tjohnson 0 Posted November 27, 2008 Share Posted November 27, 2008 On Roger's new website there is an essay about zero exponents in which he explainsFor any real number, r, a positive exponent, n, indicates that the unit 1 is to be considered as having been multiplied by r a total of n times. A negative exponent, -n, indicates that the unit 1 is considered as having been divided by r a total of n times.I just want to say this is a nice way of looking at it but it's even easier than this. One simply needs to look at the expression x^n/x^n = x^(n-n) = x^0 and x^n/x^n = 1 so x^0 = 1 for all x. Most of the "mysteries" of mathematics are based on misunderstanding of how mathematics works. Link to post Share on other sites

BaalChatzaf 60 Posted November 27, 2008 Share Posted November 27, 2008 On Roger's new website there is an essay about zero exponents in which he explainsFor any real number, r, a positive exponent, n, indicates that the unit 1 is to be considered as having been multiplied by r a total of n times. A negative exponent, -n, indicates that the unit 1 is considered as having been divided by r a total of n times.I just want to say this is a nice way of looking at it but it's even easier than this. One simply needs to look at the expression x^n/x^n = x^(n-n) = x^0 and x^n/x^n = 1 so x^0 = 1 for all x. Most of the "mysteries" of mathematics are based on misunderstanding of how mathematics works.Correct. Defining x^0 = 1 (for non-zero x) is a way of making the law of exponent addition work in general. The reason why x has to be non-zero is as follows: 0^0 = 0^n/0^n = 0/0 which is undefined. Ba'al Chatzaf Link to post Share on other sites

BaalChatzaf 60 Posted November 27, 2008 Share Posted November 27, 2008 On Roger's new website there is an essay about zero exponents in which he explainsFor any real number, r, a positive exponent, n, indicates that the unit 1 is to be considered as having been multiplied by r a total of n times. A negative exponent, -n, indicates that the unit 1 is considered as having been divided by r a total of n times.I just want to say this is a nice way of looking at it but it's even easier than this. One simply needs to look at the expression x^n/x^n = x^(n-n) = x^0 and x^n/x^n = 1 so x^0 = 1 for all x. Most of the "mysteries" of mathematics are based on misunderstanding of how mathematics works.That quote you gave is problematic. 1^1 = 1 and 1^0 = 1. One the face of it, why should they come out the same. Ba'al Chatzaf Link to post Share on other sites

tjohnson 0 Posted November 27, 2008 Author Share Posted November 27, 2008 That quote you gave is problematic. 1^1 = 1 and 1^0 = 1. One the face of it, why should they come out the same. Ba'al ChatzafYes, well Roger's explanation is an intuitive one which may help some people but it is not very rigorous. (nor was it supposed to be, I dare say) Link to post Share on other sites

anunusualname 0 Posted October 22, 2010 Share Posted October 22, 2010 (edited) I don't see the problem. Could someone explain it to me? 1^0= One multiplied by 1 raised to no power or not multiplied by itself any number of times. Whereas 1^1= One multiplied by 1 raised to a power of 1 or multiplied by itself one time.It is either 1x1x=1 or 1x1x1=1. I don't know whether or not exponents are actually defined that way, but if they are, I don't see a problem with why each of those produces a product of 1.Edited: I suppose you have to strike out an x.Edited: 7^0=7 then according to my understanding of the definition of exponents given. That is not so, is it?Edited: I see a problem.Edited: OH, you have to strike out two x's. It is either 1^0=1x1x=1 or 1^1=1x1x1 or 7^0=1x7x=1Edited: So, 7 = 7^1 = 1x7, whereas 7^0 = 1x7Edited. So, I was being confused. But I think I get it now. ^0 makes sense if you understand an expression like 7 being a simplification of 1x7. The ^0 addresses 1x7, but not 7 itself. That is interesting, but I'm not sure how valid it is. No, wait. 0^0 is undefined? It isn't 1x0 addressed in the same manner as 1x7, which would also make it produce 1? I suppose this still doesn't make any sense to me.Edited: Is math about fudging the equations until you get an answer that you can live with?Edited: Hm, I retract the question. I suppose one has to be careful with one's zeros. Edited October 22, 2010 by Brian Bahneman Link to post Share on other sites

BaalChatzaf 60 Posted October 22, 2010 Share Posted October 22, 2010 For x != 0 x^0 is -defined- to be 1 in order to make the law of adding exponents in multiplication complete general. That is the law must work for both positive and negative exponents. Consider x^n and x^(-n) which is 1/x^n. mutliple x^n by 1/x^n to get 1. But this is x^n * x^(-n) and by adding exponents for the multiplication we get x^(n + -n) which x^0. And that is x^0 = 1 for an nonzero x. For x = 0 x^0 is not defined, 0^n/0^n = 0/0 which is not defined.When I think of the level of mathematical preparedness in this country (the U.S.) I weep for the Republic.Ba'al Chatzaf Link to post Share on other sites

anunusualname 0 Posted October 23, 2010 Share Posted October 23, 2010 (edited) Yes, you're right. 7^0=the product of a number & its reciprocal.But I'd still like to expand 7^0 to an expression that doesn't make use of exponents. Is that not feasible? That is what Roger was doing, correct? What is his last name, by the way? By what was quoted in the original post, you would understand that 2^2 indicates 1x2x2. Or that when you raise two to the zeroth (or no) power, your solution is one, because what is indicated is 1 multiplied by 2 exactly zero times. Which is to say 1x2x0=1. That makes sense, but I'm not sure that it is valid. Especially when I got to 0^0. Which is to say 1x0x0=1?So, two points. What Roger is quoted as saying seems to me to be wrong. I would however like some way of expressing 7^0 in plain English and without referring to 7^1 x 7^(-1)= the product of inverses.Edit: I was taught early on that 7^2 is "some number of sevens multiplied together". That is the way it was explained, and that is the way it makes sense. However, I suppose as you derive the definition of one expression from another, one can't quite express mathematical expressions in plain English anymore. Still, I'd like to understand math in plain English insofar as possible. Edited October 23, 2010 by Brian Bahneman Link to post Share on other sites

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