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# Can you prove that 2=1 !!

metalfreek
Hay everyone can you prove mathematically that 2=1. Just give it a try. I will show you all later how it is possible.
Afaceinthematrix
It's not possible. It's mathematically incorrect. I can think of a dozen ways off the top of my head that you may use to "prove" to me that 1=2 and you should take one thing in to consideration before you post any of those ways here: you cannot divide by 0 and infinity is not an integer (those are the most common mistakes people use when trying to prove that 1=2).
Xanatos
It is mathematically impossible. In addition to those afaceinthematrix listed above, using imaginary numbers ad square roots won't work either if you had that in mind.
metalfreek
Let me prove it for you with the help of application of elementary algebra.
Let x=y ------- (1)
Multiply by 'y' on both sides we get
xy=y^2
Now substracting x^2 in both sides
xy-x^2=y^2-x^2
x(y-x)=(y+x)(y-x)
On dividing both sides by (y-x)
x=y+x
x=x+x since x=y from equation (1)
x=2x
divide both sides by 'x' we get
1=2

This completes the proof.
Xanatos
 metalfreek wrote: Let me prove it for you with the help of application of elementary algebra. Let x=y ------- (1) Multiply by 'y' on both sides we get xy=y^2 Now substracting x^2 in both sides xy-x^2=y^2-x^2 x(y-x)=(y+x)(y-x) On dividing both sides by (y-x) x=y+x x=x+x since x=y from equation (1) x=2x divide both sides by 'x' we get 1=2 This completes the proof.

Afaceinthematrix told you not to divide by zero, yet here you go doing it. You cannot divide by y-x because you stated at the beginning that y=x. If y=x then y-x=0. You cannot divide by zero.

Proof failed.
Afaceinthematrix
 metalfreek wrote: Let me prove it for you with the help of application of elementary algebra. Let x=y ------- (1) Multiply by 'y' on both sides we get xy=y^2 Now substracting x^2 in both sides xy-x^2=y^2-x^2 x(y-x)=(y+x)(y-x) On dividing both sides by (y-x) x=y+x x=x+x since x=y from equation (1) x=2x divide both sides by 'x' we get 1=2 This completes the proof.

This wasn't even a good "proof" involving a division by zero error. I've seen some "proofs" of the like in which you divide by some long line involving more advanced operators (logarithms, trig functions, etc.) where some long line equals zero... by y-z when y=z? Come on. That's not even fun to try to fool people with that.
metalfreek
Dividing by Zero is what the proof has done but its fun to see something like that and most of the people never think of the things like that and when I first saw the proof I was also a bit amazed.
infinisa
Hello metalfreek

Proofs of this kind (based on a disguised division by zero) are the oldest trick in the (mathematical) book, and do indeed surprise the first time round.

You can see more on this topic in the thread Can 1 plus 1 prove to be something else?, in particular my contribution based on the Banach-Tarski paradox.
carlospro7

ajassat
It can be proved in away which makes it seem true for those who don't properly understand the rules of algebra.

The proof is invalid. If a=b, then we can assign 0 to a, hence b is also zero. In the division of the proof 0/0 is the indeterminate form. It is undefined.
sheedatali
I have just beaten my two friends in a arm wrestle, hence proving that not only 1 = 2, 1 is Better than 2. Proved!
Afaceinthematrix
 sheedatali wrote: I have just beaten my two friends in a arm wrestle, hence proving that not only 1 = 2, 1 is Better than 2. Proved!

What? That was incoherent...
guissmo
I knew that this is going to be that old joke when dividing by zero.
Frihosters are too smart to fall for these things though.

Nice try in creating an interesting topic though, afaceinthematrix.
See you around.
sajeebr
It is related to hindus. It says that there is the god called Shiva and his wife Parwati. There is a union of both of these and forms a half man and half women. That is there is the formation of 1 from two.
That means 1+1=2.
Bikerman
 sajeebr wrote: It is related to hindus. It says that there is the god called Shiva and his wife Parwati. There is a union of both of these and forms a half man and half women. That is there is the formation of 1 from two. That means 1+1=2.

Err...I think you need to rethink that posting...
ninjakannon
Easy, 1x0 = 2x0.

Now just cancel the zeros.

Oh, wait. You can't cancel 0. Just thought I'd point out another method that doesn't require dividing by 0 but still doesn't actually work.

If you 'prove' that 1 = 2, then you shouldn't be saying "wow, maths is wrong". You should be saying "damn, I'm wrong" and check your working.
sheedatali
Afaceinthematrix wrote:
 sheedatali wrote: I have just beaten my two friends in a arm wrestle, hence proving that not only 1 = 2, 1 is Better than 2. Proved!

What? That was incoherent...

Dude that was a joke. Thread was just too mathematical, i could not help throwing in a joke.
pscompanies
Lol here's another:

Complex Numbers Method

* Step 1: -1/1 = 1/-1

* Step 2: Taking the square root of both sides:

* Step 3: Simplifying:

* Step 4: In other words, i/1 = 1/i.

* Step 5: Therefore, i / 2 = 1 / (2i),

* Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),

* Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

* Step 8: ,

* Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

* Step 10: and this shows that 1=2.

See if you can figure out in which step the fallacy lies.[img][/img]
ninjakannon
pscompanies, step 3 is wrong! That is where "the fallacy lies".

If we simplify, step 2 states that √-1 = √-1, which is correct (obviously).

Step 3 simplified is step 4, which is incorrect.
Xanatos
It is wrong because of this identity

sqrt(a/b)=sqrt(a)/sqrt(b) if and only if a and b are both positive real numbers.
infinisa
Hello pscompanies

Steps 2 & 3 are ambiguous, because every number has 2 square roots (+/-)

However, Step 4 is definitely wrong because you choose one of the possible square roots on each side, and these are not equal.

* Step 4: In other words, i/1 = 1/i.
which is the same as saying i = -i, which is obviously false and leads to the contradiction

Step 4 should be:

* Step 4: In other words, i/1 = ħ1/i.

In this case, the minus sign is appropriate:

* In fact, i/1 = -1/i.
ninjakannon
infinisa, good point there that's one I'd overlooked!

However, the plus or minus doesn't let you choose which sign is more appropriate in this case, it indicates that both are true.

So step 4 is: ħi/1 = ħ1/i

Implying both:
i/1 = 1/i
=> i = -i

and:
-i/1 = -1/i
=> -i = i

Which is not correct.

I'm sure you know that, but it 100% wasn't clear.

But you can't get to step 4 anyway because, as Xanatos states perfectly, step 3 is wrong.
infinisa
Hi ninjakannon

Let me try a slightly different approach:

Although every number (except 0!) has 2 square roots (+/-), if x > 0 then by convention √x refers to the positive square root.

Following this convention, rules such as √(xy) = √x√y and √(x/y) = √x/√y are true.

We can extend the convention so that √(-1) = i (rather than -i), but then the rules such as √(xy) = √x√y and √(x/y) = √x/√y are no longer true.

In this sense, as you said, step 2 is right and step 3 is wrong.
iman
You can do that if you divide by zero.

And well, you can say that 2 and 1 are the same in the sense that
2 = 0(mod 1) and 1 = 0(mod 1).
_AVG_
Well, I feel that the complex numbers proof of 1=2 has been cleared up by virtue of Step 3 being wrong.

However, I felt it too trivial to start a new topic for the point I am about to make so here goes ...

I have recently noticed a problem observed with the convention that √a = the positive square root of a.

Consider √1. Now we all know that √1 = 1 (which is the positive square root thereby making -1 the negative square root). However, this doesn't turn out to be the case when 1 is considered as a power of i:

Let us consider i^4 (however, we know that in reality, 1 can be represented as i^4k where k is an integer). But anyway, this problem is eliminated as √(i^4k) can be written as (√(i^4))^k.

Anyway, let us consider √(i^4). If we assume normal algebra were to apply, then √(i^4) = i^2 however i^2 = -1. And remember, this is supposed to be the positive root. So, we get the positive root of 1 as -1 i.e. a negative number?!

Now, either I have gone wrong somewhere by overlooking something, some rule, etc. BUT .... if not, then I think this is quite a serious problem ..

What do you think?
infinisa
 _AVG_ wrote: Anyway, let us consider √(i^4). If we assume normal algebra were to apply, then √(i^4) = i^2 however i^2 = -1. And remember, this is supposed to be the positive root. So, we get the positive root of 1 as -1 i.e. a negative number?!

Hello AVG

The problem is that the "normal rules of algebra" don't apply in this case.
The simple rule √x√y = √(xy) only works if x & y are non-negative real numbers.
What you have done, basically, is to apply it in the (inapplicable) case where x & y = -1, so you get:
√(-1)√(-1) = √((-1)^2)
i.e. √(-1)√(-1) = √1
i.e. -1 = 1.

Moral of the story: only apply the "normal rules of algebra" where they really apply!
infinisa
 _AVG_ wrote: Anyway, let us consider √(i^4). If we assume normal algebra were to apply, then √(i^4) = i^2 however i^2 = -1. And remember, this is supposed to be the positive root. So, we get the positive root of 1 as -1 i.e. a negative number?!

Hello AVG

The problem is that the "normal rules of algebra" don't apply in this case.
The simple rule √x√y = √(xy) only works if x & y are non-negative real numbers.
What you have done, basically, is to apply it in the (inapplicable) case where x & y = -1, so you get:
√(-1)√(-1) = √((-1)^2)
i.e. √(-1)√(-1) = √1
i.e. -1 = 1.

Moral of the story: only apply the "normal rules of algebra" where they really apply!
infinisa
Sorry about the double post, but I'm having BIG problems accessing the Frihost forums today.
After the first submission gave an error, I tried again - but in fact the first submission had actually worked.
greeneyedtaxi
1 * infinity = 2 * infinity.

divide both sides by infinity.
1 = 2.

actually, this may not make sense. you really can't divide by infinity. however it proves that in the face of infinity, all other numbers are insignificant.
_AVG_
Yeah, I think dividing infinity by infinity would be like dividing 0 by 0. Like the latter could be infinity, 1 or zero, so could the former ...
thnn
Let us use a group (Z, +) (Integers, under addition).
In group theory for something to be a group we need a identity, which is of course 0 in this case, as a + 0 = a.
We also need to define the inverse, in this case a ^ -1 = -a
The inverse is unique (I am not going to prove this)
If we then use this we can say that if 2 = 1, then -2 = 2 ^ -1 = 1^-1 = -1
Taking out 2^-1 and 1^-1 as we are assuming 2 = 1 we then get -1 = -2, which is false. But the inverse is unique, so therefore the numbers can not be the same.

On the other hand, 2 = 1 (mod 1)
Bikerman
 Quote: In group theory for something to be a group we need a identity, which is of course 0 in this case, as a + 0 = a. We also need to define the inverse, in this case a ^ -1 = -a

if f(x)=y then the inverse is f^-1(y)=x
therefore if f(a) = +0 then f^-1(0)=a
or am I missing something (set theory is really not my thing....)
thnn
Bikerman wrote:
 Quote: In group theory for something to be a group we need a identity, which is of course 0 in this case, as a + 0 = a. We also need to define the inverse, in this case a ^ -1 = -a

if f(x)=y then the inverse is f^-1(y)=x
therefore if f(a) = +0 then f^-1(0)=a
or am I missing something (set theory is really not my thing....)

I should mention that in this group the definition for inverse would be a + (a^-1) = 0
a ^ -1 is just notation, it is not used to mean 1/a
Bikerman
Ahh...OK
(told you set theory wasn't my thing )
TBSC
 Bikerman wrote: Ahh...OK (told you set theory wasn't my thing )

eday2010
 metalfreek wrote: Let me prove it for you with the help of application of elementary algebra. Let x=y ------- (1) Multiply by 'y' on both sides we get xy=y^2 Now substracting x^2 in both sides xy-x^2=y^2-x^2 x(y-x)=(y+x)(y-x) On dividing both sides by (y-x) x=y+x x=x+x since x=y from equation (1) x=2x divide both sides by 'x' we get 1=2 This completes the proof.

that's not proof. That's just an equation to make 2=1. I can make up an equation to show that 5=23. It's all meaningless. 2 only equals 1 when you include the equation to get to that point. Otherwise it's just bullshit.

You can create an equation and formula to make any number equal another number. It doesn't prove anything valid.
Bikerman
eday2010 wrote:
 metalfreek wrote: Let me prove it for you with the help of application of elementary algebra. Let x=y ------- (1) Multiply by 'y' on both sides we get xy=y^2 Now substracting x^2 in both sides xy-x^2=y^2-x^2 x(y-x)=(y+x)(y-x) On dividing both sides by (y-x) x=y+x x=x+x since x=y from equation (1) x=2x divide both sides by 'x' we get 1=2 This completes the proof.

that's not proof. That's just an equation to make 2=1. I can make up an equation to show that 5=23. It's all meaningless. 2 only equals 1 when you include the equation to get to that point. Otherwise it's just bullshit.

You can create an equation and formula to make any number equal another number. It doesn't prove anything valid.
Make your mind up. On the one hand you say that 2=1 when you include an equation to get to that point (which is false of course) and next you say it doesn't prove anything (which is also false)....

Please provide your equation to show that 5=23. I'd be most interested.
metalfreek
eday2010 wrote:
 metalfreek wrote: Let me prove it for you with the help of application of elementary algebra. Let x=y ------- (1) Multiply by 'y' on both sides we get xy=y^2 Now substracting x^2 in both sides xy-x^2=y^2-x^2 x(y-x)=(y+x)(y-x) On dividing both sides by (y-x) x=y+x x=x+x since x=y from equation (1) x=2x divide both sides by 'x' we get 1=2 This completes the proof.

that's not proof. That's just an equation to make 2=1. I can make up an equation to show that 5=23. It's all meaningless. 2 only equals 1 when you include the equation to get to that point. Otherwise it's just bullshit.

You can create an equation and formula to make any number equal another number. It doesn't prove anything valid.

I would also be interested to see how I can show 5=23. Please post it here.
LittleBlackKitten
My brain just went "PSSSSSSSSSHT! Boooooooooom."
eday2010
Let x=5
Let y=x

x=y+18-(6*3)+(72/4)
x=y+18-18+18
x=23

Since we let x=5

5=23

See how I made stuff up out of the blue? That is exactly what that other equation does as well. You can make anything equal anything else.

 Bikerman wrote: Make your mind up. On the one hand you say that 2=1 when you include an equation to get to that point (which is false of course) and next you say it doesn't prove anything (which is also false)....

I did make up my mind. 2 only equals 1 when you include that made up equation. And even when you do include it, it doesn't prove anything because there is no basis for the equation to begin with. It's all pulled from thin air.
saratdear
 eday2010 wrote: Let x=5 Let y=x x=y+18-(6*3)+(72/4)

You went wrong there itself...but if your point is to prove that anything can be equal to anything else (by incorrect steps), then you're right.
Bikerman
 eday2010 wrote: Let x=5 Let y=x x=y+18-(6*3)+(72/4) x=y+18-18+18 x=23
But this is nonsense. Once you define y=x then the rest of your 'equation' is invalid. The whole point is that you can indeed make up equations but they must be logically consistent. It isn't a case of 'making stuff up out of the blue' - each step must follow from and be consistent with the previous steps.
eday2010
saratdear wrote:
 eday2010 wrote: Let x=5 Let y=x x=y+18-(6*3)+(72/4)

You went wrong there itself...but if your point is to prove that anything can be equal to anything else (by incorrect steps), then you're right.

There are no incorrect steps when you are making up an equation to prove whatever it is you are out to prove. The openign post does it. How is it a correct step to take the squareroot of both sides in the second line? Who said to do that?

Bikerman wrote:
 eday2010 wrote: Let x=5 Let y=x x=y+18-(6*3)+(72/4) x=y+18-18+18 x=23
But this is nonsense. Once you define y=x then the rest of your 'equation' is invalid. The whole point is that you can indeed make up equations but they must be logically consistent. It isn't a case of 'making stuff up out of the blue' - each step must follow from and be consistent with the previous steps.

The opening post was nonsense too. Randomly taking to squareroot in the second step is just as nonsensical as what I did. Why did I define x and y? Because I can and it suited my needs. Why did that person take the squareroot of both sides? Because they can and it suited their needs.
Bikerman
 eday2010 wrote: There are no incorrect steps when you are making up an equation to prove whatever it is you are out to prove. The openign post does it. How is it a correct step to take the squareroot of both sides in the second line? Who said to do that?
Nobody 'said' to do it.
It is perfectly legitimate to square root both sides - as it is to do a lot of operations on both sides. The point is that by doing the same operation to each side, the equation should remain balanced. If it doesn't then you have either discovered a flaw in maths or you have made a mistake.
That is how you learn maths - not by memorising formulae, but by playing with the numbers and equations.
 Quote: The opening post was nonsense too. Randomly taking to squareroot in the second step is just as nonsensical as what I did. Why did I define x and y? Because I can and it suited my needs. Why did that person take the squareroot of both sides? Because they can and it suited their needs.

But the opening post stuck to the rules - what you do to one side you must do to the other. You just threw a load of terms around breaking the rules willy-nilly.
The opening post was working with an equation. You were working with alphabet stew.

y(x-1)=7x^2 - 5
Then there are strict rules about what I can do with it. There is no 'master' rule which tells me what to do - that will depend on the problem. Maybe I want to get a value for x, in which case I need to move things around. Maybe I want to get a value for y - same process. Maybe I can spot a really nice way of arriving at x or y without a load of tedious mucking around. You learn by doing.

In the above case, to get a value for y is easy - divide both sides by x-1. To get a value for x is a bit more tricky...we need to shift things around:
So can you get a value for x??
stop_it
 metalfreek wrote: Hay everyone can you prove mathematically that 2=1. Just give it a try. I will show you all later how it is possible.

Yes, 1 haystack + 1 haystack = 1 haystack!!!

noobcake
stop_it wrote:
 metalfreek wrote: Hay everyone can you prove mathematically that 2=1. Just give it a try. I will show you all later how it is possible.

Yes, 1 haystack + 1 haystack = 1 haystack!!!

But the factorial function isn't defined on haystack!!

Afaceinthematrix
 thnn wrote: Let us use a group (Z, +) (Integers, under addition). In group theory for something to be a group we need a identity, which is of course 0 in this case, as a + 0 = a. We also need to define the inverse, in this case a ^ -1 = -a The inverse is unique (I am not going to prove this) If we then use this we can say that if 2 = 1, then -2 = 2 ^ -1 = 1^-1 = -1 Taking out 2^-1 and 1^-1 as we are assuming 2 = 1 we then get -1 = -2, which is false. But the inverse is unique, so therefore the numbers can not be the same. I haven't though too hard about this so it is quite possibly incorrect. On the other hand, 2 = 1 (mod 1)

Of course this is even easier if you use the group Z2. Then you would have 1+1=0. But really, it's just showing that 1+1=0mod2 which is obvious....

And I see nothing incorrect with your proof that 1 does not equal 2. It seems okay as long as you prove that inverses are unique.
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