Factor of 0 in an equation?

Hi,

I have a general maths question, it's probably grade 8 level but I can't remember the answer to it.

If you have an equation like:

1(a - b) = 2(a - b)

and a = b, when you perform the operation in hte brackets on both sides of the equation, do you need to keep the remaining 0 from a - b on both sides to end up with:

1 x 0 = 2 x 0?


This is a stupid thing I came across on another forum, and my answer is that the equation is flawed because you have to assume that 1 = 2 for it to be correct, however if a = b, you can do it because 1 x 0 = 2 x 0 both sides = 0. I just am not sure on whether you have to keep the 0 factor or not. I assume you do?

what do you mean keep it? a=b is a solution to the equation, so it is not unsatisfiable if it is that you mean by flawed, and you do not have to assume that 1=2.

no, because you're dividing (a-b) by (a-b), which gives 1, not 0.

Quote:

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Originally Posted by kedaman

what do you mean keep it? a=b is a solution to the equation, so it is not unsatisfiable if it is that you mean by flawed, and you do not have to assume that 1=2.

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yes, you don't have to assume 1 = 2, in fact you can't really. i'm just talking about basic high school maths here so don't go off on a tangent where 1 can = 2 because I won't understand a bit of it.

if you actually meant you don't have to assume a = b and the above was a typo, then i am specifically and only talking about when a = b. any other possibilities for a and b i don't want to know about because it changes the outcome greatly. you must end up with a factor of 0 after performing the sum in the brackets because that's what this thread is about. the 0 you get.


and how do you get a = b for a solution? can you show working?

Quote:

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Originally Posted by Catfish

Hi,

I have a general maths question, it's probably grade 8 level but I can't remember the answer to it.

If you have an equation like:

1(a - b) = 2(a - b)

and a = b, ...

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No "and" about it.

Your equation, 1(a-b) = 2(a-b)
IS ALSO the same as a=b.

ie..

1(a-b) = 2(a-b) ==>
0 = 2(a-b)-1(a-b) ==>
0 = 1(a-b) ==>
0 = a - b ==>
b = a

which is the same thing as a = b.

Quote:

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Originally Posted by shaunyboy

no, because you're dividing (a-b) by (a-b), which gives 1, not 0.

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Ok, this has never dawned on me before. That x / x actually = 1 not 0. Right this has blown everything else out of the water and the problem above is not even an equation. Although I already knew it wasn't an equation, ending up with 1 after the division just eliminates other theories on how to do this.

This was the original post I saw:

a = b
2a + b = a + 2b
a - b = 2a - 2b
1( a - b ) = 2( a - b )
1 = 2

Explain what went wrong.

Everyone said between line 4 and 5 a division of 0 takes place and that this is not possible. However I tried it assuming that a did not equal b and you would have no division of 0 but still get a false answer. So I knew it was not division by 0's fault and now I see there IS NO division of 0 involved. My idea was that if there was a division of 0 then the real answer in that case would be 0 = 0 which would be true and the only way to solve the equation. But that's not the case.


Thanks for the help!


Edit: Wait a minute, rules of precedent state that you have to perform the operation inside the brackets before division and multiplication. therefore, you do end up with a factor of 0 and not 1. if you divide (a - b) on both sides to reove the bracketed terms, you are breaking the rules.

You have to do a - b on both sides first, which gives you 0 when a = b. So the problem remains.

If you get to this point:

1 x 0 = 2 x 0

obiously you can't divide both sides by 0, so would that be the answer to the equation or would you continue down the rules of precedent and multiply both sides by 0 to get an answer of 0 = 0 ?

lol, think about the question.

a=b

therefore a - b = 0

you can't divide by 0...

therefore the working is all wrong, and you can't arrive at the answer 1 = 2.

Quote:

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a = b
2a + b = a + 2b

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Exactly How do you go from
a = b
to
2a + b = a + 2b.

That is, using mathematical operations, not Logical Arguments.

The 5 lines were just what someone else posted. the first 3 are just to show that a = b.

But I just wanted to know what you think the final answer of the "equation" (line 4) is if a = b.

Am I correct in saying that it would be 0 = 0?

Quote:

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Originally Posted by NotLKH

Exactly How do you go from
a = b
to
2a + b = a + 2b.

That is, using mathematical operations, not Logical Arguments.

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add a+b to both sides?

Quote:

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Originally Posted by Catfish

The 5 lines were just what someone else posted. the first 3 are just to show that a = b.

But I just wanted to know what you think the final answer of the "equation" (line 4) is if a = b.

Am I correct in saying that it would be 0 = 0?

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line 4 is a tautology if a=b, but so are all other lines except line 5. This is essential for sound proof, i.e. that if you assume the premises (a=b), then the conclusion (any other line) should follow by necessity.

Quote:

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Originally Posted by kedaman

add a+b to both sides?

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:blush:

Now WHAT was I thinking!
:p

Quote:

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Originally Posted by kedaman

line 4 is a tautology if a=b, but so are all other lines except line 5. This is essential for sound proof, i.e. that if you assume the premises (a=b), then the conclusion (any other line) should follow by necessity.

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i dont understand what he/she means.

Quote:

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Originally Posted by Catfish

i dont understand what he/she means.

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they all mean the same thing.

i really don't understand why you find this such a difficult idea to comprehend.

a = b
2a + b = a + 2b
a - b = 2a - 2b
1( a - b ) = 2( a - b )
0 = 0

all numbers multiplied by 0 equal 0. This is all that that math above proves.

I knew this was one of those 1 equals something other than 1 equations/proofs.

The fact is that division by 0 is undefined. So while x/x = 1, it is only valid for for x not equal to zero
x != 0
x <> 0

So that division operation in the middle of the proof is undefined.

Quote:

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Originally Posted by Catfish

i dont understand what he/she means.

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What I mean is
If X follows by necessity of Y then X is true whenever Y is true

If A=B is true then 2a + b = a + 2b is also true.
If A=B is true then a - b = 2a - 2b is also true
If A=B is true then 2a + b = a + 2b is also true
If A=B is true then 1( a - b ) = 2( a - b ) is also true
If A=B is true then 0=0 is also true

now dividing both sides with (a-b) is only possible if a-b is not 0.

If A=B is true then 1=2 IF A=/=B

so thus we have proven that

1=2 IF A=/=B

but not that 1=2.

Yeah I was never trying to say that 1 = 2.

For some reason people can't understand I am never saying that 1 = 2. All I'm saying is that with the equation 1 (a - b) = 2 (a - b) if a = b then the simplified form of the equation is 0 = 0, NOT 1 = 2. Every forum I've had this on people seem to think I'm saying "you can do the equation and 1 = 2". Or maybe I read what they say as them saying I'm saying that. <-- if you get what I said.

If a != b like you said above, what is the problem we come across? Is it the assumption that 1x = 2x which would not literally be possible?

Well, I didn't assume you said so either, only explained why it isn't the case.
1x = 2x is not unsatisfiable, because we know 1x=2x when x=0.
The problem lies in that we divide both sides with x without taking into account that x cannot be 0.