about how to prove 1=2 algebraically

[vinay r v]

hello ,,,,everybody
i am vinay rv, i have doubt with proving 1=2 algebraically.... would anyone help me...... is it possible...???

 

[alexan_e]

yes it is , you can send me 2000 dollars and I will send you back 1000 dollars and we will be even (since 1=2 and 1000=2000) :-D

 

[vinay r v]

kk i got it........ i think there no soln fr that

 

[alexan_e]

Check this
https://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

and see what is wrong with it https://en.wikipedia.org/wiki/Mathematical_fallacy

Every such proof make a small mistake someone along the way to prove something that is not correct.

Alex

 

[bigdogguru]

1=2

Ridiculous!

However any fool knows 7x13=28.

BigDog

 

[andre_luis]

... is it possible...???

Just when assignment operator "=" works no more...


+++

 

[mishrashashi]

These are the tricky question to check anyone's logic. Solution of these questions has error.

Let me try this,
X2-X2=X2-X2
=> X(X-X)=(X+X)(X-X) ; (X-X) will omit from both sides
=> X=X+X
=> X=2X
=> 1=2 8 -O

Its just for fun. You can't prove 1=2.

---------- Post added at 10:08 ---------- Previous post was at 09:53 ----------

Just for fun,
You can prove any number equals to other number. Solution has error, but you can trick the people atleast once

Q. 1=5

A.
-5= -5 ; if x=y then take xy on bothside with -ve sign
1-6=25-30; take x2- A= y2-B= -xy
(1)2-2*1*(6/2)=(5)2-2*5*(6/2)
(1)2-2*1*(6/2)+(6/2)2=(5)2-2*5*(6/2)+(6/2)2
[1-(6/2)]2=[5-(6/2)]2
1-(6/2)=5-(6/2) ; here is the error, if x2=y2 then x=+- y, in this solution -ve sign has ignored.
1=2


have fun to trick others

 

[cubeusk]

Yes, we can prove it. Let's see:

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

There you have it. Enjoy! ;-)

 

[keith1200rs]

Line 3 is

0 = 0

so anything after that is meaningless.

Keith

 

[KerimF]

In modern Math, we can assign a function to an operator and build a new domain based on it.

Here (below) the symbol '=' is a special opeartor and '#' is equivalent to the conventional '='

x = y implies y # x + 1 or x # y - 1
1 = 2 implies 2 # 1 + 1 or 1 # 2 - 1

In this case, the left side is always true since it is a definition... exactly as saying

x = y implies y # x or x # y (reciprocal)
1 = 2 implies 2 # 2 or 1 # 1

The idea is... logic is always based on definitions. By changing any definition, we would see ourselves thinking in/for another realm.
This basic truth (axiom) is very important for the 'real' spiritual science (usually known as religion, but when it is based on faith, it cannot be seen as science anymore).

Kerim

Off topic note:
It happens that the logical answers I discovered about life (mainly the spiritual ones) are almost identical to what already pointed out by ***** ****** 'only' (claimed being said about 2000 years ago). I said 'only' because any other reference (even that claims being his) may not be as logical if not incomplete or wrong. For instance, filtering one's sayings could be achieved rather easily (always based on logic) in the same way we filter out the noise from the original data received after passing through a noisy channel

 

[cubeusk]

Line 3 is

0 = 0

so anything after that is meaningless.

Keith

All the lines are 0 = 0 since is an equation. ;-)

 

[rafaelcremm]

Line 3 is

0 = 0

so anything after that is meaningless.

Keith

lol...

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b (a-b) ->>>>> if a = b, (a-b)=0. so (a-b)/(a-b) = 0/0. Impossible division (indeterminable).
(a+b) = b >>>>> this division make 2=1.
2b = b
2 = 1

 

[keith1200rs]

All the lines are 0 = 0 since is an equation. ;-)

Not actually true. Saying a=b is not the same as saying 0=0.

Keith.

 

[cubeusk]

lol...

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b (a-b) ->>>>> if a = b, (a-b)=0. so (a-b)/(a-b) = 0/0. Impossible division (indeterminable).
(a+b) = b >>>>> this division make 2=1.
2b = b
2 = 1

See this way:

if a-b = x

lim x -> 0 => x/x = 1

Now there is no indetermination, using infinitesimal calculus. ;-)

 

[alexan_e]

I don't see the point of keeping this thread open any longer.
Thread closed.