Nope. $$ -\infty \times 0 $$ is undefined.So since you say $$\frac{-1}{0} = -\infty$$.
Does that mean $$ -\infty \times 0 = -1$$?
Nope. $$ -\infty \times 0 $$ is undefined.
Your error is in not following all the steps in altering the equation.
You should have asked whether $$\frac{0 \times -1}{0} = 0 \times -\infty $$ . Then you would have seen the undefined nature of the first term as well as the second.
Oh, and yes, they are both equally undefined.
from American Math Monthly vol 105 no 7.Richard Hamming said:In mathematics we do not appeal to authority, but rather you are responsible for what you believe.
The secret is that you just write the words "lim" in front of the equation and then you can just cancel zero's right out of the equation. For instance, the tangent line to a curve at exactly the tangent line "h" the height between two points would become zero, so you just write the word "lim" in front of it and then you can cancel out the "h" out of the equation. Then you are left with an equation that is exactly the tangent line to a curve, where "h" would be zero. Then by definition the tangent line of a curve is a line that intersects that curve at only exactly one point, not two points some real distance away from each other.So one is a dodge in that we invent tiny new numbers to divide by something "close" to zero, but never divide by zero. The other is a dodge in that we divide (anything except zero) by zero to get something which has only some of the properties of a number.
You are a danger to children. The limit of an expression as a variable takes on a value (like zero) has nothing to do with canceling or arithmetic.The secret is that you just write the words "lim" in front of the equation and then you can just cancel zero's right out of the equation. For instance, the tangent line to a curve at exactly the tangent line "h" the height between two points would become zero, so you just write the word "lim" in front of it and then you can cancel out the "h" out of the equation. Then you are left with an equation that is exactly the tangent line to a curve, where "h" would be zero. Then by definition the tangent line of a curve is a line that intersects that curve at only exactly one point, not two points some real distance away from each other.
It's the number with the smallest magnitude. It's the number that is the identity element for addition. It's the number that is its own additive inverse. It's the only number of a field without a multiplicative inverse.Is 0 a number? What kind of number is it?
That is a feature of the choice of definition of the set of natural numbers. 0 is always included in the integers or rationals or reals or complex numbers.The natural numbers exclude 0 (naturally), or include it depending on the context (in the context of algorithms there is usually a 'zeroth' step).
No. Zero is even, not odd. And zero is an integer, because $$0 = 1 \, + \, -1$$. And zero is a real number because it is the least upper bound of the set of negative numbers, which makes it a real number if you know anything about real numbers.But 0 is neither positive or negative (or it's both, if you like), it isn't odd or even (or again, it's both). So if a number is defined as having sign and being odd or even (up to some approximation of an integer), then 0 appears to lie outside the set of numbers (or at least, the integers or natural numbers).
Infinity is a concept, not a number. But in some number systems (the hyperreal numbers, the ordinal numbers, the surreal numbers), a symbol $$\infty$$ (or $$\omega$$) can take on the properties of a number with a magnitude larger than any finite number.Infinity isn't a number either, for the same reasons as well as having no definite value (at least 0 has a definite value).
You could say,$$ \lim_{a\to 0} \frac {a}{a} = 1 $$, so then you could say that as the limit of "a" approuches zero, zero divided by zero is equal to one.You are a danger to children. The limit of an expression as a variable takes on a value (like zero) has nothing to do with canceling or arithmetic.
$$\lim_{h\to 0} \frac{(x + h)^{\tiny n} \; - \; x^{\tiny n}}{h} = n x^{\tiny n-1}$$ has nothing to with the statement about numbers that division by zero cannot result in a number with all the properties of other numbers.
You could say,$$ \lim_{a\to 0} \frac {a}{a} = 1 $$, so then you could say that as the limit of "a" approuches zero, zero divided by zero is equal to one.
No, that's not how you say it.You could say,$$ \lim_{a\to 0} \frac {a}{a} = 1 $$, so then you could say that as the limit of "a" approaches zero, zero divided by zero is equal to one.
Zero is on the number line, halfway between $$-1$$ and $$1$$.
//Edit: Like here.
And how can you have an "understanding" of any system "axiomatically" unless you can either quote the axioms or point to a citation for them?
I can point to a machine-checked, browseable proof of the statement "0 is a real number" or "0 ∈ ℝ" here:
According to what axioms?Sorry, had to cut your links. New here, so can't post links yet.
How can a line be negative? A line is a line. Arbitrary conventions splitting a line into a positive and a negative half is irrational and axiomatically invalid.
It seems like you are using your own arbitrary axioms to prove what you like using your own rules?If you wish to use invalid arbitrary axioms than you can prove anything you like using your own rules.
My understanding is that a line is a line. What you make of that line is not what that line is.
It seems like you're offended by learning.Sorry if my naive understanding doesn't suit you expert opinion.
According to what axioms?
It seems like you are using your own arbitrary axioms to prove what you like using your own rules?
It seems like you're offended by learning.
The tangent line of a curve has been defined, and it assumes that $$\lim_{a \to 0} \frac{a}{a} = 1$$. In the case of $$\lim_{a \to 0, \, b \to 0} \frac{b}{a}$$, there is just not enough information in order to determine a certain value. It would be like saying that $$\frac{x}{y}$$ is not always equal to one, but that does not mean that $$\lim_{a \to 0} \frac{a}{a}$$ does not equal one. It just means that you have too many variables and not enough constants to hone in on a correct value. That is the purpose of limits, to find an answer to a problem that cannot normally be solved. It finds answers to problems closer and closer to where the point is that your looking for an answer, and then gives you the answer as to what value those problems are pointing you too. So then if you wanted to find the answer to $$\frac{0}{0}$$ then you would have to take the limit, and then after taking the limit you would find that the answer is one, that is the location that has the hole in the graph. That is how you solve for holes in equations.But $$\lim_{a \to 0} \frac{0}{a} = 0$$ and $$\lim_{a \to 0, \, b \to 0} \frac{b}{a}$$ can take on any value depending on how one approaches the limit. Thus $$\lim_{a \to 0} \frac{a}{a} = 1$$ does not cause the expression $$\frac{0}{0}$$ to cease to be undefined.
The tangent line of a curve has been defined....
I don't know what kind of tap dancing this is, but $$\lim_{a \to 0} \frac{a}{a} = 1$$. Say you picked any number, and then divided it by itself, you would get one. Then pick another number closer to zero and then divide it by itself, you get one. Do it again and again as much as you like, and you will still get one. So then the closer you get to dividing a number by itself increasingly closer to zero, you always get one. So then the limit of $$\frac{0}{0}$$ is one. So then if you wanted to find a solution to a problem and use the closer answer you could possibly find then you would replace $$\frac{0}{0}$$ with one.In the case of $$\lim_{a \to 23, \, b \to -12} \frac{b}{a} = \frac{-12}{23}$$ no matter the rest of the tapdancing you are doing because $$\frac{-12}{23}$$ is a number.
In the case of $$\lim_{a \to 0, \, b \to 0} \frac{b - 12}{23 + a} = \frac{-12}{23}$$ because $$\frac{-12}{23}$$ is a number and $$\frac{b - 12}{23 + a} = \frac{-12}{23} + \frac{12 a + 23 b}{23 a + 529}$$ (when $$a + 23 \neq 0$$)
The expression $$\frac{0}{0}$$ is meaningless by itself because there is no curve defined, therefore no concept of tangent applies and concept of an expression where a limit concept can be applied.
The expression $$\lim_{a \to 0} \frac{\sinh(a) - \sin(a)}{ a^2 \; \sin(a) }$$ is not evaluated by saying it "looks like $$\frac{0}{0}$$" and therefore must be 1.
Yes, $$\lim_{a \to 0} \frac{a}{a} = 1$$I don't know what kind of tap dancing this is, but $$\lim_{a \to 0} \frac{a}{a} = 1$$.
No, it doesn't follow.So then the limit of $$\frac{0}{0}$$ is one.
rpenner said:No. Zero is even, not odd. And zero is an integer, because . And zero is a real number because it is the least upper bound of the set of negative numbers, which makes it a real number if you know anything about real numbers.