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Can anyone prove that 2=1 based of:

a^2 -2ab +b^2 =0

Note by Erick Sumargo
4 years, 2 months ago

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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.

YIS Yulianto Indra Setiawan · 4 years, 2 months ago

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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 :)

YIS Yulianto Indra Setiawan · 4 years, 2 months ago

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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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@Yong Daniel You divided by (a-b) which equals zero, you can't divide my zero. Mohamed Abdelaaty · 4 years, 2 months ago

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@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

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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 ago

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@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

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@Farhan Pildacil So, you proved 4=5 too...

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

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@Timothy Wong No, he didn't prove it. Battista Lonardi · 4 years, 2 months ago

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@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

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@Farhan Pildacil actually I'm 13 Farhan Pildacil · 4 years, 2 months ago

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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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@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

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@Pranjal Rs You can't simplify x-x because it is equal to 0. Battista Lonardi · 4 years, 2 months ago

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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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@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

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@Erick Sumargo a=b, a-b=0

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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@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

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