Realintruder Posted February 2, 2015 Share Posted February 2, 2015 (edited) 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. Edited February 2, 2015 by Realintruder Link to comment Share on other sites More sharing options...

fiveworlds Posted February 2, 2015 Share Posted February 2, 2015 (edited) no 1^infinity=1 because no matter how many times we multiply 1 by 1 we still get 1 same with infinty√1 being=1 0^infinity=0 infinity√0=0 Edited February 2, 2015 by fiveworlds Link to comment Share on other sites More sharing options...

Realintruder Posted February 3, 2015 Author Share Posted February 3, 2015 no 1^infinity=1 because no matter how many times we multiply 1 by 1 we still get 1 same with infinty√1 being=1 0^infinity=0 infinity√0=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. Link to comment Share on other sites More sharing options...

ajb Posted February 3, 2015 Share Posted February 3, 2015 no 1^infinity=1 because no matter how many times we multiply 1 by 1 we still get 1 But the problem is that infinity is not a real number and so you have to be very careful using it as if it were. As the expression given could have more than one 'sensible' meaning it is taken to be indefinite for safety reasons. Link to comment Share on other sites More sharing options...

imatfaal Posted February 3, 2015 Share Posted February 3, 2015 Not if you multiply one by itself infinity times. (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. As when you multiply the above by any positive value it still equals the above of infinity/infinity 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. Not so - 1 is 1 in all bases. Unity in base 10 is the same as unity in base 3. Link to comment Share on other sites More sharing options...

John Posted February 3, 2015 Share Posted February 3, 2015 [math]1^{\infty}[/math] is an indeterminate form, yes, though the reasoning presented in the original post doesn't entirely work, since, as mentioned, infinity isn't simply a number we can toss in for the purposes of arithmetic.By the same token, since infinity isn't a number, when we see something like [math]1^{\infty}[/math], we should understand it as shorthand for the result of some limiting process. Furthermore, there are various ways to approach this limit, and various paths lead to various values. This is why it is considered indeterminate. Link to comment Share on other sites More sharing options...

Realintruder Posted February 10, 2015 Author Share Posted February 10, 2015 [math]1^{\infty}[/math] is an indeterminate form, yes, though the reasoning presented in the original post doesn't entirely work, since, as mentioned, infinity isn't simply a number we can toss in for the purposes of arithmetic. By the same token, since infinity isn't a number, when we see something like [math]1^{\infty}[/math], we should understand it as shorthand for the result of some limiting process. Furthermore, there are various ways to approach this limit, and various paths lead to various values. This is why it is considered indeterminate. 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. Link to comment Share on other sites More sharing options...

ajb Posted February 10, 2015 Share Posted February 10, 2015 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. The point is that infinity is not a real number and so you have to take care with expressions involving it, as your example here shows. The usual thing is to understand it in terms of limits, as already stated. If the limit is not well defined then the natural thing to do is to take the expression as being indeterminate. Link to comment Share on other sites More sharing options...

Unity+ Posted February 10, 2015 Share Posted February 10, 2015 no 1^infinity=1 because no matter how many times we multiply 1 by 1 we still get 1 same with infinty√1 being=1 0^infinity=0 infinity√0=0 Not necessarily. Although not as related, an example of the assumption fallacy is the summation of 1 to infinity. Although we think it is infinity, it actually would be -1/12(from Numberphile, a great YouTube Channel). Link to comment Share on other sites More sharing options...

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