Could the reason why no one knows what happening to the factors of a number when adding 1 be because we defined x^0 = 1 ?
>>7722003
>defined
x^0=x^(n-n)=x^n/x^n=1
>>7722054
What is the difference between this and this:
suppose that a = b
then a2 = ab
subtracting b2 from both sides gives us a2 – b2 = ab – b2
factorising gives (a + b)(a-b) = b(a-b)
dividing both sides by a-b gives a+b = b <- Here is the problem, we are not allowed to divide by zero!
But also a+b = b+b = 2b, so 2b = b. Dividing by b gives 2 = 1
We are not allowed to divide by zero because it results in shit, yet we are allowed to exponentiate n-n which also results in shit. What am I missing?
>>7722102
>>>7722098 (You)
>Are you baiting me?
I wish I would,
You said that x^0=x^(n-n)=x^n/x^n=1 , this defines some x^0 to be 1 now let y^0=y^(n-n)=y^n/y^n=1 define some y^0 to be 1 and y!=x
This means x^0=x^(n-n)=x^n/x^n = 1 = y^n/y^n = y^(n-n) = y^0 but we said y != x so this is true only if we define y^0 and x^0 = 1 in the first place.
>>7722113
what you're saying is:
a!=b
a*0=b*0 IS NOT POSSIBLE
get the fuck out
>>7722119
>a*0=b*0 IS NOT POSSIBLE
What the fuck man what has this to do with anything Ive said
>>7722098
kek, I like this
>>7722003
Nope. /thread
