#### mathman

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- product with undefined term

Let c=ab. Let b be undefined (but finite). If a=0, is c undefined or does c=0?

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- product with undefined term

Let c=ab. Let b be undefined (but finite). If a=0, is c undefined or does c=0?

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"0 times any real number = 0" is a property of 0.

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What does ##b## finite but undefined even mean??

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Or what does c=ab mean? For the conclusion 0b=0 we need the distributive law and a loop for a.

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It is 0, afaik. Using limits, let ##\epsilon \rightarrow 0## assume a is not really zero but very small and you can make the product very small thanks to the bound on c: ## ab < Max(b) \frac {\epsilon}{ Max(b)} ## for all ##\epsilon##.Summary:product with undefined term

Let c=ab. Let b be undefined (but finite). If a=0, is c undefined or does c=0?

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Not necessarily. Even if we assume ##0## as neutral element of addition,It is 0, afaik.

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Perhaps it means that the vlerth level of b is less than 16, but we don't know what a vlerth level is.What does ##b## finite but undefined even mean??

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See my previous. This is used e.g. to conclude that lim h-->0 hSin(1/h)=0 . The expression is bounded above by ##Max(a)\frac{\epsilon}{Max(a)}## for all ##\epsilon##.Not necessarily. Even if we assume ##0## as neutral element of addition,andthe unwritten operation as a multiplication, it is not clear if we have the distributive law, or another combination of the two operations. Furthermore, we need an additive inverse and an element ##a\neq 0##. If we lack any of these, then I don't know a valid proof for ##0\cdot b=0##.

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Yes, but you already make a hidden assumption, namely that we are talking about ordinary numbers.See my previous. This is used e.g. to conclude that lim h-->0 hSin(1/h)=0 . The expression is bounded above by ##Max(a)\frac{\epsilon}{Max(a)}## for all ##\epsilon##.

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How are they not ordinary numbers? Arent they all assumed to be Real numbers? Whatever value the bounded undefined takes, the product will be bounded above by epsilon.Yes, but you already make a hidden assumption, namely that we are talking about ordinary numbers.

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Who knows? I always think in terms of algebras, in which multiplication is not a natural given extension of school counting. But I admit that distribution is one of the very basic and here crucial properties, which is usually assumed.How are they not ordinary numbers? Arent they all assumed to be Real numbers?

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Fair enough, I was assuming numbers were Real. If not, it is a whole different situation. When we talk about bounded I guess we assume a modulus or valuation.Who knows? I always think in terms of algebras, in which multiplication is not a natural given extension of school counting.

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I have found another proof, which only uses the distributive law, and characteristic not two. Again, everything depends on the distributive law. Given that we can conclude with ##a-a=0##, either for an ##a\neq 0## or for ##a=0## with ##\operatorname{char} \neq 2##.But in what setting outside of the Reals does a question like this make sense? Maybe @mathman can clarify the setting?

However, if distribution is not given, then everything is possible.

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But there was a notion of boundedness used, so there is a norm or valuation too.

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Instead of making assumptions about what is being asked here, why don't we hold off until @mathman returns to clarify what sorts of things a, b, and c are?Let c=ab. Let b be undefined (but finite). If a=0, is c undefined or does c=0?

Clearly if a, b, and c are members of the real or complex field, the question is trivial and undeserving of an 'A' tag.

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Then just let x approach 0 as ##\epsilon/(Max(Sin(1/x))= \epsilon/1 ## so that ## xsin(1/x) < \epsilon \cdot 1 = \epsilon ## for all ## \epsilon##

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An interest variation is about ##f(x)=x^2sin(1/x)##, with ##f'(x)=2xsin(1/x)-cos(1/x)## for ##x\ne 0##, which is undefined as ##x\to 0##, but ##f'(0)=0##

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Then the operation of multiplication is undefined. Axioms/rules only stipulate multiplication within the Real numbers.Then just let x approach 0 as ##\epsilon/(Max(Sin(1/x))= \epsilon/1 ## so that ## xsin(1/x) < \epsilon \cdot 1 = \epsilon ## for all ## \epsilon##

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f'(x) is defined as ##\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}##, so ##f'(x)## doesn't exist at a value ##x## where ##f(x)## doesn't exist. So I assume you are defining ##f(0)## to be ##\lim_{x \rightarrow 0}x^2 sin(1/x) = 0##.An interest variation is about ##f(x)=x^2sin(1/x)##, with ##f'(x)=2xsin(1/x)-cos(1/x)## for ##x\ne 0##, which is undefined as ##x\to 0##, but ##f'(0)=0##

I don't understand how this example illustrates the question in your original post.

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Yes, since 1/0 is undefined, then so is sin(1/0), so the product 0 * sin(1/0) is undefined.However there seems to be a school of thought which asserts that sin(1/0) being undefined, multiplying by 0 results in an undefined product.

You can't even say that b (= sin(1/0) here) is finite.Let c=ab. Let b be undefined (but finite). If a=0, is c undefined or does c=0?

However, you can work with limits to establish that ##\lim_{x \to 0} x \cdot \sin(\frac 1 x)## exists and is 0, which is easy to show without resorting to the limit definition. I.e., ##\forall x \gt 0, -x \le x \cdot \sin(\frac 1 x) \le x ##. Since both bounds approach 0 as x approaches 0, so must the expression in the middle.

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But computer floating point math isn't relevant here because of the inherent limitations of representing very small or very large numbers in a finite number of bits, particularly floating point numbers as described by IEEE-754, which describes different situations in which NANs (both signalling and nonsignalling) are produced. I'm aware that there are libraries that support larger numbers and higher precision than the 2-, 4-, 8, or 10-byte floating point numbers described in IEEE-754, but I'm not familiar with the limitations of those libraries. In any case, if these libraries store numbers have less than infinite precision, and signal events such as underflow and the like, then they too have limitations.

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My apologies for sidetracking the thread.

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