Algebraic Proof That a*0 = 0Date: 11/02/2003 at 20:34:35 From: Katie Subject: algebraic properties of real numbers If "a" is any element of all real numbers, then a*0 = 0. I would like to understand the specific theorem or proof behind this idea. Date: 11/03/2003 at 00:23:15 From: Doctor Rob Subject: Re: algebraic properties of real numbers Thanks for writing to Ask Dr. Math, Katie! The basic property of 0 is that, for any real number b, 0 + b = b. In particular, if b = 1, 0 + 1 = 1. Now multiply both sides by any real number a. Then a*(0 + 1) = a*1. Now recall that a*1 = a, by the definition of 1. Apply the Distributive Law to the left side: a*0 + a*1 = a*1, a*0 + a = a. Now add the negative of a to both sides: (a*0 + a) + (-a) = a + (-a). Use the Associative Law of Addition on the left side: a*0 + (a + (-a)) = a + (-a). Now use the fact that a number plus its negative is 0, which is the definition of the negative of a real number, so a + (-a) = 0: a*0 + (a + (-a)) = a + (-a), a*0 + 0 = 0, a*0 = 0. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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