Nope, and you know why? \(2 \neq 1\) :)
–
Tim Vermeulen
·
4 years, 2 months ago

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@Tim Vermeulen
–
yes. it's because we assume that the symbol \(2\) as two thing, \(1\) as one thing, also \(\neq\) as a relation that explain everything between was inequal. the word thing mean something which we usually called it as the unit of measurement that always equal something which must be has same size (can be anything characteristic, not just size in meter for example).

so, we must obey them as International Mathematics Standardization to minimize missunderstanding between people those using Mathematics Theorem in this world.

if you assume that \(a=b\) then you divide \(a-b\) with \(a-b\)

what do you get?
you get zero divide by zero.

many mathematical fallacy appear because divide by zero.
if zero divide by zero, then what do you get?

you will get unlimited count (it's not same as infinity, but infinity is included to the list) of number or simply you can say that is indeterminate. because the result is everything, not just infinity.

this fallacy can make \(2=1\) or \(9=1\) or \(-8=8\) and simply say that \(everything=everything\).

in reality, \(everything=everything\) is not yet possible because we still have independent variable of universe which we called it Time and Space.

2 would be equal to 1 in the set of integers modulo 1 (where every number is equal to every other number).
In the usual set of integers, 2=1 can't be proven because in our convention 2 is different from 1.
–
Battista Lonardi
·
4 years, 2 months ago

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let
a=b
a2=ab
a2+a2=a2+ab
2a2=a2+ab
2a2-2ab=a2+ab-2ab
2a2-2ab=a2-ab
2(a2-ab)=1(a2-ab)
so.2=1.
now enjoy ur day.
–
Aman Singh
·
4 years, 2 months ago

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a=b
a^2=ab
a^2-b^2=ab-b^2
(a-b)(a+b)=b(a-b)
a+b=b
2b=b
2=1
–
Yong Daniel
·
4 years, 2 months ago

@Mohamed Abdelaaty
–
what? We're trying to prove that 2=1 and you're concerned about dividing by zero?
It's a creativity exercise, we're not concerned about rules.
–
Soren Henrichsen
·
4 years, 2 months ago

@Farhan Pildacil
–
\[ \frac{81}{4} + 4^2 - 4 \times 9 = \frac{81}{4} + 5^2 - 5 \times 9 \]
Then you made a mistake with the signs
\[ { \biggl( \frac{9}{2} - 4 \biggr) }^2 = { \biggl( 5 - \frac{9}{2} \biggr) }^2 \]
\[ { \biggl( \frac{1}{2} \biggr) }^2 = { \biggl( \frac{1}{2} \biggr) }^2 \]
–
Battista Lonardi
·
4 years, 2 months ago

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@Farhan Pildacil
–
The step from equation 4 to equation 5 has no sense at all. It would have it if you wrote (4 - 9/2 )^{2} = (5 - 9/2 )^{2}, but then you can´t get equation 6, because numbers have two different square roots (a positive and a negative one)
–
Marco Flores
·
4 years, 2 months ago

@Timothy Wong
–
Funny thing: It is actually true that if n = n+1 for some integer n, then all integers are the same.
–
Marco Flores
·
4 years, 2 months ago

IT'S NOT IMPOSSIBLE TO PROVE 1=2 OR 3=4 OR ANY CONSTANT TO ANOTHER. SOME PEOPLE DO THIS BY MULTIPLYING BOTH SIDES BY ANY NUMBER LIKE X^2-X^2 AND SQUAREING AND MULTIPLYING MANY MORE BUT IT WE PUT ZERO OR A NON ZERO NUMBER IN ANY OF THOSE CASES IT DOES NOT SATISFY ITS ORIGINAL EQUATION OR IT BECOME INDETERMINENT(0/0) OR BECOMES NOT DEFINED(1/0).
–
Sachin Kumar Singh
·
4 years, 2 months ago

@Pranjal Rs
–
absolutely wrong . it is equivalent to a crime in mathematics. you have multiplied both side by zero as x^2 - x^2= x^2 - x^2....,,,which is equal to zero.
–
Sachin Kumar Singh
·
4 years, 2 months ago

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@Sachin Kumar Singh
–
see is x = x ?? Yes it is so.
subtracting \(x^{2}\) from both sides.

\(x^{2}\) - \(x^{2}\) = \(x^{2}\) - \(x^{2}\)
–
Pranjal Rs
·
4 years, 2 months ago

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@Sachin Kumar Singh
–
That's the point. It's a trick. The rules don't matter because we're TRYING to break the rules. 1=2 is a clear violation of the rules, and isn't true in real life, so the procedure to get it has to include breaking the rules.
–
Soren Henrichsen
·
4 years, 2 months ago

http://satishkumarnagarajan.blogspot.in/2013/02/prove-1-2.html
–
Satish Kumar
·
4 years, 2 months ago

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http://satishkumarnagarajan.blogspot.in/2013/02/prove-1-2.html
–
Satish Kumar
·
4 years, 2 months ago

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Well, maybe it's just a joking. But through the algebra law, it can be proved. Here we are:

Through the question, we know that (a - b)^2 = 0 and it equals to a = b

Then, both side we multiple it by a, so that: a^2 = ab

Both side again, add it by a^2 so that: 2a^2 = a^2 + ab

Final, substract both side by 2ab and we get: 2a^2 - 2ab = a^2 - ab

Factorize those side and we get: 2(a^2 -ab) = (a^2-ab)

Then it means 2 = 1 .
–
Erick Sumargo
·
4 years, 2 months ago

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@Erick Sumargo
–
Amusing..But that proof is incorrect. The correct solution to the system \((a-b)^2=0\) and \(a=b\) is \(a=0\) and \(b=0\). And because of this,while the initial steps of the proof are correct the final step is incorrect,because you divided both sides of the equation by \(a^2-ab\) which is equal to zero(\(a=0\),\(b=0\)). And it is well known fact that dividing by zero is indeterminate in mathematics. Hence the error that led to the specious result.
–
Thaddeus Abiy
·
4 years, 2 months ago

@Erick Sumargo
–
Here a=b is considered. So a^2-ab=0.
So the expression 2(a^2 -ab) = (a^2-ab) is actually 2(0)=1(0).
2(0)=0 1(0)=0
It does not implies 2=1.
–
Manaswini Pandit
·
4 years, 2 months ago

## Comments

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TopNewestNope, and you know why? \(2 \neq 1\) :) – Tim Vermeulen · 4 years, 2 months ago

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thing, \(1\) as onething, also \(\neq\) as a relation that explain everything between was inequal. the wordthingmean something which we usually called it as the unit of measurement that always equal something which must be has same size (can be anything characteristic, not just size in meter for example).so, we must obey them as

International Mathematics Standardizationto minimize missunderstanding between people those usingMathematics Theoremin this world.– Yulianto Indra Setiawan · 4 years, 2 months agoYISLog in to reply

if you assume that \(a=b\) then you divide \(a-b\) with \(a-b\)

what do you get? you get zero divide by zero.

many mathematical fallacy appear because divide by zero. if zero divide by zero, then what do you get?

you will get unlimited count (it's not same as infinity, but infinity is included to the list) of number or simply you can say that is indeterminate. because the result is everything, not just infinity.

this fallacy can make \(2=1\) or \(9=1\) or \(-8=8\) and simply say that \(everything=everything\).

in reality, \(everything=everything\) is not yet possible because we still have independent variable of universe which we called it

Time and Space.happy critical thinking :)

– Yulianto Indra Setiawan · 4 years, 2 months agoYISLog in to reply

2 would be equal to 1 in the set of integers modulo 1 (where every number is equal to every other number). In the usual set of integers, 2=1 can't be proven because in our convention 2 is different from 1. – Battista Lonardi · 4 years, 2 months ago

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let a=b a2=ab a2+a2=a2+ab 2a2=a2+ab 2a2-2ab=a2+ab-2ab 2a2-2ab=a2-ab 2(a2-ab)=1(a2-ab) so.2=1. now enjoy ur day. – Aman Singh · 4 years, 2 months ago

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a=b a^2=ab a^2-b^2=ab-b^2 (a-b)(a+b)=b(a-b) a+b=b 2b=b 2=1 – Yong Daniel · 4 years, 2 months ago

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– Mohamed Abdelaaty · 4 years, 2 months ago

You divided by (a-b) which equals zero, you can't divide my zero.Log in to reply

– Soren Henrichsen · 4 years, 2 months ago

what? We're trying to prove that 2=1 and you're concerned about dividing by zero? It's a creativity exercise, we're not concerned about rules.Log in to reply

watch this

-20 = -20

16 - 36 = 25 - 45

4^{2} - 4 * 9 = 5^{2} - 5 * 9

81/4 + 4^{2} - 4 * 9 = 81/4 + 5^{2} - 5 * 9

(4 + 9/2 )^{2} = (5 + 9/2 )^{2}

4 + 9/2 = 5 + 9/2

4 = 5

4 - 3 = 5 - 3

1 = 2

Can you understand

this ?– Farhan Pildacil · 4 years, 2 months agoLog in to reply

– Battista Lonardi · 4 years, 2 months ago

\[ \frac{81}{4} + 4^2 - 4 \times 9 = \frac{81}{4} + 5^2 - 5 \times 9 \] Then you made a mistake with the signs \[ { \biggl( \frac{9}{2} - 4 \biggr) }^2 = { \biggl( 5 - \frac{9}{2} \biggr) }^2 \] \[ { \biggl( \frac{1}{2} \biggr) }^2 = { \biggl( \frac{1}{2} \biggr) }^2 \]Log in to reply

– Marco Flores · 4 years, 2 months ago

The step from equation 4 to equation 5 has no sense at all. It would have it if you wrote (4 - 9/2 )^{2} = (5 - 9/2 )^{2}, but then you can´t get equation 6, because numbers have two different square roots (a positive and a negative one)Log in to reply

and 1=2, 2=3, 3=4, 4=5, etc..... – Timothy Wong · 4 years, 2 months ago

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– Battista Lonardi · 4 years, 2 months ago

No, he didn't prove it.Log in to reply

– Marco Flores · 4 years, 2 months ago

Funny thing: It is actually true that if n = n+1 for some integer n, then all integers are the same.Log in to reply

– Farhan Pildacil · 4 years, 2 months ago

actually I'm 13Log in to reply

its nt possible – Prateek Thakur · 4 years, 2 months ago

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IT'S NOT IMPOSSIBLE TO PROVE 1=2 OR 3=4 OR ANY CONSTANT TO ANOTHER. SOME PEOPLE DO THIS BY MULTIPLYING BOTH SIDES BY ANY NUMBER LIKE X^2-X^2 AND SQUAREING AND MULTIPLYING MANY MORE BUT IT WE PUT ZERO OR A NON ZERO NUMBER IN ANY OF THOSE CASES IT DOES NOT SATISFY ITS ORIGINAL EQUATION OR IT BECOME INDETERMINENT(0/0) OR BECOMES NOT DEFINED(1/0). – Sachin Kumar Singh · 4 years, 2 months ago

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here is what I can make out.

Let x = 1,

\(x^{2}\) - \(x^{2}\) = \(x^{2}\) - \(x^{2}\)

x(x - x) = (x+x)(x - x)

x = (x + x)

Putting x = 1 in above eqn.

1 = (1 + 1)

1 = 2

Enjoy!! – Pranjal Rs · 4 years, 2 months ago

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– Sachin Kumar Singh · 4 years, 2 months ago

absolutely wrong . it is equivalent to a crime in mathematics. you have multiplied both side by zero as x^2 - x^2= x^2 - x^2....,,,which is equal to zero.Log in to reply

\(x^{2}\) - \(x^{2}\) = \(x^{2}\) - \(x^{2}\) – Pranjal Rs · 4 years, 2 months ago

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– Soren Henrichsen · 4 years, 2 months ago

That's the point. It's a trick. The rules don't matter because we're TRYING to break the rules. 1=2 is a clear violation of the rules, and isn't true in real life, so the procedure to get it has to include breaking the rules.Log in to reply

– Battista Lonardi · 4 years, 2 months ago

You can't simplify x-x because it is equal to 0.Log in to reply

We know that, 1-3 = 4-6

Adding (9/4) on both side,

1-3+(9/4) = 4 – 6 + (9/4)

12 – 3 + (3/2)2 = 22 – 6 + (3/2)2

Equation in the form, a2 + 2ab + b2 = (a+b)2

Therefore, {1-(3/2)}2 = {2-(3/2)}2

Taking square root on both side,

1-(3/2) = 2 – (3/2)

1= 2.

Hence the proof

There is a Mistake involved in it,

Mystery unfolds

{1-(3/2)}2 = {2-(3/2)}2

(-1/2)2 = (1/2)2

Taking square root on both side,

(-1/2) ≠ (1/2).

http://satishkumarnagarajan.blogspot.in/2013/02/prove-1-2.html – Satish Kumar · 4 years, 2 months ago

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http://satishkumarnagarajan.blogspot.in/2013/02/prove-1-2.html – Satish Kumar · 4 years, 2 months ago

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Well, maybe it's just a joking. But through the algebra law, it can be proved. Here we are:

Through the question, we know that (a - b)^2 = 0 and it equals to a = b

Then, both side we multiple it by a, so that: a^2 = ab

Both side again, add it by a^2 so that: 2a^2 = a^2 + ab

Final, substract both side by 2ab and we get: 2a^2 - 2ab = a^2 - ab

Factorize those side and we get: 2(a^2 -ab) = (a^2-ab)

Then it means 2 = 1 . – Erick Sumargo · 4 years, 2 months ago

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– Thaddeus Abiy · 4 years, 2 months ago

Amusing..But that proof is incorrect. The correct solution to the system \((a-b)^2=0\) and \(a=b\) is \(a=0\) and \(b=0\). And because of this,while the initial steps of the proof are correct the final step is incorrect,because you divided both sides of the equation by \(a^2-ab\) which is equal to zero(\(a=0\),\(b=0\)). And it is well known fact that dividing by zero is indeterminate in mathematics. Hence the error that led to the specious result.Log in to reply

a(a-b)=0*a a^2-ab=0

in the line "Factorize those side and we get: 2(a^2 -ab) = (a^2-ab)" you multiply both sides by 1/(a^2-ab)

2(0)/0 = 0/0

0/0 is indeterminate

that's why 2 is not equal to 1 =D – Mark Kevin Aguilar · 4 years, 2 months ago

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(0)=1(0).2

(0)=0 1(0)=0It does not implies 2=1. – Manaswini Pandit · 4 years, 2 months ago

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