4 ms·
If you are rigorous in your expansion of "it follows that x is not two", you will find your contradiction. Here is a more rigorous form of what you've said the
by gamegoblin 12y ago
If you are rigorous in your expansion of "it follows that x is not two", you will find your contradiction.
Here is a more rigorous form of what you've said there:
Let X be an integer with a prime factorization in the form 2^(2k) * p_1 * p_2 ... * p_n where k is a positive integer and p_i for any i is a prime not equal to 2.
Proposition: X is not 2
Proof:
Assume X is 2.
The prime factorization of X is:
X = 2
By the definition of X, we have:
2 = 2^(2k) * p_1 * p_2 ... * p_n
Since we defined k to be a positive integer, we have
min(2^(2k)) = 2^(2(min(k)) = 2^(2*1) = 4
So we have:
2^(2k) >= 4
So we have:
2^(2k) > 2
Since p_i for all i are all positive integers, we have:
2 < 2^(2k) * p_1 * p_2 ... * p_n
So we have:
2 != 2^(2k) * p_1 * p_2 ... * p_n
Thus:
2 != X
And we have arrived at a contradiction.
- ikeboy 12y agoFirst 3 lines are same as yours. Consider k. Either it is 0, or greater. Consider 0. In that case, X is not divisible by 2, hence cannot be 2. Now consider k positive. (My original proof didn't show k wa positive, so I needed to deal seperately.) Use your proof to show that 2^(2k) > 2, and then 2!=X. As far as I can tell you never actually used the assumption that 2=x, so that can be deleted from the proof and it remains valid.
- ebola1717 12y agoI think "Proof by cases" uses proof by contradiction. You're proving that given A -> C and !A -> C, then C, which requires the axiom A v !A, i.e. proof by contradiction. In your proof, you've proven k=0 -> X != 2, and k > 0 -> X !=2, but it doesn't follow that X != 2 without contradiction.
- ikeboy 12y agoI proved k was a whole number previously. I don't need that axiom, because I can prove it in the case of k.