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22 Jun 18
1 edit
D. It isn't a big deal. Your little '2 as a prime factor on one side is odd, on the other is even ' is neater. It isn't revolutionary so.
You WOULD have lost marks in a UK 16 yr old Maths exam by your careless use of rational rather than irrational in your conclusion, which may have been deemed to show 'choice of answer' (Winky face)
22 Jun 18
The post that was quoted here has been removedstill stuck on our current number system. poor duchess. It's just a tool to explain what we see around us....what happens when that tool can't explain what we see?
A better number theory would start with the premise that there is only 1 number that exists and that number is 1. All other numbers just tell you how many more than 1 you have. Start with that premise as your basis for a new number theory and imagine the possibilities......
22 Jun 18
The post that was quoted here has been removedI think that Blood on the Tracks is referring to the following:
Suppose sqrt(2) is a rational number, then there exist co-prime integers A and B s.t. sqrt(2) = A/B.
We then have that:
A^2 = 2B^2
The right hand side is explicitly even, this means that the left hand side is even. Since the square of an odd number is odd and A^2 is even A must be even. So we can find a = A/2, substituting this into the above and simplifying we now have that:
B^2 = 2a^2.
This shows that B^2 must be even and so B contains a factor of 2. So both A and B have factors of 2 contradicting our statement that they be co-prime and disproving that root two is rational. Therefore the square root of two is irrational.
Note for non-mathematicians: co-prime means that the two numbers do not share any non-trivial factors, so 15 and 8 are co-prime because 15 has factors 5 and 3 and 8 has factors of 2 and 4. Non-trivial, in this context, means not 1.
22 Jun 18
Originally posted by @uzless"Food for thought."
It's possible to find the square root of 2 to be a rational but there is one big caveat. you can't use our current number theory to prove it. You have to use non-numerical theory, or cognitive intuition if you will.
First consider that 2 is a complete number with no decimals or fractions. It is just 2. Easy enough.
Now comes the tricky part.
Con ...[text shortened]... ool. Our tool is not designed to answer, What is the square root of 2.
Food for thought. 😉
if you are not picky about the taste of thought food. or how sick it's going to make you. or how much will high schoolers with basic math training laugh at you.
22 Jun 18
Originally posted by @uzlesscognitive intuition = guessing?
It's possible to find the square root of 2 to be a rational but there is one big caveat. you can't use our current number theory to prove it. You have to use non-numerical theory, or cognitive intuition if you will.
First consider that 2 is a complete number with no decimals or fractions. It is just 2. Easy enough.
Now comes the tricky part.
Con ...[text shortened]... ool. Our tool is not designed to answer, What is the square root of 2.
Food for thought. 😉
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22 Jun 18
Originally posted by @deepthoughtindeed I was
I think that Blood on the Tracks is referring to the following:
Suppose sqrt(2) is a rational number, then there exist co-prime integers A and B s.t. sqrt(2) = A/B.
We then have that:
A^2 = 2B^2
The right hand side is explicitly even, this means that the left hand side is even. Since the square of an odd number is odd and A^2 is even A must ...[text shortened]... 15 has factors 5 and 3 and 8 has factors of 2 and 4. Non-trivial, in this context, means not 1.
it is certainly a less neat way than D64's argument involving prime factors
22 Jun 18
2 edits
Originally posted by @blood-on-the-tracksThe catch is that although Duchess64's proof is correct, as presented here it silently invokes a couple of theorems which are not entirely obvious. The first is the fundamental theorem of arithmetic, that there is a unique prime factorization of any given number. The second is that the ratio of products of non-equal primes is non-integer. Write the prime factorization of A and B as follows:
indeed I was
it is certainly a less neat way than D64's argument involving prime factors
A = (2^a_2)(3^a_3)...(p^a_p)...
B = (2^b_2)(3^b_3)...(p^b_p)...
Note that these products terminate as A and B are required to be finite. Since we can always choose A and B to be co-prime if a_n is non-zero then b_n is zero and vice versa. We have that:
2 = 2^2(a_2 - b_2)3^2(a_3 - b_3)...p^2(a_p - b_p)...
Where in each term one of a_x or b_x is zero. Now one has to argue that the right hand side is not an integer (it isn't). The ratio of two primes is not an integer, but one has to show that the ratio of products of non-equal primes are necessarily fractional, which isn't entirely trivial. This is why you were shown the even/odd proof I presented above, it doesn't rely on additional theorems.
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22 Jun 18
Yes, I had already seen the extension there, replace '2' with 'k' (k is prime) and the contradiction still stands.
It also can be used to show that root ( any positive integer which is not a square number) is also irrational. They all, upon prime factor decomposition must have a (or more) prime factors that occur an 'odd' number of times, thus lending themselves to the argument.
We have had our differences, but I have found this exchange refreshing.
23 Jun 18
The post that was quoted here has been removedI've got my head around this, your proof relies only on the fundamental theorem, since it states that the prime factorization is unique. Consequently the left and right hand side cannot have different numbers of factors of any of the primes. Even so, the proof of the fundamental theorem of arithmetic involves the Chinese Remainder Theorem - which isn't the most straightforward and why Blood's teachers gave him the other proof.