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Algebraic Proof That a*0 = 0

Date: 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 
Associated Topics:
Elementary Multiplication
High School Basic Algebra

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