Anura 9 Posted December 23, 2010 Share Posted December 23, 2010 I understand that anything other than zero to the zero power equals one. But this doesn't seem to make sence to me. Could someone explain how this is, rather than that it is? I'd like as much feedback on this as possible. Thank you. Link to post Share on other sites

Samm 2 Posted December 23, 2010 Share Posted December 23, 2010 (edited) I like to think of it this way: x^{n}=x^{(n+1)}/x Although the above equation is pretty obvious, it helps to illustrate the idea. Let x be 4 and n be 0. 4^{0}=4^{(0+1)}/4=1 You could think of it another way. When multiplying or dividing by indices, you add or subtract the indices. So: 4^{1}/4^{1}=1 Which could be rewritten as: 4^{0}=1 I hope it makes more sense now. Edited December 23, 2010 by Samm 1 Link to post Share on other sites

Anura 9 Posted December 23, 2010 Author Share Posted December 23, 2010 I like to think of it this way: x^{n}=x^{(n+1)}/x Although the above equation is pretty obvious, it helps to illustrate the idea. Let x be 4 and n be 0. 4^{0}=4^{(0+1)}/4=1 You could think of it another way. When multiplying or dividing by indices, you add or subtract the indices. So: 4^{1}/4^{1}=1 Which could be rewritten as: 4^{0}=1 I hope it makes more sense now. not yet Link to post Share on other sites

D H 1371 Posted December 23, 2010 Share Posted December 23, 2010 (edited) not yet So let's start with the basics, the integers. Exponentiation extends multiplication just as multiplication extends addition. Multiplication in the integers is defined as repeated addition. For example, 4*3 = 4+4+4. Recursively, the base case is[math] a*1 = a[/math] and [math]a*n= a*(n-1) + a[/math] is the reduction rule. Exponentiation is similarly defined as repeated multiplication. The base case is [math]a^1=a[/math], but now the reduction rule uses multiplication rather than addition: [math]a^n=a^{n-1}*a[/math]. This definition of raising a number to the power of a positive integer still applies when one is working in the rationals or the reals: [math]x^n = x^{n-1}*x[/math] So, let's naively substitute n=1: [math]x^1 = x^0*x[/math] Now solve for [math]x^0[/math], and use the fact that by definition [math]x^1=x[/math]: [math]x^0 = \frac{x^1}{x} = \frac {x}{x} = 1[/math] This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined. Edited December 23, 2010 by D H Link to post Share on other sites

Anura 9 Posted December 23, 2010 Author Share Posted December 23, 2010 (edited) So let's start with the basics, the integers. Exponentiation extends multiplication just as multiplication extends addition. Multiplication in the integers is defined as repeated addition. For example, 4*3 = 4+4+4. Recursively, the base case is[math] a*1 = a[/math] and [math]a*n= a*(n-1) + a[/math] is the reduction rule. Exponentiation is similarly defined as repeated multiplication. The base case is [math]a^1=a[/math], but now the reduction rule uses multiplication rather than addition: [math]a^n=a^{n-1}*a[/math]. This definition of raising a number to the power of a positive integer still applies when one is working in the rationals or the reals: [math]x^n = x^{n-1}*x[/math] So, let's naively substitute n=1: [math]x^1 = x^0*x[/math] Now solve for [math]x^0[/math], and use the fact that by definition [math]x^1=x[/math]: [math]x^0 = \frac{x^1}{x} = \frac {x}{x} = 1[/math] This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined. Mabye im looking for a simplified answer that just aint there. Does this fact need to be taken on fath? Is it just true by definition? Edited December 23, 2010 by Anura Link to post Share on other sites

D H 1371 Posted December 23, 2010 Share Posted December 23, 2010 My answer wasn't simple enough? Sigh. Try this one more time. Suppose x is not zero. That means we can divide x raised to some power by x. For example, [math]x^4/x = x^3[/math], [math]x^3/x = x^2[/math] and [math]x^2/x = x^1=x[/math]. Note that [math]x^n/x = x^{n-1}[/math]. So what is [math]x^1/x[/math], both expressed as a numeric value and as a power of x? Link to post Share on other sites

michel123456 547 Posted December 23, 2010 Share Posted December 23, 2010 This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined. Put that into a diagram. Link to post Share on other sites

Anura 9 Posted December 23, 2010 Author Share Posted December 23, 2010 (edited) My answer wasn't simple enough? Sigh. Try this one more time. Suppose x is not zero. That means we can divide x raised to some power by x. For example, [math]x^4/x = x^3[/math], [math]x^3/x = x^2[/math] and [math]x^2/x = x^1=x[/math]. Note that [math]x^n/x = x^{n-1}[/math]. So what is [math]x^1/x[/math], both expressed as a numeric value and as a power of x? I understand that all of what your saying is true. D H. Is it possible im learning in the wrong order ? I dont have all of first year algabra figured out yet. And if thats the case would someone like to tutor me? Im not doing this for school but my own personal gain. Edited December 23, 2010 by Anura Link to post Share on other sites

D H 1371 Posted December 23, 2010 Share Posted December 23, 2010 (edited) [math]x^0 = \frac{x^1}{x} = \frac {x}{x} = 1[/math] This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined. Put that into a diagram. What are you complaining about? My use of "this"? My use of "for all x not equal to zero"? Is this better: [math]x^0 = \frac{x^1}{x} = \frac {x}{x} = 1[/math] This result, x^{0}=1, is valid for all x except for x=0. It is not valid for x=0 because that would involve computing 0/0, which isn't defined. I understand that all of what your saying is true. D H. Is it possible im learning in the wrong order ? I dont have all of first year algabra figured out yet. And if thats the case would someone like to tutor me? Im not doing this for school but my own personal gain. First year algebra in high school? If that is the case, perhaps you are learning things a bit out of order. Edited December 23, 2010 by D H Link to post Share on other sites

michel123456 547 Posted December 23, 2010 Share Posted December 23, 2010 (edited) (...)This result, x^{0}=1, is valid for all x except for x=0. It is not valid for x=0 because that would involve computing 0/0, which isn't defined. The horizontal line represent the result of x^0 for all numbers. For an infinitesimal close to zero on the positive side, the result is 1. The same occurs on the negative side. The line is like the trajectory of a bullet. Do we have to wonder what happen to this line at point zero? Does a notion of continuity exist in math? Edited December 23, 2010 by michel123456 Link to post Share on other sites

D H 1371 Posted December 23, 2010 Share Posted December 23, 2010 (edited) The horizontal line represent the result of x^0 for all numbers. For an infinitesimal close to zero on the positive side, the result is 1. The same occurs on the negative side. The line is like the trajectory of a bullet. Do we have to wonder what happen to this line at point zero? Does a notion of continuity exist in math? That is but one view of determining the value of 0^0. Let's try a couple more. 1. For all positive values of x, 0^x=0. Thus [math]\lim_{x\to 0} 0^x = 0[/math]. So perhaps we should have 0^0 be zero? 2. Even worse, given any number a one can find a pair of functions f(x) and g(x) such that both f(x) and g(x) approach zero as x approaches some value x_{0} and such that [math]\lim_{x\to x_0} f(x)^{g(x)} = a[/math]. Bottom line: 0^0 can only be defined in terms of limits, but different limits can be found that let 0^0 take on any value whatsoever. It is an indeterminate form. That said, it is convenient to define 0^0 to be 1 as an abuse of notation. This lets us write power series such as [math]f(x) = \sum_{n=0}^{\infty} a_n x^n[/math] Without this convenient abuse of notation, this series would not make sense at x=0. Edited December 23, 2010 by D H Link to post Share on other sites

Mr Skeptic 1154 Posted December 23, 2010 Share Posted December 23, 2010 It helps to know how to do math with exponents. When you multiply the same number with an exponent by the same number with a different exponent, you add up the exponents. When you divide by a number with an exponent, it is the same as having a negative exponent. Feel free to test this out on as many real numbers as you want. [math]3^1 = 3[/math], [math]3^2 = 3*3=9[/math], [math]3^3 = 3*3*3 = 27[/math], [math]3^3*3^4 = (3*3*3)*(3*3*3*3) = 3^7[/math] Use that knowledge to solve for division; for example [math]3^5/3^2 = (3*3*3*3*3)/(3*3) = 3^3 = 3^{(5-2)}[/math], which if you test works for all numbers. But when the exponents are equal, [math]3^n/3^n = 3^{(n-n)} = 3^0[/math], but also [math][/math], [math]3^n/3^n = 1 = 3^0[/math]. This is not a formal proof since it uses specific numbers, but if you test it with any number you like (other than zero, because division by zero is undefined), then you get the same. And yes, you're learning things out of order but that can be fun too. 2 Link to post Share on other sites

Anura 9 Posted December 23, 2010 Author Share Posted December 23, 2010 Here's what i figured. When you multiply any number (n) by itself zero times, Whatever that number is then becomes one unit (n/n), thus becoming one. Is this correct? Link to post Share on other sites

Samm 2 Posted December 24, 2010 Share Posted December 24, 2010 Here's what i figured. When you multiply any number (n) by itself zero times, Whatever that number is then becomes one unit (n/n), thus becoming one. Is this correct? That's the reasoning I used. But then, I'm some silly high school student. Link to post Share on other sites

michel123456 547 Posted December 24, 2010 Share Posted December 24, 2010 (edited) That is but one view of determining the value of 0^0. Let's try a couple more. 1. For all positive values of x, 0^x=0. Thus [math]\lim_{x\to 0} 0^x = 0[/math]. So perhaps we should have 0^0 be zero? You mean this: I see no problem. IMHO it is an algebraic notation issue: the horizontal axis is the input, the vertical axis is the result, they are different. So IMHO there are 2 zero's, with a different meaning. If you had written the formula's with red & green numbers, you would have seen that red zero is not the same as green zero. Just as if the graph was a projection of a 3D situation where the horizontal axis is 100 meters above the vertical one. (note for the reader, this is not what you will learn in standard math) Edited December 24, 2010 by michel123456 Link to post Share on other sites

Shadow 67 Posted December 24, 2010 Share Posted December 24, 2010 2. Even worse, given any number a one can find a pair of functions f(x) and g(x) such that both f(x) and g(x) approach zero as x approaches some value x_{0} and such that [math]\lim_{x\to x_0} f(x)^{g(x)} = a[/math]. Does this theorem have a name? Link to post Share on other sites

D H 1371 Posted December 24, 2010 Share Posted December 24, 2010 (edited) (note for the reader, this is not what you will learn in standard math) Right. I guess that exemplifies why nobody teaches the non-standard michel123456 math. What you will learn in standard math is that In vicinity of the origin of the half plane x>0, x^{y} takes on all values from 0 to infinity. This means from the perspective of a limit (or continuity), 0^0 is indeterminate. However, it is extremely convenient if we define it to be one, at least in certain contexts such as the binomial expansion and power series. --------------------------------------------------------------------------------------------- 2. Even worse, given any number a one can find a pair of functions f(x) and g(x) such that both f(x) and g(x) approach zero as x approaches some value x_{0} and such that [math]\lim_{x\to x_0} f(x)^{g(x)} = a[/math]. Does this theorem have a name? 0^0 is indeterminate. http://www.wolframal.../input/?i=0%5E0 As far as finding such a pair of functions, here is one such pair: f(x)=x, g(x)=ln(a)/ln(x). Note that with f(x) and g(x) defined in this manner that f(x)^{g}^{(}^{x}^{)} is identically equal to a for all x>0. Note also that both f(x) and g(x) approach zero as x→0^{+} as required. Edited December 24, 2010 by D H Link to post Share on other sites

khaled 13 Posted January 20, 2011 Share Posted January 20, 2011 (edited) If we take things based on the basic definition of what is the power, since power is multiplicity of self, its definition was made recursive, [latex]a^{-p} = \frac{1}{a^p} = { \left( \frac{1}{a} \right) }^{p}[/latex] [latex]a^p = a^{p-1} \times a[/latex] and [latex]a^p = \frac{a^{p+1}}{a}[/latex] with base case is for: [latex]a^{0} = 1[/latex], [latex]{0}^x = 0[/latex], and [latex]0^{0} = [/latex] undefined OR 1, [latex]0^{+\infty} = [/latex] undefined OR 0, [latex]0^{-\infty} = [/latex] undefined OR 0 where undefined might be 1 for [latex]0^{0}[/latex] and 0 for [latex]-\infty[/latex] [latex]+\infty[/latex] according to some mathematicians that's all I know ... Edited January 20, 2011 by khaled Link to post Share on other sites

throng 12 Posted January 26, 2011 Share Posted January 26, 2011 If x = 1 any power = 1 so x to the zero power = 1 If x is not equal to 1 any power is not equal to one except the power of zero 2^x doubles so 2 to the power of zero is half of two = 1 5^x is 5, 25 125 so increases five fold so 5to the power of 0 is 5/5 = 1 Link to post Share on other sites

alpha2cen 18 Posted February 1, 2011 Share Posted February 1, 2011 (edited) i^{0}=1 i^{0}=1^{0} i=1 i= all real number, all imaginary number Can we use x^{0}as a term in the equation? Edited February 4, 2011 by alpha2cen Link to post Share on other sites

D H 1371 Posted February 1, 2011 Share Posted February 1, 2011 i^{0}=1 i^{0}=1^{0} i=1 Can we use x^{0}as a term in the equation? In what equation? As I have said many times now, you cannot prove that 0^{0} is one because x^{y} takes on all values for (x,y) in the vicinity of the origin. 0^{0} is indeterminate. However, it is often extremely convenient to define x^{0} to be 1 for all x, including x=0. This is the interpretation of x^{0} in the binomial theorem, power series, etc. Link to post Share on other sites

alpha2cen 18 Posted February 2, 2011 Share Posted February 2, 2011 (edited) In what equation? As I have said many times now, you cannot prove that 0^{0} is one because x^{y} takes on all values for (x,y) in the vicinity of the origin. 0^{0} is indeterminate. However, it is often extremely convenient to define x^{0} to be 1 for all x, including x=0. This is the interpretation of x^{0} in the binomial theorem, power series, etc. From #20 i is an imaginary number, i.e., sqrt(-1) Edited February 2, 2011 by alpha2cen Link to post Share on other sites

Mr Skeptic 1154 Posted February 2, 2011 Share Posted February 2, 2011 i^{0}=1 i^{0}=1^{0} i=1 Can we use x^{0}as a term in the equation? You cannot undo a relation that maps multiple numbers to one number. Thus your line of reasoning fails at the third step. Link to post Share on other sites

alpha2cen 18 Posted February 4, 2011 Share Posted February 4, 2011 x^{0}=1 x=all real number, all imaginary number possible 2^{0}, 3^{0}, -2^{0}, -3^{0}, -i^{0}, i^{0},........................= x^{0} Can we use x^{0}as a term in the equation? Link to post Share on other sites

DrRocket 337 Posted February 10, 2011 Share Posted February 10, 2011 I understand that anything other than zero to the zero power equals one. But this doesn't seem to make sence to me. Could someone explain how this is, rather than that it is? I'd like as much feedback on this as possible. Thank you. It is a convention designed to make the usual algebra of exponents work [math] 1 = x/x = x^1 x^{-1} = x^{1-1} = x^0 [/math] Usually [math] 0^0 =1 [/math] by definition. Link to post Share on other sites

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