- #1

Poolala

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- 0

1 / 0 = x, so x * 0 = 1

But 0 does not equal 1, so x / 0 is unsolvable.

Oh, and I'm new to forums so if this shouldn't be here you can delete it.

You should upgrade or use an alternative browser.

- Thread starter Poolala
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- #1

Poolala

- 2

- 0

1 / 0 = x, so x * 0 = 1

But 0 does not equal 1, so x / 0 is unsolvable.

Oh, and I'm new to forums so if this shouldn't be here you can delete it.

- #2

RamaWolf

- 95

- 2

Perhaps you want to look at:

[itex]\underbrace{lim}_{i->0} \frac{Sin(x)}{x} = 1[/itex]

[itex]\underbrace{lim}_{i->0} \frac{Sin(x)}{x} = 1[/itex]

- #3

Simon Bridge

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$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$$

... mathematicians can set up numbers to do all kinds of things.

- #4

Hurkyl

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1 / 0 = x, so x * 0 = 1

But 0 does not equal 1, so x / 0 is unsolvable.

Oh, and I'm new to forums so if this shouldn't be here you can delete it.

This argument makes the assumption that, whenever x/y is defined, the equation y(x/y) = x holds. It also makes the assumption that x0 is always defined, and it is 0.

Both of these assumptions are, indeed, true for the real number system, and the complex number system. But other important number systems do not have these properties.

RamaWolf is talking about something different (although some people have trouble distinguishing the two).

- #5

HallsofIvy

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Yes, that's a perfectly valid proof that the number 0 (or the additive identity in any field with more than one member) does not have an inverse.

1 / 0 = x, so x * 0 = 1

But 0 does not equal 1, so x / 0 is unsolvable.

Oh, and I'm new to forums so if this shouldn't be here you can delete it.

True but has nothing to do with the topic here.Perhaps you want to look at:

[itex]\underbrace{lim}_{i->0} \frac{Sin(x)}{x} = 1[/itex]

- #6

Poolala

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- 0

Oh, and I've been thinking: does 0 / 0 = 1, or is it undefined?

- #7

ramsey2879

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In most number systems 0/0 is undefined. Only a few exotic number systems have defined it and that is outside my scope.Oh, and I've been thinking: does 0 / 0 = 1, or is it undefined?

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