If 0^0=1 then 0^0=0/0
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haha
Although, 0-1 = -1, bit 0 in math
So 0^(0-1) actually = 0^-1, which is actually 1/0, so that number isnt real, because anything divided by 0 isnt a real number
Oct 24, 2019
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#5
DavidNorman435 wrote:
So 0^(0-1) actually = 0^-1, which is actually 1/0, so that number isnt real, because anything divided by 0 isnt a real number
Yes that is true as well. But my point being that 0^0 is not a real defined number either; as is 1.
Oct 24, 2019
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#6
If n/n=n^(1-1)=n^0 why should there be an acception made for 0, just for "conveinience" of operations?
Ok, so i see what you mean,
3/3 = 1,
And 3^0 = 1,
So 3/3 = 3^0
But in order to understand that, we mist know how anything to the power of 0 is equal to 1 before we can refute/defend anything on that subject, but i forgot the explanation.
Would you mind enlighting me?
3/3 = 1,
And 3^0 = 1,
So 3/3 = 3^0
But in order to understand that, we mist know how anything to the power of 0 is equal to 1 before we can refute/defend anything on that subject, but i forgot the explanation.
Would you mind enlighting me?
Must*
I think it was something like n^0 = 0/n, so, in this case n=0 (the number of acception), then 0/0 =1 because anything divided by itself equals 1
So, technically it is an acception, but it kind of makes sense because i guess, like you're saying, anything divided by 0 isnt a real number, so it kind of doesnt make sense at the same time
And i think "acception" is spelled "exception"
And also, 0 (speaking literally here) is the equivalent to nothing, and nothing (the thing, not the nothingness to it) normally gets the exception
Oct 24, 2019
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#13
If we know that the 0th power of 4 is 1, because 4^(1-1)=(4*1)/4=1 and like wise 1*4=4. Then we know that 3 is the 0th power of 0 because 0^(1-1)=(0*3)/0=3
To say 0^0 equals just 1 can be proven to be false. Yet Google Calculator and schools across the world, insist it to be true.
Oct 24, 2019
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#14
We know that infinity^0=any real positive number n, because n times infinity still equals infinity for any real that is positive.
Infinity^0=infinity^(1-1)=infinity/infinity equals any positive real n, because any real positive n times infinity is still infinity.
So why should there be an exceptional rule for 0?
Answer, by the examples shown above there isn't.
Oct 24, 2019
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#15
0/0 and infinity^0 are both indeterminates. In other words, they do not have a defined value, which means their value can technically be any real number. That is calculus at its finest.
And nothing can equal anything, so technically, he's right
Oct 25, 2019
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#18
Reminds me of a "proof" from my high school days (too many years ago) that "proved 1=2. The sequence began with a=b and proceeded to add and multiple variables to both sides. Mathematically, all of the equations seemed to make sense. It was only upon closer inspection that one of the multiplicands was seen to be (a-b) and since the premise started with a=b, then a-b was zero (0).
Of course, anything multiplied by 0 is, well, 0. So at the midway of this "proof" we had 0 = 0. However, since the theorem was using the letters a and b, this was not readily noticeable so when it finally "proved out" we had something like 2a=b, and since a=b; 2a=a or 2=1.
PS. As I remember it, division by zero is "undefined", so cannot be done. (I said "as I remember it", we are going back nearly half a century!!)
Oct 26, 2019
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#19
m_connors escribió:
Reminds me of a "proof" from my high school days (too many years ago) that "proved 1=2. The sequence began with a=b and proceeded to add and multiple variables to both sides. Mathematically, all of the equations seemed to make sense. It was only upon closer inspection that one of the multiplicands was seen to be (a-b) and since the premise started with a=b, then a-b was zero (0).
Of course, anything multiplied by 0 is, well, 0. So at the midway of this "proof" we had 0 = 0. However, since the theorem was using the letters a and b, this was not readily noticeable so when it finally "proved out" we had something like 2a=b, and since a=b; 2a=a or 2=1.
PS. As I remember it, division by zero is "undefined", so cannot be done. (I said "as I remember it", we are going back nearly half a century!!)
Dividing by zero is likely one of the worst nightmares in math students, I've certainly fallen for it!
m_connors wrote:
Reminds me of a "proof" from my high school days (too many years ago) that "proved 1=2. The sequence began with a=b and proceeded to add and multiple variables to both sides. Mathematically, all of the equations seemed to make sense. It was only upon closer inspection that one of the multiplicands was seen to be (a-b) and since the premise started with a=b, then a-b was zero (0).
Of course, anything multiplied by 0 is, well, 0. So at the midway of this "proof" we had 0 = 0. However, since the theorem was using the letters a and b, this was not readily noticeable so when it finally "proved out" we had something like 2a=b, and since a=b; 2a=a or 2=1.
PS. As I remember it, division by zero is "undefined", so cannot be done. (I said "as I remember it", we are going back nearly half a century!!)
Let a = b.
a² = ab
a² - b² = ab - b²
(a+b)(a-b) = b (a-b)
a+b = b
2b = b
2 = 1
If 0^0=1 and only 1, then it follows that
0^(0-1) shall equal (0^0)/0=1/0 and likewise
0^(0+1) shall equal (0^0)*0=1*0=0
Therefore (1/0)*0=0/0 and equals 0^0
If mathematics is a consistent science we must follow it's rules of operation.