Taking the author's argument that a/b = the number x, for which x * b = a, it's easy to draw this out to it's logical conclusion:
For numerators (that is: a) other than 0, you get impossible equations. That is: 1 / 0 = "the number which, when multiplied by 0 is 1." In equation form: x * 0 = 1 Well no such number exists. Therefore x can be said to be "no number."
But for the case where a and b are both zero, the equation becomes: x * 0 = 0. And in this case, the answer is any number, since any number multiplied by zero = 0.
So we can think of 0 / 0 as "any number" and non-0 / 0 as "no number".
That means assigning each instance of 0/0 to a variable is a very good solution. Because we use a letter to represent an unknown number in basic algebra, but in linear algebra we also use a letter to represent a value which could be any number. As in the equation for a line: y=mx+b, where x and y can both take on any value.
Taking the author's argument that a/b = the number x, for which x * b = a, it's easy to draw this out to it's logical conclusion:
For numerators (that is: a) other than 0, you get impossible equations. That is: 1 / 0 = "the number which, when multiplied by 0 is 1." In equation form: x * 0 = 1 Well no such number exists. Therefore x can be said to be "no number."
But for the case where a and b are both zero, the equation becomes: x * 0 = 0. And in this case, the answer is any number, since any number multiplied by zero = 0.
So we can think of 0 / 0 as "any number" and non-0 / 0 as "no number".
That means assigning each instance of 0/0 to a variable is a very good solution. Because we use a letter to represent an unknown number in basic algebra, but in linear algebra we also use a letter to represent a value which could be any number. As in the equation for a line: y=mx+b, where x and y can both take on any value.