48÷2(9+3) = ???

Originally Posted by DeadsetAce

3390182310_f86c82cb95.jpg
roll.gif
@ 17 pages.
 
Originally Posted by OptimusADL

Originally Posted by Boys Noize

Originally Posted by usainboltisfast

you are assuming the problem is 48 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.
There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated. 

as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.

and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))

Hopefully that clears it up once and for all.
 
Originally Posted by OptimusADL

Originally Posted by Boys Noize

Originally Posted by usainboltisfast

you are assuming the problem is 48 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.
There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated. 

as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.

and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))

Hopefully that clears it up once and for all.
 
Originally Posted by do work son

Originally Posted by balloonoboy

Originally Posted by DeadsetAce

Originally Posted by balloonoboy

That's not necessarily true. Distribution is only making a(b+c) = ab + ac. Not going a step further and finding the sum.
laugh.gif


i dont know what else to tell you. you're reading the chart too literally. the only reason it doesnt go further is because there is nothing left to do in the example they provided. if there was another part of the equation in the example, you would combine before moving forward.
Word?

Taken from Wikipedia:


Given a set S and two binary operations · and + on S, we say that the operation ·

is left-distributive over + if, given any elements x, y, and z of S,

x · (y + z) = (x · y) + (x · z);



is right-distributive over + if, given any elements x, y, and z of S:

(y + z) · x = (y · x) + (z · x);



is distributive over + if it is both left- and right-distributive.

Notice that when · is commutative, then the three above conditions are logically equivalent.


They didn't combine anything or leave them in parenthesis.
nerd.gif

wait wait wait, you didn't just quote wikipedia did you?

Spoiler [+]
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.
 
Originally Posted by do work son

Originally Posted by balloonoboy

Originally Posted by DeadsetAce

Originally Posted by balloonoboy

That's not necessarily true. Distribution is only making a(b+c) = ab + ac. Not going a step further and finding the sum.
laugh.gif


i dont know what else to tell you. you're reading the chart too literally. the only reason it doesnt go further is because there is nothing left to do in the example they provided. if there was another part of the equation in the example, you would combine before moving forward.
Word?

Taken from Wikipedia:


Given a set S and two binary operations · and + on S, we say that the operation ·

is left-distributive over + if, given any elements x, y, and z of S,

x · (y + z) = (x · y) + (x · z);



is right-distributive over + if, given any elements x, y, and z of S:

(y + z) · x = (y · x) + (z · x);



is distributive over + if it is both left- and right-distributive.

Notice that when · is commutative, then the three above conditions are logically equivalent.


They didn't combine anything or leave them in parenthesis.
nerd.gif

wait wait wait, you didn't just quote wikipedia did you?

Spoiler [+]
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.
 
Originally Posted by usainboltisfast

Originally Posted by K2theAblaM

Originally Posted by eddiehouse5

The answer is 2.

BB was going wild with this, Neogaf is also.

Also- http://answers.yahoo.com/question/index?qid=20110408055505AA0F9In

http://www.neogaf.com/forum/showpost.php?p=26993361&postcount=1030

READ THIS

The distributive property of multiplication CLEARLY states that the 2(9+3) is an entire statement and CANNOT be broken up.
Jesus Christ people get your math game up.

Answer is 2
you are assuming the problem is 488 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.


No man. I'm not assuming anything. I got the problem from the same source you did, and applied the above priciple like I was taught.

2(9+3) is an entire statement that can't be broken up. after this is calculated, the PEDMAS applies.
 
Originally Posted by usainboltisfast

Originally Posted by K2theAblaM

Originally Posted by eddiehouse5

The answer is 2.

BB was going wild with this, Neogaf is also.

Also- http://answers.yahoo.com/question/index?qid=20110408055505AA0F9In

http://www.neogaf.com/forum/showpost.php?p=26993361&postcount=1030

READ THIS

The distributive property of multiplication CLEARLY states that the 2(9+3) is an entire statement and CANNOT be broken up.
Jesus Christ people get your math game up.

Answer is 2
you are assuming the problem is 488 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.


No man. I'm not assuming anything. I got the problem from the same source you did, and applied the above priciple like I was taught.

2(9+3) is an entire statement that can't be broken up. after this is calculated, the PEDMAS applies.
 
Originally Posted by OptimusADL

Originally Posted by Boys Noize

Originally Posted by usainboltisfast

you are assuming the problem is 48 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.
There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
No, you can't distribute a fraction, man. What do you mean without parenthesis?!? There are parenthesis right there... hence the distributive property.
 
Originally Posted by Boys Noize

Originally Posted by OptimusADL

Originally Posted by Boys Noize

There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated. 

as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.

and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))

Hopefully that clears it up once and for all.


48 / 2(9 + 3)
 IS NOT the original equation...

48÷2(9+3) IS the original equation.

-waystinthyme

  
 
Originally Posted by Boys Noize

Originally Posted by OptimusADL

Originally Posted by Boys Noize

There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated. 

as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.

and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))

Hopefully that clears it up once and for all.


48 / 2(9 + 3)
 IS NOT the original equation...

48÷2(9+3) IS the original equation.

-waystinthyme

  
 
Originally Posted by OptimusADL

Originally Posted by Boys Noize

Originally Posted by usainboltisfast

you are assuming the problem is 48 divided by 2 times the solution of 9 and 3. It is actually 48 divided by 2 times the solution of 9+3 the way it is written in the calculator.
There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
No, you can't distribute a fraction, man. What do you mean without parenthesis?!? There are parenthesis right there... hence the distributive property.
 
Originally Posted by balloonoboy

Originally Posted by do work son

Originally Posted by balloonoboy

Word?

Taken from Wikipedia:


Given a set S and two binary operations · and + on S, we say that the operation ·

is left-distributive over + if, given any elements x, y, and z of S,

x · (y + z) = (x · y) + (x · z);



is right-distributive over + if, given any elements x, y, and z of S:

(y + z) · x = (y · x) + (z · x);



is distributive over + if it is both left- and right-distributive.

Notice that when · is commutative, then the three above conditions are logically equivalent.


They didn't combine anything or leave them in parenthesis.
nerd.gif

wait wait wait, you didn't just quote wikipedia did you?

Spoiler [+]
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.
after you distribute, you combine the like terms derived from the distribution. then you continue onwards.
 
Originally Posted by balloonoboy

Originally Posted by do work son

Originally Posted by balloonoboy

Word?

Taken from Wikipedia:


Given a set S and two binary operations · and + on S, we say that the operation ·

is left-distributive over + if, given any elements x, y, and z of S,

x · (y + z) = (x · y) + (x · z);



is right-distributive over + if, given any elements x, y, and z of S:

(y + z) · x = (y · x) + (z · x);



is distributive over + if it is both left- and right-distributive.

Notice that when · is commutative, then the three above conditions are logically equivalent.


They didn't combine anything or leave them in parenthesis.
nerd.gif

wait wait wait, you didn't just quote wikipedia did you?

Spoiler [+]
Yea but it agrees with the chart I posted and other distributive property rules. Take that %+%@ somewhere else. No where does it state you HAVE to resolve what's in the parenthesis. Only that you have to distribute.
after you distribute, you combine the like terms derived from the distribution. then you continue onwards.
 
Originally Posted by K2theAblaM

Originally Posted by OptimusADL

Originally Posted by Boys Noize

There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
No, you can't distribute a fraction, man. What do you mean without parenthesis?!? There are parenthesis right there... hence the distributive property.

is this real life?

a fraction is just another way to write a number...

so if 2, was written as 4/2 instead...you couldn't distribute it?

please tell me you're trollin' right now...

-waystinthyme
  
 
Originally Posted by waystinthyme

Originally Posted by Boys Noize

Originally Posted by OptimusADL

So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated. 

as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.

and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))

Hopefully that clears it up once and for all.


48 / 2(9 + 3)
 IS NOT the original equation...

48÷2(9+3) IS the original equation.

-waystinthyme

  
/ and Ã· are the same thing.
 
Originally Posted by waystinthyme

Originally Posted by Boys Noize

Originally Posted by OptimusADL

So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
You can. But as the question is written, you can't.
If it were written (48 / 2)(9 + 3) it would be 288. You can not make that assumption because it was not explicitly stated. 

as it stands, 48 / 2(9 + 3) MUST be solved by distributing the 2 amongst (9 + 3) first.

and again 2(9 + 3) is equivalent to (2 * 1(9 + 3))

Hopefully that clears it up once and for all.


48 / 2(9 + 3)
 IS NOT the original equation...

48÷2(9+3) IS the original equation.

-waystinthyme

  
/ and Ã· are the same thing.
 
Originally Posted by K2theAblaM

Originally Posted by OptimusADL

Originally Posted by Boys Noize

There is no assumption. That is just how you solve the problem if you follow the distributive property. If you ignore the distributive property, you can get 288.
So you saying you cant distribute a fraction? Without parentheses you cant assume just 2 is the distribution. If you distribute 48/2 you get 288.
No, you can't distribute a fraction, man. What do you mean without parenthesis?!? There are parenthesis right there... hence the distributive property.

is this real life?

a fraction is just another way to write a number...

so if 2, was written as 4/2 instead...you couldn't distribute it?

please tell me you're trollin' right now...

-waystinthyme
  
 
Back
Top Bottom