Hay everyone can you prove mathematically that 2=1. Just give it a try. I will show you all later how it is possible.
Can you prove that 2=1 !!
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).
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.
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.
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.
| 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.
| 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.
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.
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.
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.
I think this image may help you understand what you have done with your proof. 
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.
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.
I have just beaten my two friends in a arm wrestle, hence proving that not only 1 = 2, 1 is Better than 2. Proved!
| 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...
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.
Frihosters are too smart to fall for these things though.
Nice try in creating an interesting topic though, afaceinthematrix.
See you around.
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.
That means 1+1=2.
| 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...
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.
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.
| Afaceinthematrix wrote: | ||
What? That was incoherent... |
Dude that was a joke. Thread was just too mathematical, i could not help throwing in a joke.
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]
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]
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.
If we simplify, step 2 states that √-1 = √-1, which is correct (obviously).
Step 3 simplified is step 4, which is incorrect.
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.
sqrt(a/b)=sqrt(a)/sqrt(b) if and only if a and b are both positive real numbers.
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.
Your step 4 is
* 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 = ±1/i.
In this case, the minus sign is appropriate:
* In fact, i/1 = -1/i.
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.
Your step 4 is
* 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 = ±1/i.
In this case, the minus sign is appropriate:
* In fact, i/1 = -1/i.
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: ±i/1 = ±1/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.
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: ±i/1 = ±1/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.
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.
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.
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