Bedmas Questions: Why does (3*5*10)/3 =/= 3*3^-1 * 5*3^-1 * 10*3^-1

[Amark]

Why does (3*5*10)/3 =/= 3*3^-1 * 5*3^-1 * 10*3^-1
or if it is in the form 3/3 * 5/3 * 10/3 why does multiplying the numerator then dividing by the denominator get the correct answer but doing the division to get 1 * 1.67 * 3.33 then multiplying, I thought Multiplying and dividing were interchangeable but must be done in order from left to right Is the numerator always before the denominator in terms of Left to right?

and another somewhat related question
why can you distribute 5(2+3) to (10+15) or finish whats inside the bracket to get 5(5) then multiply and get the same answer but if it was 5(2*3) the bracket must be done first because distributing the 5 would yield (10*15)

When simplifying equations eg moving values around are there any rules that i should follow to prevent me from making mistakes such as 16=5*3+1 i understand that i should bring the 1 over to the other side then divide by 5 to get 15/5=x
But what rule should i follow so that it will stop me from dividing 16/5 then bring the one over to get -1 +16/5 = x
I also prefer to simplify fractions in to decimals when simplifying equations, is this something i should not do as its an operation taking place before every other operation like how i simplified in my first question
to get rid of the 3 in the denominators


Thanks ahead of time for any help


EDIT: additional question: I have the equation 1/(2*_^sqrt(x²+1)) which i rewrote as 0.5(x²+1)^-1/2 I wanted to simplify by distribution the 0.5 in to the brackets resulting in
(0.5x²+0.5)^-1/2 then solving for x which yields different results than "0.5(x²+1)^{-1/2 "} due to the order of operations, Is there a Rule i can follow so I don't end up in this situation when simplifying

 

[Deleted member 4993]

Why does (3*5*10)/3 =/= 3*3^-1 * 5*3^-1 * 10*3^-1
or if it is in the form 3/3 * 5/3 * 10/3 why does multiplying the numerator then dividing by the denominator get the correct answer but doing the division to get 1 * 1.67 * 3.33 then multiplying, I thought Multiplying and dividing were interchangeable but must be done in order from left to right Is the numerator always before the denominator in terms of Left to right?

and another somewhat related question
why can you distribute 5(2+3) to (10+15) or finish whats inside the bracket to get 5(5) then multiply and get the same answer but if it was 5(2*3) the bracket must be done first because distributing the 5 would yield (10*15)

When simplifying equations eg moving values around are there any rules that i should follow to prevent me from making mistakes such as 16=5*3+1 i understand that i should bring the 1 over to the other side then divide by 5 to get 15/5=x
But what rule should i follow so that it will stop me from dividing 16/5 then bring the one over to get -1 +16/5 = x
I also prefer to simplify fractions in to decimals when simplifying equations, is this something i should not do as its an operation taking place before every other operation like how i simplified in my first question
to get rid of the 3 in the denominators


Thanks ahead of time for any help

Do you know that:

5 * (2 * 3 * 4) \(\displaystyle \displaystyle{\ne}\) (5 * 2) * (5 * 3) * (5 * 4) ?

Same logic - "multiplication" does NOT DISTRIBUTE among other multiplied factors.

 

[sinx]

Why does (3*5*10)/3 =/= 3*3^-1 * 5*3^-1 * 10*3^-1
or if it is in the form 3/3 * 5/3 * 10/3 why does multiplying the numerator then dividing by the denominator get the correct answer but doing the division to get 1 * 1.67 * 3.33 then multiplying, I thought Multiplying and dividing were interchangeable but must be done in order from left to right Is the numerator always before the denominator in terms of Left to right?

and another somewhat related question
why can you distribute 5(2+3) to (10+15) or finish whats inside the bracket to get 5(5) then multiply and get the same answer but if it was 5(2*3) the bracket must be done first because distributing the 5 would yield (10*15)

When simplifying equations eg moving values around are there any rules that i should follow to prevent me from making mistakes such as 16=5*3+1 i understand that i should bring the 1 over to the other side then divide by 5 to get 15/5=x
But what rule should i follow so that it will stop me from dividing 16/5 then bring the one over to get -1 +16/5 = x
I also prefer to simplify fractions in to decimals when simplifying equations, is this something i should not do as its an operation taking place before every other operation like how i simplified in my first question
to get rid of the 3 in the denominators


Thanks ahead of time for any help

Instead of remembering specific rules, when you are stumped, do a simple example you know the answer to.
(3*5*10)/3 =?? do a simple example,
e.g. (2*3)/3)=2, it is not 2/3, so you can't divide each number by 3 and multiply the result.
so you don't divide each multiplied term by the denominator.
however, (2+3)/3 does=2/3+3/3, obviously (2+3)/3=5/3, or 1²/₃
so you do divide each added term by the denominator.
again, you use simple examples when you are in doubt. (i do it all the time)


16=5*3+1
you can divide by 3 first if you want, but you have to divide both sides by 3, and all term(s), where "terms" are seperated by a + or-.
16/3=5*3/3+1/3
the rule here is 'you have to do the same thing to both sides of the equation.' [basic rule of algebra]

you can convert fractions into decimals and get the right answer, but you need to learn to work with fractions because there are times when you can't convert a fraction to a decimal. Also it is often simpler and more accurate to work with the fractions.
e.g. 1/6+1/3= ?
It is both simpler and more accurate to find a common denominator and add, and the answer is 3/6=1/2.
If you try to write down the decimals and add you will get 0.499........

 

[Amark]

Instead of remembering specific rules, when you are stumped, do a simple example you know the answer to.
(3*5*10)/3 =?? do a simple example,
e.g. (2*3)/3)=2, it is not 2/3, so you can't divide each number by 3 and multiply the result.
so you don't divide each multiplied term by the denominator.
however, (2+3)/3 does=2/3+3/3, obviously (2+3)/3=5/3, or 1²/₃
so you do divide each added term by the denominator.
again, you use simple examples when you are in doubt. (i do it all the time)


16=5*3+1
you can divide by 3 first if you want, but you have to divide both sides by 3, and all term(s), where "terms" are seperated by a + or-.
16/3=5*3/3+1/3
the rule here is 'you have to do the same thing to both sides of the equation.' [basic rule of algebra]

you can convert fractions into decimals and get the right answer, but you need to learn to work with fractions because there are times when you can't convert a fraction to a decimal. Also it is often simpler and more accurate to work with the fractions.
e.g. 1/6+1/3= ?
It is both simpler and more accurate to find a common denominator and add, and the answer is 3/6=1/2.
If you try to write down the decimals and add you will get 0.499........

Thanks for the tip i actually do write down simple examples to work through whenever Im unsure of something , its how I've survived this long.

"so you don't divide each multiplied term by the denominator." could you go a bit more in-depth with this, I understand it intuitively but having a formal rule to remember would help.

As for the 16=5*3+1 example i caught my error when i re-read it, But is this a rule that applies always that a term is something between +/- signs and should only be interacted as a "unit" such as x (2x+1)= 2x^2+x and not 2x * x^2+ x , the 2*x should be considered one number not two separate numbers. does this 2*x stay "bound" together while i manipulate the equation until i divide the term by 2?

 

[Denis]

Are you a student attending math classes?

 

[Amark]

Are you a student attending math classes?

Yes, I am. I want to make Ive got a solid grasp on the foundations so I don't make simple errors in the future. Why ask?

 

[Steven G]

Are you a student attending math classes?

Student and attending class do not always go together.
.

 

[Amark]

Instead of remembering specific rules, when you are stumped, do a simple example you know the answer to.
(3*5*10)/3 =?? do a simple example,
e.g. (2*3)/3)=2, it is not 2/3, so you can't divide each number by 3 and multiply the result.
so you don't divide each multiplied term by the denominator.
however, (2+3)/3 does=2/3+3/3, obviously (2+3)/3=5/3, or 1²/₃
so you do divide each added term by the denominator.
again, you use simple examples when you are in doubt. (i do it all the time)


16=5*3+1
you can divide by 3 first if you want, but you have to divide both sides by 3, and all term(s), where "terms" are seperated by a + or-.
16/3=5*3/3+1/3
the rule here is 'you have to do the same thing to both sides of the equation.' [basic rule of algebra]

you can convert fractions into decimals and get the right answer, but you need to learn to work with fractions because there are times when you can't convert a fraction to a decimal. Also it is often simpler and more accurate to work with the fractions.
e.g. 1/6+1/3= ?
It is both simpler and more accurate to find a common denominator and add, and the answer is 3/6=1/2.
If you try to write down the decimals and add you will get 0.499........

Reviving this thread since it took multiple days for my reply to appear

 

[Steven G]

Reviving this thread since it took multiple days for my reply to appear

Can you please explain where you are stuck/confused after reading the replies?

 

[JeffM]

Why does (3*5*10)/3 =/= 3*3^-1 * 5*3^-1 * 10*3^-1
or if it is in the form 3/3 * 5/3 * 10/3 why does multiplying the numerator then dividing by the denominator get the correct answer but doing the division to get 1 * 1.67 * 3.33 then multiplying, I thought Multiplying and dividing were interchangeable but must be done in order from left to right Is the numerator always before the denominator in terms of Left to right?

As you say later, we define dividing by a as equivalent to multiplying by 1/a so I am going to answer this question in terms of multiplication.

20 = 13 + 7
20 = 13 + 7
20 = 13 + 7

60 = 39 + 21 or
3 * 20 = (3 * 13) + (3 * 7). But 20 = (13 + 7) so

3 * (13 + 7) = (3 * 13) + (3 * 7).

If we consider arithmetic as an experimental science, we posit a rule that

\(\displaystyle a(b + c) = (a * b) + (a * c).\)

When we develop arithmetic axiomatically, we make that an axiom.

10 = 2 * 5
10 = 2 * 5
10 = 2 * 5

30 = 3 * 10 = 3 * (2 * 5). 30 does not equal (3 * 2) * (3 * 5) = 6 * 15 = 90.

When we consider arithmetic as an experimental science, we do not posit a rule that

\(\displaystyle a * (b * c) = (a * b) * (a * c)\)

because we have seen that rule fails. Experimenting, however, lets us posit that

\(\displaystyle a * (b * c) = (a * b) * c.\)

So we end up with three axioms dealing with operations involving three numbers.

\(\displaystyle a + (b + c) = (a + b) + c,\ a * (b * c) = (a * b) * c, \text { and } a * (b + c) = (a * b) + (a * c).\)

and another somewhat related question
why can you distribute 5(2+3) to (10+15) or finish whats inside the bracket to get 5(5) then multiply and get the same answer but if it was 5(2*3) the bracket must be done first because distributing the 5 would yield (10*15)

It's the same question and so already answered.

I also prefer to simplify fractions in to decimals when simplifying equations, is this something i should not do as its an operation taking place before every other operation like how i simplified in my first question to get rid of the 3 in the denominators.

The main reason that we avoid decimals as substitutes for fractions in algebra and calculus is that decimals are frequently wrong. The only fractions that have exact decimal equivalents are fractions the denominators of which can be expressed as a positive power of 2 or 5 or a product of positive powers of 2 and 5. All other decimal representations are inexact approximations. Furthermore, what is the decimal equivalent of x / y?

There is a much better way to simplify equations containing fractions: multiply both sides of the equation by the least common multiple of the denominators. That technique is called "clearing fractions."

EDIT: additional question: I have the equation 1/(2*_^sqrt(x²+1)) which i rewrote as 0.5(x²+1)^-1/2 I wanted to simplify by distribution the 0.5 in to the brackets resulting in
(0.5x²+0.5)^-1/2 then solving for x which yields different results than "0.5(x²+1)^{-1/2 "} due to the order of operations, Is there a Rule i can follow so I don't end up in this situation when simplifying

You are mixing up order of operations with the axioms of arithmetic. (By the way, what you are talking about is an expression.) You are operating by frequently false analogy instead by thinking. Here your preference for decimals has led you astray.

It is true that \(\displaystyle \dfrac{1}{2\sqrt{x^2 + 1}} = 0.5(x^2 + 1)^{-(1/2)}\)

but it is NOT easier to work with the latter than with the former in an algebraic context.

\(\displaystyle \dfrac{1}{2\sqrt{x^2 + 1}}= \dfrac{1}{4} \implies\\

4 = 2\sqrt{x^2 + 1} \implies 2 = \sqrt{x^2 + 1} \implies 4 = x^2 + 1 \implies \\

x^2 = 3 \implies x = \pm \sqrt{3}.\)

Let's check.

\(\displaystyle \dfrac{1}{2\sqrt{(\pm \sqrt{3})^2 + 1}} = \dfrac{1}{2\sqrt{3 + 1}} = \dfrac{1}{2\sqrt{4}} = \dfrac{1}{2 * 2} = \dfrac{1}{4}.\)

 

[sinx]

Thanks for the tip i actually do write down simple examples to work through whenever Im unsure of something , its how I've survived this long.

"so you don't divide each multiplied term by the denominator." could you go a bit more in-depth with this, I understand it intuitively but having a formal rule to remember would help.

As for the 16=5*3+1 example i caught my error when i re-read it, But is this a rule that applies always that a term is something between +/- signs and should only be interacted as a "unit" such as x (2x+1)= 2x^2+x and not 2x * x^2+ x , the 2*x should be considered one number not two separate numbers. does this 2*x stay "bound" together while i manipulate the equation until i divide the term by 2?

you don't divide each multiplied term by the denominator

what i mean is (2*4*6)/2=1*4*6;
[ not 1*2*3]
on the other hand (2+4+6)/2=1+2+3
[not 1+4+6]

It would have been more clear for me to say; 2*4*6 is one term,
but (2+4+6) is 3 terms.
then, you multiply (or divide) each term only once.

x(2x+1)= 2x^2+x and not 2x * x^2+ x , the 2*x should be considered one number not two separate numbers. does this 2*x stay "bound" together while i manipulate the equation until i divide the term by 2?

in a sense, yes.
you follow the same rules as with division. numbers or variables that are 'bound together' are one term.
e.g. In 2x, the 2 and x are multipled together (or 'bound together) as 2*x,
2x is one term
so 2(2x)=4x; and x(2x)=2x²
and your example;
x(2x+1)= 2x^2+x and not 2x*x^2+ x.

I don't know how to state the rules formally,
but you have the right idea, and thinking about this like you are will lead you to understand it better.
the more you practice the easier it gets.
just work a lot of problems, and when you get confused, use simple examples like you are already doing.
practice is the key.