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1/0 is undefined because you are basically asking what number times 0 or how many zeroes should you add so that you get 1, even if you add infinite zeroes the answer would be 0.
What about this? Assume 00 is undefined. Then 01=∞, but 0×∞=1, since to obtain this, we would have to multiply 01 by 0, resulting in 00×1, which would then be undefined.
@Zakir Husain
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This is true, but only if we assume that there is both a positive and negative value of infinity. What if we have a single point of infinity where everything ends, similar to 0, where everything begins. Then we would get the situation I described in this discussion. Please note this is all speculation on my part. Definitely fun to think about!
Well, I just watched a video by Eddie Woo on 00=? and he solves the question by using limits. He says that it is undefined, but it seems as if its one, so we have agreed upon that value, until we find a better solution. He shows that limx→0xx=1 by calculating x^x for decimals and showing that is approaches 1.
Easy Math Editor
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Percy, here's my opinion on the value of 01, and here's my general opinions on working with ∞ (my comment, not the discussion).
01=∞ as then ⇒ 1=0×∞ ⇒ 1=0
This is true for all numbers divided by 0, so division by 0 is just undefined......
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If 01 "is" =∞, then we have an answer for ∞×,0 which will be 1
@Percy Jackson
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But we know that 0×x is always 0 with any value of x.
@Hamza Anushath
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@Percy Jackson
Only with finite values of x
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x is a NUMBER and ∞ is NOT a NUMBER ∴ x alway have to be finite
x→∞lim(x×0) then, x→∞limx×0=0
And if you are talking about :1/0 is undefined because you are basically asking what number times 0 or how many zeroes should you add so that you get 1, even if you add infinite zeroes the answer would be 0.
@Hamza Anushath
If limn→0nn=0⇒00=1
The result will be different if you include complex numbers (something like lima→0limb→0(a+bi)a+bi or lima→0(a+ai)a+ai or lima→0(ai)a, etc ).
∴00 is undefined.
Also if limx→af(x)=k⇒f(a)=k
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@Hamza Anushath see this if you don't believe me.
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Link Isn't the topic still debatable?
What about this? Assume 00 is undefined. Then 01=∞, but 0×∞=1, since to obtain this, we would have to multiply 01 by 0, resulting in 00×1, which would then be undefined.
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But why 01 must be ∞?, why not UNDEFINED
Also f(x)=x1 is discontinuous at x=0 so you can't give f(0) any value
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0, where everything begins. Then we would get the situation I described in this discussion. Please note this is all speculation on my part. Definitely fun to think about!
This is true, but only if we assume that there is both a positive and negative value of infinity. What if we have a single point of infinity where everything ends, similar toPercy, 0^0 is not equal to 1...
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Yes that's just the*best *answer
Well, I just watched a video by Eddie Woo on 00=? and he solves the question by using limits. He says that it is undefined, but it seems as if its one, so we have agreed upon that value, until we find a better solution. He shows that limx→0xx=1 by calculating x^x for decimals and showing that is approaches 1.
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(x→alimf(x)=A)⇒(f(a)=A) (f(a)=A)⇒(x→alimf(x)=A)
Also you are looking only for Real limits not Complex limits
Contradiction in Probability
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Haha, nice @Zakir Husain!!!