0 = 1? (1 Viewer)

[jackmurray1989]

I saw my friends do this before, but I forgot how they did it. I think it had something to do with index laws and a function to the power zero and stuff. Does anyone know how to do it?

 

[jackmurray1989]

I just looked it up, and I found it algebraically, but the way I saw it was way better. It started with 0 = ...... and then solved it as a proof, and ended up with = 1.

1. a = b + 1
2. (a-b)a = (a-b)(b+1)
3. a2 - ab = ab + a - b2 - b
4. a2 - ab -a = ab + a -a - b2 - b
5. a(a - b - 1) = b(a - b - 1)
6. a = b
7. b + 1 = b
8. Therefore, 1 = 0

 

[mr_brightside]

Proof that 0 = 1

Start with the addition of an infinite succession of zeros

0 = 0 + 0 + 0 + ...

Then recognize that 0 = 1 − 1

0 = (1 - 1) + (1 - 1) + (1 - 1) + ...

Applying the associative law of addition results in

0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...

Of course − 1 + 1 = 0

0 = 1 + 0 + 0 + 0 + ...

And the addition of an infinite string of zeros can be discarded leaving

0 = 1



is that it?

 

[ellen.louise]

what's the point of that?

 

[201055]

Proof that 0 = 1

Start with the addition of an infinite succession of zeros

0 = 0 + 0 + 0 + ...

Then recognize that 0 = 1 − 1

0 = (1 - 1) + (1 - 1) + (1 - 1) + ...

Applying the associative law of addition results in

0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...

Of course − 1 + 1 = 0

0 = 1 + 0 + 0 + 0 + ...

And the addition of an infinite string of zeros can be discarded leaving

0 = 1



is that it?

I can see where you're coming from

Although when you group the (-1 + 1) = 0 under the associate law of addition, at the end of the infinity equation (hypothetically) wouldn't there be a -1 at the end?

So: 0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... ... ... (-1 + 1) -1

0 = 1 -1....

Good try though, although this is interesting

 

[watatank]

I just looked it up, and I found it algebraically, but the way I saw it was way better. It started with 0 = ...... and then solved it as a proof, and ended up with = 1.

1. a = b + 1
....
5. a(a - b - 1) = b(a - b - 1)
6. a = b
7. b + 1 = b
8. Therefore, 1 = 0

you can't go from 5 to 6 because you cant divide both sides by zero.

you can only do this if (a - b - 1) =/= 0

but you've said (from step 1) a = b + 1 which rearranges to (a - b - 1) = 0 so you can't do that division.

so yeah

 

[aussiechica7]

I just looked it up, and I found it algebraically, but the way I saw it was way better. It started with 0 = ...... and then solved it as a proof, and ended up with = 1.

1. a = b + 1
2. (a-b)a = (a-b)(b+1)
3. a2 - ab = ab + a - b2 - b
4. a2 - ab -a = ab + a -a - b2 - b
5. a(a - b - 1) = b(a - b - 1)
6. a = b
7. b + 1 = b
8. Therefore, 1 = 0

Yeah, the problem is dividing by 0 (a-b-1 = 1-0-1 = 0) which of course stuffs up the proof.

 

[mr_brightside]

I can see where you're coming from

Although when you group the (-1 + 1) = 0 under the associate law of addition, at the end of the infinity equation (hypothetically) wouldn't there be a -1 at the end?

So: 0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... ... ... (-1 + 1) -1

0 = 1 -1....

Good try though, although this is interesting

hypothetically.

I stole that from wikipedia anyway.
Here's the link to lots of invalid proofs

http://en.wikipedia.org/wiki/Invalid_proof

 

[ellen.louise]

could we just accept here that 1 and 0 are different numbers? The proofs just don't work. deal.

 

[Dumsum]

lol infinite series are easy ones to mess around with.

s = 1 + 2 + 3 + 4 + ...
= 1 + ((1 + 1) + (1 + 2) + (1 + 3) + ...)
= (1 + 1 + ...) + (1 + 2 + 3 + ...)
= (1 + 1 + ...) + s
0 = 1 + 1 + 1 + ...

THARFOAR INFINITEY = ZERO

 

[jackmurray1989]

Yeah, I didn't make my one up. I just got it from some site when I was trying to look for it.

But the one I saw my friends do (4u accelerants) was sweet. It was heaps long and used index laws and worked really well. I'm pretty sure it used the fact that any number to the power of 0 = 1. From what I could understand of it, that was crux of the proof.

 

[jackmurray1989]

prove that .99999(repeats) = 1 using two unit methods

x = 0.99999999....
10x = 9.99999999....
9x = 9
x = 1.


Also, if you think about it,

1/3 = 0.333333.....
2/3 = 0.666666.....
3/3 = ________

 

[jb_nc]

x = 0.99999999....
10x = 9.99999999....
9x = 9
x = 1.


Also, if you think about it,

1/3 = 0.333333.....
2/3 = 0.666666.....
3/3 = ________

limits are better

what happened to my post there??

 

[Dumsum]

0.999... = 9/10 + 9/100 + ...
= (9/10)/(1-1/10) (sum of geometric series)
= 1

(2u does do geometric series, right? I can't remember that far back)

 

[jackmurray1989]

You know, zero didn't exist for a hell of a long time. It was just invented to fill the gaping chasm between -1/inf and 1/inf.

If I have 3 apples, and I give 2 to Sally and 1 to Jeff, how many apples do I have left?

 

[watatank]

0.999... = 9/10 + 9/100 + ...
= (9/10)/(1-1/10) (sum of geometric series)
= 1

(2u does do geometric series, right? I can't remember that far back)

yeah they do. thats the two unit method.

 

[mr_brightside]

If I have 3 apples, and I give 2 to Sally and 1 to Jeff, how many apples do I have left?

-1 + 1 apples left?

 

[jackmurray1989]

I'm suprised this hasn't been mentioned yet. (Unless it has, in which case I'm blind).

x⁰ = 1
This applies to all Real x.

Thus, it is concievable that

0⁰ = 1

Is is it not?

Yeah, that's what I've been saying. It had 1^0 in there. Except there was lots of working out and logs and limits and the end result was 0 = 1.

 

[Trebla]

0⁰ is undefined. x⁰ = 1 is only true for non-zero x. When you apply logarithms, you cannot take the logarithm of zero. When you apply index laws in an attempt to obtain it, you'll find that zero is undefined with a negative index, hence when the index is zero, it is undefined, because we cannot apply index laws to subtract indices from positive indices when the base is zero.

The 0.99999.... = 1 is actually true if you have a definition of 'infinity' as a point. So 'at infinity', 0.99999.... = 1, but if it is finite it is never equal to 1. So it would be true to say that ALL whole numbers are infinite limiting sums and do not "exist" in reality. All measurements in reality supposedly have infinite significant figures which specify the approximate end position, hence we only approximate as close as we can to the real value in reality. So in a way, you could say that nothing can ever be exactly and precisely specified by measurement in reality! Now that's something to think about! lol

An interesting thing to note is that some mathematicians regard infinity as a well defined 'point' such that "1/∞ = 0" (almost implying that infinity is the "last" number of the real number system), while other mathematicians regard infinity as undefined which just represents a huge number increasing forever.

 

[jackmurray1989]

0<SUP>0</SUP> is undefined. x<SUP>0</SUP> = 1 is only true for non-zero x. When you apply logarithms, you cannot take the logarithm of zero. When you apply index laws in an attempt to obtain it, you'll find that zero is undefined with a negative index, hence when the index is zero, it is undefined, because we cannot apply index laws to subtract indices from positive indices when the base is zero.

The 0.99999.... = 1 is actually true if you have a definition of 'infinity' as a point. So 'at infinity', 0.99999.... = 1, but if it is finite it is never equal to 1. So it would be true to say that ALL whole numbers are infinite limiting sums and do not "exist" in reality. All measurements in reality supposedly have infinite significant figures which specify the approximate end position, hence we only approximate as close as we can to the real value in reality. So in a way, you could say that nothing can ever be exactly and precisely specified by measurement in reality! Now that's something to think about! lol

An interesting thing to note is that some mathematicians regard infinity as a well defined 'point' such that "1/∞ = 0" (almost implying that infinity is the "last" number of the real number system), while other mathematicians regard infinity as undefined which just represents a huge number increasing forever.

Yeah, that's what I was going to say.