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.