$4=3$ .

Kimmie wants to prove thatShe starts off with the equation $a-b=c$ which is equivalent to

$\large 4a-3a-4b+3b = 4c-3c$

She moved the variables to obtain $4a-4b-4c = 3a-3b-3c$ and factorized it to

$\large 4(a-b-c) = 3(a-b-c)$

Finally, she cancelled both sides to obtain $4=3$ .

What is the error in Kimmie's work?

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I like how you used 8 words to come with a solution that explained the whole problem. Nice work!

Kevin Mo
- 7 years, 4 months ago

a-b-c=0 We can not divide by zero

ARUNA MANE
- 7 years ago

Nice job!

Valerian Pratama
- 7 years, 4 months ago

That's why they call it Maths! :D So, trivial yet so brain-wrecking!

Thanks for the problem Kevin. ;)

Kou$htav Chakrabarty
- 7 years, 3 months ago

Good Paradox

Adhiraj Mandal
- 7 years, 3 months ago

Great Explanation!!

Dhiren Shah
- 7 years, 2 months ago

good!!!

Rajarshi Majumder
- 7 years, 2 months ago

a-b=c ....then a-b-c=0 .... and something divided by 0 is not possible

Asif Xaman
- 7 years, 1 month ago

Yah!

Mardokay Mosazghi
- 7 years, 4 months ago

Softly Busted :D

Karim Monir
- 7 years, 4 months ago

simple yet interesting to solve...!!!!

sunitha p
- 7 years, 3 months ago

aawsum..!!!!

Adhi Anand
- 7 years, 2 months ago

according to Bodmas rule it must be 4th option as this would be the eqn 4a-3a-7b=4c-3c can any one xplain me in this?

Sandeep Vishwamithra
- 6 years, 10 months ago

Log in to reply

Mdas may be circumvented as multiplication and division orders don't have a consequence, neither do addition and subtraction. But to make this easier to see, think of the equation as (4a)+(-3a)+(-4b)+(3b)=(4c)+(-3b)

Seth Boden
- 5 years, 10 months ago

WHA......! Great solution!

Ruhan Habib
- 7 years, 3 months ago

The starting equation states that ----

$a-b=c \implies a-b-c=0$ .

Now, when Kimmie cancelled both the sides in her work, what she actually did was dividing both sides of the equation $4(a-b-c)=3(a-b-c)$ by $(a-b-c)$ so as to get $4=3$ but since $(a-b-c)=0$ , so she actually divided both sides of the equation by $0$ which is mathematically incorrect.

So, her main error in the work was
**
Dividing both sides by 0.
**

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You know, lots of people without formal algebra education (i.e. knowledge that you can't divide by zero) make this mistake when trying to prove impossible things. Another mistake that is often made is taking the square root of negative number without using the formal definition of i (square root of minus 1).

Justin Chan
- 7 years, 3 months ago

Since she has assumed a-b=c,but in the end 4(a-b-c)=3(a-b-c) makes 4(0)=3(0) therefore both sides multiplied by 0 (her assumption is wrong to prove 4=3. K.K.GARG,INDIA

Krishna Garg
- 7 years, 3 months ago

Not main error but the only error

Kushagra Sahni
- 7 years, 3 months ago

good thinking

Abhay Rajput
- 7 years, 4 months ago

Cool (Y)

Nilesh Mali
- 7 years, 4 months ago

good solution

Ashwin Upadhyay
- 7 years, 4 months ago

ooooooo.....

Tootie Frootie
- 7 years, 3 months ago

good solution

KHANDAKER TAWHEED
- 7 years, 3 months ago

good dear ..

Sambit Sahu
- 6 years, 9 months ago

Good soln.

Ashley Shamidha
- 7 years, 1 month ago

**
strictly
**
prohibited, which causes no solution already. No solution means that 4 != 3.

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AH! I was stuck at this the whole day. But great question, Kevin. Really stimulates the mind into working.

Arjun Bahuguna
- 7 years, 3 months ago

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lets start with -

a - b = c

a - b - c = 0

4 ( a - b - c ) = 4 ( 0 ) = 0 --- eqn (1)

3 ( a - b - c ) = 3 ( 0 ) = 0 ---eqn (2)

dividing eqn. 2 by eqn 1 ,

4 ( a- b - c ) / 3 ( a - b - c ) = 0 / 0

Here , Kimmie took 0 / 0 to be equal to 1, thereby getting 4 = 3 as the result.

However , as we know, 0 / 0 is not defined , this was a wrong step. Hence the answer.

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a-b=c so, a-b-c=0

4(a-b-c)=3(a-b-c) because a-b-c = 0, then 4 (a-b-c) = 3 (a-b-c) should not be divided by (a-b-c) or in other words should not be divided by 0.

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niceanswer

Aswad Hariri Mangalaeng
- 7 years, 4 months ago

*
Okay jump to the last equation
*

**
4(a - b - c) = 3(a - b - c)
**

**
4(a - b - c) - 3(a - b - c) = 0
**

**
(a - b - c)( 4- 3 ) = 0
**

*
Therefore a - b - c has to be = 0
*

*
* And therefore in the last step we cannot cancel by a - b - c as it would be equivalent to dividing by 0 *
*

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$a-b=c$ $a-b-c=0$ Since you cannot divide by 0, the conclusion is proven.

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Aman Jaiswal is right. Obviously, $a-b-c=0$ .

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a major mistake, a-b=c then a-b-c=0 , and any constant divided by 0 equals undefined (or infinity)

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*
0=3
*
0 which gives 0=0, she can't divide both sides by (a-b-c) for cancellation, as it is 0. It was the main error that she divided by 0.

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We have a - b = c <=> a - b - c = 0. Kimmie cannot both sides 4(a - b - c) = 3(a - b - c).

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as a-b=c the a-b-c=0 so it becomes 0/0 which is incorrect

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If we solve this the usual algebraic way, we won't find anything unusual. But this Mathematical fallacy obviously calls for a little more careful analysis.

If you see the first step itself (
**
a - b = c
**
) and shift c to the left side, it becomes
**
a - b - c = 0
**

So practically in the final step, you are
**
dividing by 0!
**
And that also without realizing because we always try to eliminate as many unknowns as possible.

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a-b-c=0 because a-b=c, so moving c to the left makes a-b-c=0.

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she assumed a-b=c →a-b-c=0 in the final step she cancelled (a-b-c) from both sides. but (a-b-c)=0

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she did the mistake at first line. because.

a - b = c

and,

a - b - c =0

when she try to clear both side with (a - b - c) , actually she divided with 0. and 0/0 is no valid.

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a-b=c hence a-b-c=0 therefore mistake is that it was divided by zero

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a-b-c=0 so divide by 0 got it error

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you cannot cancel a-b-c as that is zero

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if a-b = c a-b-c = 0 so, it cannot be cancelled out.

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Given: (a-b)=c. Replace (a-b) with c before cancelling out in last step. We get 4(0)=3(0). :P

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a - b = c so a - b - c = c - c = 0

Nothing can be divided by 0. So it is the main error of Kimmie,s.

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The starting equation states that ----

.

Now, when Kimmie cancelled both the sides in her work, what she actually did was dividing both sides of the equation by so as to get but since , so she actually divided both sides of the equation by which is mathematically incorrect.

So, her main error in the work was Dividing both sides by 0.

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We start with a-b = c;

This means a-b-c = 0.

In last step both sides are divided by a-b-c, which is zero division and in valid

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since a-b=c implies a-b-c=0 how can she divide an equation with zero?? thats the error....

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a-b=c

a-b-c=0

4(a-b-c)=3(a-b-c)

4(0)=3(0)

0=0

She devided by 0

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She started off with the equation a-b=c =>a-b-c=0

but at the end she got 4(a-b-c)=3(a-b-c) =>4x0=3x0 =>0=0

But before that she cancelled a-b-c from both side which is impossible. Because for this she need to divide by 0 in both sides, which is impossible. So this is wrong.

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she shouldnt have cancelled in the last step!!

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Since a-b=c, Then (a-b-c) = 0

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Here a-b=c,so,a-b-c=0,so.we can not cancel both side (a-b-c) in last step.

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If a-b=c then it is obvious that a-b-c=0. For example x-3y=6 implies that x-3y-6=o