{"id":672,"date":"2010-02-25T14:56:16","date_gmt":"2010-02-25T14:56:16","guid":{"rendered":"http:\/\/mathscitech.org\/articles\/?p=672"},"modified":"2024-01-01T12:02:44","modified_gmt":"2024-01-01T12:02:44","slug":"zero-to-zero-power","status":"publish","type":"post","link":"https:\/\/mathscitech.org\/articles\/zero-to-zero-power","title":{"rendered":"Why Zero Raised to the Zero Power is defined to be One"},"content":{"rendered":"<p><font color=\"red\">Updated! February 5, 2017<\/font><\/p>\n<p>The value of zero raised to the zero power, <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%280%5E0%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(0^0)' title='(0^0)' class='latex' \/>, has been discussed since the time of Euler in the 18th century (1700s). There are three reasonable choices: 1,0, or &#8220;indeterminate&#8221;.  Despite consensus amongst mathematicians that the correct answer is one, computing platforms seem to have reached a variety of conclusions: Google, R, Octave, Ruby, and Microsoft Calculator choose 1; Hexelon Max and TI-36 calculator choose 0; and Maxima and Excel throw an error (indeterminate).  In this article, I&#8217;ll explain why, for discrete mathematics, the correct answer cannot be anything other than 0^0=1, for reasons that go beyond consistency with the Binomial Theorem (Knuth&#8217;s argument).<br \/>\n<!--more--><\/p>\n<hr>\n<p><h4>Context of the Debate: Continuous Mathematics<\/h4>\n<p>The three choices for the value of <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0' title='0^0' class='latex' \/> appear because <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%5Ey&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^y' title='x^y' class='latex' \/>, as a function of two continuous variables, is discontinuous at (0,0) and takes three different values depending on the direction of approach to the discontinuity:<\/p>\n<ol>\n<li> Fixing y=0, we have <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%5E0%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^0=1' title='x^0=1' class='latex' \/> for all <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x+%5Cneq+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x \\neq 0' title='x \\neq 0' class='latex' \/>.  (Proof: <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%5E0+%3D+x%5E%7B%281-1%29%7D+%3D+x%5E1+x%5E%7B-1%7D+%3D+x%2Fx+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^0 = x^{(1-1)} = x^1 x^{-1} = x\/x = 1' title='x^0 = x^{(1-1)} = x^1 x^{-1} = x\/x = 1' class='latex' \/>, each statement holding for all <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x+%5Cneq+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x \\neq 0' title='x \\neq 0' class='latex' \/>).  Indeed, <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%5Ey+%5Crightarrow+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^y \\rightarrow 1' title='x^y \\rightarrow 1' class='latex' \/> as <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x+%5Crightarrow+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x \\rightarrow 0' title='x \\rightarrow 0' class='latex' \/>, approaching from left or right, with y=0.  (This was Euler&#8217;s reason.)\n<li> Fixing x=0, we have <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5Ey%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^y=0' title='0^y=0' class='latex' \/> for <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=y+%3E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y &gt;0' title='y &gt;0' class='latex' \/>.  (When y &lt; 0 we have division by zero which is undefined in the reals and <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%2B%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='+\\infty' title='+\\infty' class='latex' \/> in the extended reals).  Taking limits, <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%5Ey+%5Crightarrow+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^y \\rightarrow 0' title='x^y \\rightarrow 0' class='latex' \/> as <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=y+%5Csearrow+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y \\searrow 0' title='y \\searrow 0' class='latex' \/>, approaching from above only, with x=0.\n<li> Fixing x=0, we have an undefined value when y &lt; 0 due to division by zero.\n<\/ol>\n<p>Notice that the discontinuity is not a simple (point) discontinuity, but rather a pole discontinuity due to the approach from below.  (Exercise: what happens as the origin is approached from 45 degrees?)<\/p>\n<h4>Principles for a Decision in Mathematics: Extension and Consistency<\/h4>\n<p>In mathematics, when there is more than one choice, a decision is typically made by extending an existing precedent to maintain consistency with the evidence that is already accumulated and accepted.  <\/p>\n<p>An elementary example is the way ordinary multiplication is extended from two positive numbers to a positive and a negative number, then to two negative numbers, i.e. <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%28-1%29%28-1%29+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(-1)(-1) = 1' title='(-1)(-1) = 1' class='latex' \/>.<\/p>\n<blockquote><p> &#8220;Minus times minus is plus.<br \/>The reason for this we need not discuss!&#8221;<br \/> &#8212; W.H. Auden<\/p><\/blockquote>\n<p>Empirically, multiplication of two positive numbers has a well-defined, tangible meaning as repeated addition.  This meaning holds when one of the numbers is negative.  But when both are negative, the empirical meaning fails.<\/p>\n<p>For the mathematician, declaring something to be undefined (throwing an error) means a loss of efficiency because every instance now has to be checked for the undefined case, and this must be treated separately.  If a definition could be found that remains consistent with all other empirically obtained rules, and if that definition means that calculation can proceed indifferent to the decision, then that is a big win.  <\/p>\n<p>The consistency in this particular case is the distributivity of multiplication over addition, a law which, for positive numbers, can be accepted on entirely empirical grounds.  (See the footnote for the full argument.<sup class='footnote'><a href='#fn-672-1' id='fnref-672-1' onclick='return fdfootnote_show(672)'>1<\/a><\/sup>.)<\/p>\n<h4>Turning to Discrete Mathematics &#8211; Consistency with the Binomial Theorem<\/h4>\n<p>In discrete mathematics, there is no notion of &#8220;approaching&#8221; &#8212; one is either <em>at<\/em> <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0' title='0^0' class='latex' \/> or away from it, in which case <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=1%5E0+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1^0 = 1' title='1^0 = 1' class='latex' \/> or <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E1+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^1 = 0' title='0^1 = 0' class='latex' \/>.<\/p>\n<p>The case of <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0' title='0^0' class='latex' \/> can be decided on consistency grounds with respect to the binomial theorem, i.e. loss of computational efficiency to have to treat this case separately.    This is the argument of Knuth (of <a href=\"http:\/\/en.wikipedia.org\/wiki\/The_Art_of_Computer_Programming\">The Art of Computer Programming<\/a>, and <a href=\"http:\/\/en.wikipedia.org\/wiki\/TeX\">TeX<\/a> fame), based on maintaining consistency with the binomial theorem <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%28a%2Bb%29%5Ex&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(a+b)^x' title='(a+b)^x' class='latex' \/> when x=0, due to its fundamental place in both discrete and continuous mathematics:<\/p>\n<blockquote><p>&#8220;Some textbooks leave the quantity <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0' title='0^0' class='latex' \/> undefined, because the functions <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5Ex&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^x' title='0^x' class='latex' \/> and <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^0' title='x^0' class='latex' \/> have different limiting values when <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' \/> decreases to 0. But this is a mistake. We must define <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%5E0%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^0=1' title='x^0=1' class='latex' \/> for all <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' \/> , if the binomial theorem is to be valid when <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%3D0%2C+y%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=0, y=0' title='x=0, y=0' class='latex' \/> , and\/or <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=x%3D-y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=-y' title='x=-y' class='latex' \/>. The theorem is too important to be arbitrarily restricted!  By contrast, the function <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5Ex&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^x' title='0^x' class='latex' \/> is quite unimportant.&#8221;<br \/>\n    &#8211; from Concrete Mathematics, p.162, R. Graham, D. Knuth, O. Patashnik, Addison-Wesley, 1988\n<\/p><\/blockquote>\n<p><h4>Different Conventions Among Mathematical Computing Platforms<\/h4>\n<p>Given the universality of the <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0=1' title='0^0=1' class='latex' \/> convention amongst mathematicians, it is surprising to find that various computing platforms have implemented different values:<\/p>\n<ul>\n<li>Value one: <a href=\"https:\/\/www.google.com\/search?q=calculate+0%5E0&#038;ie=utf-8&#038;oe=utf-8&#038;client=firefox-b\" target=\"_blank\" rel=\"noopener\">Google Calculator<\/a>, R, Octave, Ruby, and Microsoft&#8217;s Calculator all give <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0 = 1' title='0^0 = 1' class='latex' \/>.\n<li>Value zero: Hexalon Max (calculator) and a physical TI-36 hand calculator give <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0 = 0' title='0^0 = 0' class='latex' \/>.\n<li>Value indeterminate: Maxima and Microsoft Excel (2000) give <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0' title='0^0' class='latex' \/> is indeterminate, i.e. an error is thrown.\n<\/ul>\n<p><h4>An Alternative Decision Criteria &#8211; tangible computation with verifiable count that requires the answer<\/h4>\n<p>While Knuth&#8217;s argument of convenient extension works, the finite summation of integer powers  provides us with a real, tangible result (a finite sum), whose value (an empirically determinable fact) depends unavoidably on the chosen value of <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0' title='0^0' class='latex' \/>.  So here we have a consistency argument that does not rely on efficiency.<\/p>\n<p>The crucial step in this argument occurs in the derivation of <a href=\"#star1b\">(*1b)<\/a> from <a href=\"#star1a\">(*1a)<\/a> in <em><a href=\"http:\/\/www.mathscitech.org\/articles\/finite-summations-2\">Finite Summation of Integer Powers, Part 2<\/a><\/em>.<\/p>\n<p>Extracting the relevant part of that derivation, we have: <\/p>\n<p align=center><a name=\"star1a\"><\/a><img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%5Csum_%7Bk%3D1%7D%5E%7BN-1%7D+%28N-K%29%28K%2B1%29%5E%7Bp-1%7D%5C+%5C+%5C+%5C+%5C+%28%2A1a%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\\sum_{k=1}^{N-1} (N-K)(K+1)^{p-1}\\ \\ \\ \\ \\ (*1a)' title='\\sum_{k=1}^{N-1} (N-K)(K+1)^{p-1}\\ \\ \\ \\ \\ (*1a)' class='latex' \/><\/p>\n<p>After expanding the binomial power using the binomial formula and further manipulation, we arrive at:<\/p>\n<p align=center><a name=\"starstarstar\"><\/a><br \/>\n <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%3D+%5Csum_%7Bj%3D0%7D%5E%7Bp-1%7D+%5Cbinom%7Bp-1%7D%7Bj%7D+%5Csum_%7Bk%3D0%7D%5E%7BN-1%7D+%5Cleft%5BNK%5Ej+-K%5E%7Bj%2B1%7D%5Cright%5D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= \\sum_{j=0}^{p-1} \\binom{p-1}{j} \\sum_{k=0}^{N-1} \\left[NK^j -K^{j+1}\\right]' title='= \\sum_{j=0}^{p-1} \\binom{p-1}{j} \\sum_{k=0}^{N-1} \\left[NK^j -K^{j+1}\\right]' class='latex' \/><br \/>\n (Pull the <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=K%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='K=0' title='K=0' class='latex' \/> term out of both summations. Note: <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0+%3D+1%5C+%5C+%5C+%5C+%5C+%5Cmbox%7B%28%2A%2A%2A%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0 = 1\\ \\ \\ \\ \\ \\mbox{(***)}' title='0^0 = 1\\ \\ \\ \\ \\ \\mbox{(***)}' class='latex' \/>)<br \/>\n <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%3D+N+%2B+%5Csum_%7Bj%3D0%7D%5E%7Bp-1%7D+%5Cbinom%7Bp-1%7D%7Bj%7D+%5Csum_%7Bk%3D1%7D%5E%7BN-1%7D+%5Cleft%5BNK%5Ej+-K%5E%7Bj%2B1%7D%5Cright%5D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= N + \\sum_{j=0}^{p-1} \\binom{p-1}{j} \\sum_{k=1}^{N-1} \\left[NK^j -K^{j+1}\\right]' title='= N + \\sum_{j=0}^{p-1} \\binom{p-1}{j} \\sum_{k=1}^{N-1} \\left[NK^j -K^{j+1}\\right]' class='latex' \/><br \/>\n (which, after additional manipulation, yields)<br \/>\n <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%3D+N+%2B+N%28N-1%29+-+S_p%28N-1%29+%2B+%5Csum_%7Bj%3D1%7D%5E%7Bp-1%7D+S_j%28N-1%29%5Cleft%5BN%5Cbinom%7Bp-1%7D%7Bj%7D+-+%5Cbinom%7Bp-1%7D%7Bj-1%7D%5Cright%5D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= N + N(N-1) - S_p(N-1) + \\sum_{j=1}^{p-1} S_j(N-1)\\left[N\\binom{p-1}{j} - \\binom{p-1}{j-1}\\right]' title='= N + N(N-1) - S_p(N-1) + \\sum_{j=1}^{p-1} S_j(N-1)\\left[N\\binom{p-1}{j} - \\binom{p-1}{j-1}\\right]' class='latex' \/><br \/>\n <a name=\"star1b\"><\/a><img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%3D+N%5E2+-+S_p%28N-1%29+%2B+%5Csum_%7Bj%3D1%7D%5E%7Bp-1%7D+S_j%28N-1%29%5Cleft%5BN%5Cbinom%7Bp-1%7D%7Bj%7D+-+%5Cbinom%7Bp-1%7D%7Bj-1%7D%5Cright%5D%5C+%5C+%5C+%5C+%5C+%5Cmbox%7B%28%2A1b%29%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= N^2 - S_p(N-1) + \\sum_{j=1}^{p-1} S_j(N-1)\\left[N\\binom{p-1}{j} - \\binom{p-1}{j-1}\\right]\\ \\ \\ \\ \\ \\mbox{(*1b)}' title='= N^2 - S_p(N-1) + \\sum_{j=1}^{p-1} S_j(N-1)\\left[N\\binom{p-1}{j} - \\binom{p-1}{j-1}\\right]\\ \\ \\ \\ \\ \\mbox{(*1b)}' class='latex' \/><\/p>\n<p>The key step happens in (***) above: we peel off the <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=K%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='K=0' title='K=0' class='latex' \/> term of the inner summation to get: <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=N0%5Ej+-+0%5E%7Bj%2B1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='N0^j - 0^{j+1}' title='N0^j - 0^{j+1}' class='latex' \/>. Peeling this out of the outer summation requires considering the expression for <em>all<\/em> <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=j&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='j' title='j' class='latex' \/>. Now, 0 raised to any <em>positive<\/em> power is 0, so we can dispel the case of <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=j%3E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='j&gt;0' title='j&gt;0' class='latex' \/>.  But what is <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0' title='0^0' class='latex' \/>?  A decision must be made: it is either <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0' title='0' class='latex' \/> or <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1' title='1' class='latex' \/>. Indeterminacy is not an option, since the situation is real and is required to continue the simplification. <\/p>\n<p><h4>The Argument for <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0 = 1' title='0^0 = 1' class='latex' \/><\/h4>\n<p>What are the consequences of choosing the other definition, i.e. <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0 = 0' title='0^0 = 0' class='latex' \/>?  In this case, the final formula for <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=S_p%28N%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='S_p(N)' title='S_p(N)' class='latex' \/> is off by a linear constant <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=N&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='N' title='N' class='latex' \/>, while the choice <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0%5E0+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0^0 = 1' title='0^0 = 1' class='latex' \/> leads to the exact formula and a computed value that matches a brute force summation.  <\/p>\n<p>For <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=S_5%2810%29+%3D+1%5E5%2B2%5E5%2B3%5E5%2B4%5E5%2B5%5E5%2B6%5E5%2B7%5E5%2B8%5E5%2B9%5E5%2B10%5E5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='S_5(10) = 1^5+2^5+3^5+4^5+5^5+6^5+7^5+8^5+9^5+10^5' title='S_5(10) = 1^5+2^5+3^5+4^5+5^5+6^5+7^5+8^5+9^5+10^5' class='latex' \/>, the difference is between 220,825 (the correct, verifiable answer), and 220,815 (verifiably NOT correct).  The correct definition is clear: 0^0 = 1 is for empirical reasons that have to do with counting and summing.  While it is the binomial theorem that provides the detail, the argument is one of verifiable necessity and not one of consistency.<\/p>\n<p>For discrete mathematics, the empirical evidence shows that 0^0=1 is required:<sup class='footnote'><a href='#fn-672-2' id='fnref-672-2' onclick='return fdfootnote_show(672)'>2<\/a><\/sup><\/p>\n<hr>\n<p>\n<b>References<\/b><br \/>\n<a href=\"http:\/\/mathforum.org\/dr.math\/faq\/faq.0.to.0.power.html\">The Math Forum<\/a><\/p>\n<hr>\n<p>(<em>If you&#8217;re a software developer of a mathematical package, I&#8217;d be interested in how you arrived at your decision.  You can send me an email using the Comments link below.<\/em>)<\/p>\n<p>\n<em>If you enjoyed this article, feel free to <a href=\"https:\/\/mathscitech.org\/articles\/feed\">click here to subscribe to my RSS Feed.<\/a><\/em><\/p>\n<hr>\n<pr>\n<h4>Footnotes<\/h4>\n<div class='footnotes' id='footnotes-672'>\n<div class='footnotedivider'><\/div>\n<ol>\n<li id='fn-672-1'> Considering that any quantity times zero is zero, and that one times any quantity is the quantity, we have no hesitation in granting <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=-1+%5Ctimes+0+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='-1 \\times 0 = 0' title='-1 \\times 0 = 0' class='latex' \/>.  But then observe that we way write <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0+%3D+%28-1+%2B+1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0 = (-1 + 1)' title='0 = (-1 + 1)' class='latex' \/>, which means, combining the two expressions, we have <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=-1+%281+%2B+%28-1%29%29+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='-1 (1 + (-1)) = 0' title='-1 (1 + (-1)) = 0' class='latex' \/>.  If we accept the law of distribution of multiplication over addition for positive whole numbers, purely on empirical grounds, and if we wish negative numbers to behave in the same manner as our <em>empirically accepted<\/em> positive whole numbers, then we want the distributive law to hold as well.  And therefore we have <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=0+%3D+%28-1%29%281%29+%2B+%28-1%29%28-1%29+%3D+-1+%2B+%28-1%29%28-1%29.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0 = (-1)(1) + (-1)(-1) = -1 + (-1)(-1).' title='0 = (-1)(1) + (-1)(-1) = -1 + (-1)(-1).' class='latex' \/>  Which means that <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%28-1%29%28-1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(-1)(-1)' title='(-1)(-1)' class='latex' \/> must be the oppositive (additive inverse) of <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%28-1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(-1)' title='(-1)' class='latex' \/>, and hence <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%28-1%29%28-1%29+%3D+%2B1.&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(-1)(-1) = +1.' title='(-1)(-1) = +1.' class='latex' \/> <span class='footnotereverse'><a href='#fnref-672-1'>&#8617;<\/a><\/span><\/li>\n<li id='fn-672-2'> The implications for continuous mathematics are a consideration for another discussion.  The statement that a discontinuity exists at the origin <img loading=\"lazy\" src='https:\/\/s0.wp.com\/latex.php?latex=%28x%2Cy%29+%3D+%280%2C0%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(x,y) = (0,0)' title='(x,y) = (0,0)' class='latex' \/> isn&#8217;t quite enough. <span class='footnotereverse'><a href='#fnref-672-2'>&#8617;<\/a><\/span><\/li>\n<\/ol>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Updated! February 5, 2017<\/p>\n<p>The value of zero raised to the zero power, , has been discussed since the time of Euler in the 18th century (1700s). There are three reasonable choices: 1,0, or &#8220;indeterminate&#8221;. Despite consensus amongst mathematicians that the correct answer is one, computing platforms seem to have reached a variety of conclusions: [Read More&#8230;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"nf_dc_page":"","footnotes":""},"categories":[31,119],"tags":[25,38,126,23,32,16,33,24],"coauthors":[112],"class_list":["post-672","post","type-post","status-publish","format-standard","hentry","category-mathematics","category-technical","tag-discrete-mathematics","tag-excel","tag-mathematics","tag-maxima","tag-octave","tag-phenomenology","tag-r","tag-ruby","odd"],"views":56899,"_links":{"self":[{"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/posts\/672","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/comments?post=672"}],"version-history":[{"count":77,"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/posts\/672\/revisions"}],"predecessor-version":[{"id":10525,"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/posts\/672\/revisions\/10525"}],"wp:attachment":[{"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/media?parent=672"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/categories?post=672"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/tags?post=672"},{"taxonomy":"author","embeddable":true,"href":"https:\/\/mathscitech.org\/articles\/wp-json\/wp\/v2\/coauthors?post=672"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}