1+0 is odd, therefore 0 is even.It depends how closely you want to define parity; in the arithmetic sense, if 0 was odd then 1 + 0 would be even, making it even lets 1 + 0 be odd.
Is there any sense in which zero can be considered an odd number?
1+0 is odd, therefore 0 is even.It depends how closely you want to define parity; in the arithmetic sense, if 0 was odd then 1 + 0 would be even, making it even lets 1 + 0 be odd.
I guess not.Is there any sense in which zero can be considered an odd number?
No, it doesn't follow.
"the limit of $$\frac{0}{0}$$" is not a correct description of $$\lim_{a \to 0} \frac{a}{a}$$..
The only correct way to say $$\lim_{a \to 0} \frac{a}{a}$$ is "The limit of a/a as a approaches zero"
Consider these limits. Would you say that they are also "the limit of $$\frac{0}{0}$$"?
$$\lim_{a \to 0} \frac{0}{a}$$
$$\lim_{a \to 0} \frac{a}{0}$$
$$\lim_{a \to 0} \frac{2a}{a}$$
The phrase "the limit of $$\frac{0}{0}$$" doesn't really mean anything. It's ambiguous, not well defined, and should be avoided.
They do always say that quadradic's cannot be equal to zero, even if there is not division in it by zero. Never got why they cared to do that until now, it seemed kind of redundant in everything. I just thought that they really didn't like the number zero.I think the problem is you are treating $$a^2 - b^2 = 0 $$ like it is a quadradic equation. A quadradic equation is in the form $$ax^2 + bx + c $$. The quadradic can of course also be in the form $$ax^2 - b^2 = 0$$.
But $$a^2 - b^2 = 0 $$ is not in the right form it is essentially $$b^2 - b^2 = 0 $$ or $$x^2 - x^2 = 0 $$ so yu cannot expand it using the rules for quadradics.
The "trick" is in moving from step 4 to step 5. Division by (a-b) is equivalent to division by zero, and the point is that dividing by zero leads to contradictions.I think the problem is you are treating $$a^2 - b^2 = 0 $$ like it is a quadradic equation. A quadradic equation is in the form $$ax^2 + bx + c $$. The quadradic can of course also be in the form $$ax^2 - b^2 = 0$$.
But $$a^2 - b^2 = 0 $$ is not in the right form it is essentially $$b^2 - b^2 = 0 $$ or $$x^2 - x^2 = 0 $$ so yu cannot expand it using the rules for quadradics.
The "trick" is in moving from step 4 to step 5. Division by (a-b) is equivalent to division by zero, and the point is that dividing by zero leads to contradictions.
Almost. It is the dividing of zero by zero that leads to the contradiction.The "trick" is in moving from step 4 to step 5. Division by (a-b) is equivalent to division by zero, and the point is that dividing by zero leads to contradictions.
The "trick" is in moving from step 4 to step 5. Division by (a-b) is equivalent to division by zero, and the point is that dividing by zero leads to contradictions.
Surely you jest!No, Kiteman. In the given example, there is no division of zero by zero.
So, in other words, your comment amounts to: "if we take this well-defined concept, and redefine it to mean something else, then it can mean something else."It depends how closely you want to define parity;...
The only time $$ a^{2} - b^{2} = ab - b^{2} $$ is when $$ a = b $$, I would be willing to bet that in almost every math book that teaches these concepts, says "a" does not equal "b". $$ a^{2} - b^{2} $$ could never equal $$ ab - b^{2} $$, so then the statement has already gone wrong before there was even actual division by zero. You would be saying that you are factoring out a zero out of each side of the equation, before you even divided by zero. The division by zero is just another error on top of errors already made. If you assumed that "a" and "b" is actually zero, then it would be the only time it is correct, 0+0=0. How do you factor out a zero out of something that is non-zero?Surely you jest!
(a+b)(a-b)=b(a-b)
Then the actual next step, which was tricked past is
(a+b)(a-b)/(a-b)=b(a-b)/(a-b)
Where (a-b)/(a-b) in any other case would be 1 but in this case is undefined. Duh!
The only time $$ a^{2} - b^{2} = ab - b^{2} $$ is when $$ a = b $$.
Well ya, and that would have been the first error in the equations.Right. We're talking about Emil's post, which starts from a=b:
Prof.Layman, there's nothing wrong with proclaiming that a=b. The subsequent equations are derived from that fact.Well ya, and that would have been the first error in the equations.
I would say that it is mathmatical proof that there is something wrong with claiming a=b. I never seen any algebra books that claimed that $$ a^{2} = ab $$ from then on it is just nonsense.Prof.Layman, there's nothing wrong with proclaiming that a=b. The subsequent equations are derived from that fact.
If you write a program that defines 0 as having no parity, then 0 doesn't 'behave' arithmetically. That doesn't mean the program can't do arithmetic.funkstar said:So, in other words, your comment amounts to: "if we take this well-defined concept, and redefine it to mean something else, then it can mean something else."
Such insight.