Realintruder

We know that a^0=a/a as it is exponently one less than a^1 - the number needed to multiply number x to equal a^1 where x equals any number. This is why 0^0=any number can equal x. because with x^0 the question is what when mutiplied by a equals x^1 with 0 this can equal any number.

infinity^0=infinity^(1-1)=infinity/infinity equals any positive number.

The answer is yes! infinity*0= infinity (1-1)=infinity-infinity, which equals any number. because infinity-infinity-3 is absorbed in infinity like a blackhole. and still equals infinity-infinity, likewise infinity-infinity-5 equals the same thing.

Thank you. But the software will not let me edit my post. What is the standard proof for the idea?

If we assume that (-1) (-1)=-1 and (-2)(-2)=-4 but know that (0) (0)=0 Then (1+-1) (1+-1) must equal 1-1-1-1 =-3 and not the sum of (0) (0) which it should equal 0 unless we have assumed otherwise but have not. Likewise (-1+2) (-1+2) =(-1)(-1) must equal -1+-2+-2+4=-7 and not the product of (-1+2) (-1+2)=(-1)(-1) which should equal -1, which is inconsistent and does not equal -7 if we assume that (-1)(-1)=-1. But if we assume that (-1)(-1)= 1 then (1+-2)(1+-2) should equal 1+-2+-2+4=1 likewise (1+-1)(1+-1)=(0) (0)=1-1-1+1=0 which it does

Thank You for the well written response. I think the question has been answered just fine despite all the needless replies.

If a/b is greater than 1 than b/a must greater than 0 but less than 1.<br><br> For every number greater than 1 there is a multiplicative inverse greater than 0 but less than 1. <br><br> Examples 3/4-4/3, 7/8-8/7, 9/10-10/9, 1/3-3/1, 3/5-5/3 <br><br> But is there a proof for this?<br><br>

Not so. Infinity times 0 equals any number because infinity (1-1) which it is equal to, is equal to anything (infinity-infinity) equals any number. Thus infinity is not treated as just a concept but as an interminately large quanity when added any finite number to one side does not make a difference.

Not so, as discussed earlier. For if 0^0 indeniably equals 0^(1-1)=(0^1)/(0^1)=0/0 Except in practical applications, which only exist in "man made' settings, do we have 0^0=1, otherwise it exists only, as the number when multiplyed by 0 equals 0, such that is exponently less than 0 that is 0. If x is 0 then when: x^0 times x^1 must equal base 0 as well. When x^0 equals 0 that holds true for any number complex or real for x^0.

<br><br> Not if you multiply one by itself infinity times.<br><br> (1^0)[which is redundant]^infinity equals 1^(0*infinity)=1^(infinity-infinity) which is indeterminate. Since infinity minus infinity can equal any real number.<br><br> But if change the base to 3, and substitute 3 for 1 in the above equation it shall not at all change the above value and we get 3^(infinity-infinity) which we can further reduce to <br><br> (3^infinity)/(3^infinity)=infinity/infinity which equals any positive value.<br><br> As when you multiply the above by any positive value it still equals the above of infinity/infinity<br><br> What about (3^0)^infinity=1...to the infini-ith that is base <b>3 to the 0</b> to the infinity equal <b>1</b> you have a comprehension problem with is beyond me.

I meant for this to go on a seperate block but the software would not let me. I cannot edit it as it has timed out that I can no longer edit recent posts on this page, so I must repost it!:<br><br> Likewise if 0^0=1 then (0^0)^-1 should equal 1 since 1^(-1) is equal to 1. But it does not (0^0)^-1=(0^-1)/(0^-1)=(1/0)/(1/0)=(1/0)*(0/1)=0/0

What is infinity times the imaginary unit, that is also the square root of negative 1, i?

If we allow b to equal 1 and 0^b=0^1=0=0^1 Then 0^(b-1)=0^0=0^(1-1)=(0^1)/(0^1)=0/0 Let's hear your argument against that. Likewise if 0^0=1 then (0^0)^-1 should equal 1 since 1^(-1) is equal to 1. But it does not (0^0)^-1=(0^-1)/(0^-1)=(1/0)/(1/0)=(1/0)*(0/1)=0/0

Does anyone know?

Yes my therom or theory has been disproven wrong. Thank You I wonder if it hold true for all counting numbers less than 1,000. Have you tested that out? That would be something.

Yes If we let 3^0 then; (3^0)^infinity=3^(0*infinity)=3^((1-1)*infinity))=3^(infinity-infinity)(3^infinity)/(3^infinity)=infinity/infinity) <br><br> which equals any positive number.

When solving for 0^0 the question remains what number when multiplyed by 0^1 equals 0. The answer is any number. Just as 1^0 is the number when multyplied by the base 1 equals 1^1, the answer is only 1.<br><br> Or more conscisely what does 0^(1-1)=0^0 equal. The answer is (0^1)/(0^1)=(0/0)=0^0 which is not 1. The teaching in schools all across the world that 0^0=1 is a conspiracy.

My formula has not been proven wrong yet. What works to produce the same result is of (-1)(-1) does not equal (-1) but 1 is m^2-2mn-2(mn)^2 as stated in my forum on that link in the bottom of my post. But if you use m^2-2mm-n^2 as suggested by you for 1,000 you will get 2004001 which is a prime (or relative prime) number. And has no factors but itself and 1.

Thank tou for the advice I will use it.

Something that should be punched into college computing systems although out the nation:<br><br> If m is a counting number, of the set 1,2,3etc and m=n-1 And we assume that (-1) (-1) always equals = -1 (and is not equal to 1) Then (m-n)(m-n) shall always equal the negative opposite of a prime or relitively prime number! A prime number being one that is only divisible by itself and the number one! A relative prime being the number that can only be expressed as the product of two primes and no more. Examples: (1-2)(1-2)=1-2-2-4=-7 (prime) (2-3)(2-3)=4-6-6-9=-13 (prime) (3-4)(3-4)=9-12-12-16=-31 (prime) (4-5)(4-5)=16-20-20-25=-49 (relative prime (-7)*7) (5-6)(5-6)=25-30-30-36=-71 (prime) (6-7)(6-7)=36-42-42-49=-97 (prime) (7-8)(7-8)=49-56-56-64=-127 (prime) (8-9)(8-9)=64-72-72-81=-161 (relative prime (-7)*23) (9-10)(9-10)=81-90-90-100=-199 (prime) (10-11)(10-11)=100-110-110-121=-241 (prime) (11-12)(11-12)=121-132-132-144=-287 (relative prime (-7)*41 (12-13)(12-13)=144-156-156-169=-337 (prime) (13-14)(13-14)=169-182-182-196=-391 (relative prime (-17)*23) So far all the relative primes are divisible by 7 with the other factor being a prime. except for the 13 result. Unfortunately the 4th product and 8th product are relative while the 12th not being so, but this happening on the 11th. The 13th result being -391 which is equal to 17*23 not being a factor of 7 either. What pattern the relatives fall under the primes do not, is not yet answerable. If you can compute futher, please post the results below here. http://theprimenumberformula.wordpress.com