>>3
wrong,
1 = 0 + 1 assumes nothing about 1 being greater or less than 0
a = b + a
0 = b
a could be anything. i.e. -2 + 0 = -2, -2 is not greater than 0
but the argument does fail if 1 doesn't have to be positive, which is what the proof is attempting.
but 1 is the smallest positive integer, so it is > 0 by definition.
>>11
Huh in English like this?
Step 1 - 1 does not equal 0.
Step 2 - 1 does not equal 0 if and only if not 1 equals 0.
Step 3 - Using Step 1 and Step 2, not 1 equals 0.
Step 4 - 1 is an element of the real number set.
Step 5 - Using Step 4, not 1 equals 0 implies 0 is less than 1 multiplied by 1.
Step 6 - Using Step 3 and Step 5, 0 is less than 1 multiplied by 1.
Step 7 - 1 is an element of the complex number set.
Step 8 - Using Step 7, 1 multiplied by 1 equals 1
Step 9 - Using Step 6 and Step 8, 0 is less than 1
The proof is rather simple. Originally when I saw the problem I thought that you would go to go back to logical definitions of the natural number set and the ordering operator, potentially messy.
Name:
Anonymous2006-05-19 10:48
didnt get it... why not 1 equals 0 implies that 0 is less than 1x1?
Haha dumbasses, mathematics is simply a model of real life occurances. 1 > 0 is something that has to be observed to be understood, or it is an "a posteriori" concept in unneccesarily complex speak.
Name:
Anonymous2006-05-21 9:16
1 > 0 is all about axi-oms,
Name:
Anonymous2006-05-23 10:07
what aboot 24623572457^468 being bigger than 0,000000000000000000000000000000000000000000000000000001 then?
Name:
Anonymous2006-05-23 11:49
probably a consequence of 1 being greater than 0
Name:
Anonymous2006-05-23 13:06
What is the smallest positive number
Name:
Anonymous2006-05-23 16:56
0={} is contained in 1={{}}, but {{}} is not contained in {} therefore 0<1. qed.