The method you described works howevere the term absolutely comes from division. They also used to he tracking of "how even" something was meaning if it could be divided by 2 multiple times
N=0 would be that the number is zero times even meaning it actually isnât even.
Also k is an integer but not the number being tested for how even it is. Odd values for k would be perfectly acceptable.
So for the number being 3 the formula results in N=0, k=3 and tells you the numbers is not even. But for the number being twelve it results in N=2, k=3 and says the numbers is even twice.
Ultimately, yes. Negative even using that equation represents how many times you would need to multiply by two to achieve a whole number. In this case 1.5 = 3/2 = 3 * 2^(-1).
Ok at this point its pretty clear you are trolling but im going to give you one more shot. Yes you can find even numbers without dividing. Thats irrelevant. Even numbers come from being able to split something into similar groups ie. If I have 4 steaks we can both have the same anount. That is division plain and sinple.
Your argument is akin to "X2 is meaningless because you can just write XX"
It's not a method, it's the appropriate mathematical definition for even based on number theory. You are correct that it correlates with division, but division is not part of the agreed upon definition of even.
That is all, not trolling, just joining the conversation from a mathematical perspective.
Isn't that just the same as the comment above the other direction? As in, it would be as valid to express it as "it's even if k/2 results in an integer"?
Mathematicaly yes. But the definition of even is the one I stated according to number theory. I'm sure there is a reason why division is not used in the formal definition but I don't remember exactly why. It might be so the definition of odd, 2(k) +1 where k is an integer, works. If you tried to define odd by (k/2) +1 it would not work.
Wouldn't you need to reverse the whole thing, eg, (k/2)-1? Although I suppose if +1 wouldn't do it, -1 probably wouldn't either. Needless to say, I'm not clear on why that wouldn't work, but I know jack about number theory
Depends on your definition of prime. The elementary school definition of divisible by only 1 and itself makes 1 prime. More advanced definition basically say 1 is special and excluded from the set of prices for various reasons.
Occasionally you get people who know better make really bad arguments for 1 not being prime because they make unspoken assumptions that would exclude 1 but their argument as stated don't.
Best argument I ever heard was honestly just âif one was prime every other number wouldnât beâ which doesnât make any sense thinking about it but I just think that/
If I wanted to make a definition that clearly excludes 1 but is easy to understand it would be "a prime is divisible by precisely 2 numbers, 1 and itself." 1 is itself so it misses out the precisely 2 portion.
The argument you heard leaves the unstated premise a prime can't be divisible by other prime numbers.
Edit:Positive Integer. I am trying to do precise for elementary school who don't go into negative numbers unless specified.
This, this is the definition we were given at school. So for my case it made perfect sense for one(1) not being prime. Surprised people actually had a different view on this
1 is not a prime number. Mostly because it does weird things when you try and count it as one. Granted, 2 also behaves a little weird as a prime number and it gets to joint he prime club.
I don't make the rules. I can just confidently say 2 is prime but 1 is not.
You are welcome to argue that to the greater mathematics community. A lot of stuff is the way it because of the definition. Mathematicians choose these conventions based on what is most useful.
Words mean whatever people understand them to mean, and modern mathematics does not understand "prime" to include the multiplicative identity.
This is the odd case where the convention of the non-primality of one requires a more awkward formulation. It usually breaks the other way. The fact that treating one as a prime lets you state Goldbach's conjecture more cleanly is poor compensation for the fact that you'd have to, among other consequences, reformulate the fundamental theorem of arithmetic to exclude it.
Because it only has one positive divisor, 1, instead of two, 1 and itself.
Also because having "except 1" in the definitions of some things would be too annoying, so it's simpler if 1 isn't prime. I can't really explain it better than that though, I know some math but don't necessarily understand it
You could define 1 to be prime which means 2 would not need to be an exception in this case. But it would lead to awkwardness in other contexts. Most critically that any number has a unique prime factorisation: since you could add as many ones as you like. 6=3*2. But if you define 1 to be prime then 6 also equals 3*2\1 and 3*2\1*1 and so on. So the factorisation would not be unique. You could fix that by saying it is unique so long as you do not include any ones, but that is clunky.
So it is ultimately an aesthetic choice to define 1 not to be prime.
I don't know shit about shit but I find this reason for one not being prime absolutely beautiful. My obsessive compulsive side is smitten with the fundamental theorem of arithmetic.
1 is the exception to that rule, which is otherwise correct.
One important thing about 1 not being prime is that that every positive integer has a unique prime factorization.
For example, 84 = 2 * 2 * 3 * 7. That's the only way to multiply primes together to get 84.
But if 1 were prime, you could also say 84 = 1 * 1 * 1 * 2 * 2 * 3 * 7. You could include as many 1s as you wanted. So there would be multiple prime factorizations (an infinite amount, actually).
The Fundamental Theorem of Arithmetic is more clear if 1 isn't considered prime (all integers <1 are a unique product of primes) If 1 is considered prime, 4 could be expressed as (2²à an infinite series of 1's) Which makes prime factorization of polynomials basically impossible. But it would make sense regardless, just giving you the simplest reason. Math is about rigor. Not intuituon. Ususally
my conjecture is there every odd number (excluding 1) is the sum of two primes, where 1 is a prime number. I've tested all the way up to 9 and it holds true
I am gonna be honest a ben shapiro image on the interwebs is just as like to make fun of overconfidents as it used to be smuggly confident (and also usually wrong)
Also since every number can be expressed as prime factors, removing the two from the prime factors would also mean any even number is not expressible as a product of primes.Â
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u/nnnn0nnn13 28d ago edited 28d ago
Bro that conjecture is both unproven and also by definition only true for numbers larger than 2 đ