately preceding operator.

The common convention may turn out particularly useful when writing for example polynomials: When expressions such as $$ 3x^4-2x^3+x^2-1$$ are common, it is helpful that these do not need to be written as $$ (((3(x^4))-(2(x^3)))+(x^2))-1.$$

Answer 3

Nice question! To be completely honest, it comes down to convention. However, here's how I make sense of it. (We'll acknowledge that division is simply the inverse of multiplication.) Say you have the problem $\frac{3}{7} + 5 × \frac{2}{4} = x$. Now, following the order of operations, $x = \frac{41}{14}$. If we add $\frac{3}{7}$ to $5$ first, $x = \frac{19}{7}$. When it comes to solving problems that involve rational numbers, it's usually easiest to multiply than it is to add. Even if it's argued that it's easier to add than multiply with a problem like $13 + 5 × 13 = x$, that's only when all numbers share the same denominator, in this case, $1$. Of course, the majority of rational numbers have a denominator that is greater than $1$.

If you want it to be absolutely clear that $x = \frac{41}{14}$ is the solution, you can write $\frac{3}{7} + 5 (\frac{2}{4}) = x$. For $x = \frac{19}{7}$, you can write $\frac{2}{4}(5 + \frac{3}{7}) = x.$

This is just a way that I "rationalize" multiplying before adding. This is not the reason why multiplication is done first, though. I hope this helps!