Comments on: The Powers of Two http://www.exploringbinary.com/the-powers-of-two/ Binary Numbers, Binary Code, and Binary Logic Mon, 30 Jan 2012 01:37:52 +0000 hourly 1 http://wordpress.org/?v=3.2.1 By: sami morsi http://www.exploringbinary.com/the-powers-of-two/comment-page-1/#comment-5387 sami morsi Thu, 22 Sep 2011 11:18:31 +0000 http://www.exploringbinary.com/?p=148#comment-5387 dear Messers Your presentation is very fascinating, easily absorbed especially for those people who do feel with mathematic phobia. Thank you so much, and much greetings from the first cairo, egypt. dear Messers
Your presentation is very fascinating, easily absorbed especially for those people who do feel with mathematic phobia. Thank you so much, and much greetings from the first cairo, egypt.

]]> By: Rick Regan http://www.exploringbinary.com/the-powers-of-two/comment-page-1/#comment-70 Rick Regan Thu, 20 Nov 2008 20:28:11 +0000 http://www.exploringbinary.com/?p=148#comment-70 <a href="http://www.exploringbinary.com/about-the-math-notation/" title="Read Article “About the Math Notation”" rel="nofollow">I haven't enabled math in comments</a> so we'll have to do this without LaTeX.... Your answer IS better; his is circular. ln(1)=0 is the same as saying e^0=1, and e^0=1 is the same as saying ln(1)=0. But why is ln(1)=0, or equivalently, why is e^0=1? Because it's defined that way. I said parenthetically in the article ``so that exponent arithmetic works out.'' Let me elaborate (I'll put it in terms of base 2, but any base will do). <a href="http://www.exploringbinary.com/composing-powers-of-two-using-the-laws-of-exponents/" title="Read Article “Composing Powers of Two Using The Laws of Exponents”" rel="nofollow">The laws of exponents</a> say 2^a / 2^b = 2^(a-b). To make this rule work for a=b, you have to allow 2^0=1: 2^a/2^a = 1, so 2^(a-a) = 1, so 2^0 = 1. I haven’t enabled math in comments so we’ll have to do this without LaTeX….

Your answer IS better; his is circular. ln(1)=0 is the same as saying e^0=1, and e^0=1 is the same as saying ln(1)=0. But why is ln(1)=0, or equivalently, why is e^0=1? Because it’s defined that way. I said parenthetically in the article “so that exponent arithmetic works out.” Let me elaborate (I’ll put it in terms of base 2, but any base will do).

The laws of exponents say 2^a / 2^b = 2^(a-b). To make this rule work for a=b, you have to allow 2^0=1: 2^a/2^a = 1, so 2^(a-a) = 1, so 2^0 = 1.

]]> By: Steve Pixton http://www.exploringbinary.com/the-powers-of-two/comment-page-1/#comment-68 Steve Pixton Thu, 20 Nov 2008 02:01:57 +0000 http://www.exploringbinary.com/?p=148#comment-68 I recall in one of my calculus classes, when we were studying logs, natural logs, and the number e, the professor asked "why does e^0=1?". I was the only one to raise my hand. I said "because any number raised to the zero = 1." He was a bit taken aback, and said "true, but not what I was looking for." Then he waited, until someone else said "because ln(1)= 0." He said "right". (You see, he had previously explained how if ln(x)=b, then e(ln(x))=e(b), then x=e(b). e to the ln of something "cancels"). I still like my answer better. So in binary, another answer to why is 2^0=1 is because log(base 2)(1)=0. I recall in one of my calculus classes, when we were studying logs, natural logs, and the number e, the professor asked “why does e^0=1?”. I was the only one to raise my hand. I said “because any number raised to the zero = 1.” He was a bit taken aback, and said “true, but not what I was looking for.” Then he waited, until someone else said “because ln(1)= 0.” He said “right”. (You see, he had previously explained how if ln(x)=b, then e(ln(x))=e(b), then x=e(b). e to the ln of something “cancels”). I still like my answer better. So in binary, another answer to why is 2^0=1 is because log(base 2)(1)=0.

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