Not As Simple As It Seems
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Or is it...? Because it really is. Lol.
After looking at a proof for the simple equation 1 + 1 = 2 I found this.
http://humor.beecy.net/misc/principia/
Was like... "What the hell are those symbols...?"
Here's a more friendly version of 1 + 1 = 2
To prove this, one must first what natural number N is. N is the smallest set that satisfies these five postulates (P). Peano's Postulates to be exact.
P1. 1 is in N.
P2. If x is in N, then x', its "successor", is also in N.
P3. There is no x' such that x' = 1.
P4. If x ≠1 then there is a y in N such that y' = x.
P5. If a statement holds for 1 and it holds for x as well as x', then it also holds for N.
Now you have also defined addition recursively.
Let a and b be in N. If b = 1, then a + b = a' using P1 and P2. If b ≠1, then c' = b with c in N using P4 and defined is a + b = a + c'.
Using this definition, you have now defined the number 2.
2 = 1'
2 is in N by P1, P2, and the definition of the number 2.
The Theorem is 1 + 1 = 2.
The Proof by using the definition of addition with a = b = 1, 1 + 1 = 1' = 2. Q.E.D.
After looking at a proof for the simple equation 1 + 1 = 2 I found this.
http://humor.beecy.net/misc/principia/
Was like... "What the hell are those symbols...?"
Here's a more friendly version of 1 + 1 = 2
To prove this, one must first what natural number N is. N is the smallest set that satisfies these five postulates (P). Peano's Postulates to be exact.
P1. 1 is in N.
P2. If x is in N, then x', its "successor", is also in N.
P3. There is no x' such that x' = 1.
P4. If x ≠1 then there is a y in N such that y' = x.
P5. If a statement holds for 1 and it holds for x as well as x', then it also holds for N.
Now you have also defined addition recursively.
Let a and b be in N. If b = 1, then a + b = a' using P1 and P2. If b ≠1, then c' = b with c in N using P4 and defined is a + b = a + c'.
Using this definition, you have now defined the number 2.
2 = 1'
2 is in N by P1, P2, and the definition of the number 2.
The Theorem is 1 + 1 = 2.
The Proof by using the definition of addition with a = b = 1, 1 + 1 = 1' = 2. Q.E.D.
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Nashrakh
Little White Butterflies Staff
So, what's your point?
You take college/university math courses? If not, it's no miracle you don't know many of those (I wonder why you would look up a a proof, then).
Too bad the scan is so crappy. But most of the symbols are covered in the first semester...
You take college/university math courses? If not, it's no miracle you don't know many of those (I wonder why you would look up a a proof, then).
Too bad the scan is so crappy. But most of the symbols are covered in the first semester...
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Nashrakh wrote...
So, what's your point?You take college/university math courses? If not, it's no miracle you don't know many of those (I wonder why you would look up a a proof, then).
Too bad the scan is so crappy. But most of the symbols are covered in the first semester...
It's for fun. *shrugs*
Was thinking about discussing mathematical theorems and postulates and turning the topic into one of intellectual discourse concerning the origins of mathematics through long complicated proofs.
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When the Peano axioms were first proposed, Bertrand Russell and others agreed that these axioms implicitly defined what we mean by a "natural number". Henri Poincaré was more cautious, saying they only defined natural numbers if they were consistent; if there is a proof that starts from just these axioms and derives a contradiction such as 0 = 1, then the axioms are inconsistent, and don't define anything.
That's basically it.
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Nashrakh
Little White Butterflies Staff
Too bad I don't know math lingo in English, we just had a proof like that in class - why 0.9999... is 1.
One other proof I once read is quite simple. For 1 not to be 0.999..., there has to be an x which suffices 1-0.999...=x
Since the chain of 9s is eternal, there is no such x=!0, which means 1=0.999...
One other proof I once read is quite simple. For 1 not to be 0.999..., there has to be an x which suffices 1-0.999...=x
Since the chain of 9s is eternal, there is no such x=!0, which means 1=0.999...