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Equivalence of equations

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Equivalence of equations

I was wondering whether everyone agrees that 0 is not in the solution set of the equation

(1) x*(1/x)=1

and that, hence, (1) is not equivalent to

(2) 1=1


This is more of an epistemological question IMHO.

For instance, if 1 is removed from the domain how can anyone say 2 = 2 when necessarily 2 * 1 = 2 / 1? The underlying component that makes the statement true doesn't exist and furthermore if the identity is gone no cancellation is possible. Put another way 2 * 3 != 6 because 6 != 6 since 6 / 6 != 1. Meaning 6 is not a component of itself, because 1 is undefined. So unless 1 holds no number holds. This is somewhat different than saying the number 7 is removed from the domain. For instance,

6*5 = 30 =>
30 = 30 =>
30 / 30 = 1

The statement holds no matter how you break it up, even if you evaluate various constituent factors (ie/ 15 * 2, 10 * 3, ...). So even though 7 can be a factor of 30 it isn't a _necessary_ factor for the operation to hold when rebalancing the equation. 1 is a necessary factor for all values > and < 1 to equal themselves. The only number this isn't true of is 0.

Ontologically the question can be rephrased, "is 1, as the multiplicative identity, a special property of multiplication or a special property of the number?"

This argument can also be used for the additive identity which seems to be at the heart of Philidor's question. Though the way Philidor phrased it (x*(1/x)=1) he's effectively making 0 a function of 1 which is why we see the hole at 0 because intuitively 1 is a product of 0 not the other way around.

We have two functions here.
f(x) = (1/x)*x where x is in R-{0}
g(x) = 1 where x is in R

The short answer is that f(x)=g(x) almost everywhere.

We did not "remove" 0 from the domain of f. It was never there because, in the field of real numbers, 0 does not have a multiplicative inverse.

We can note that g(x) is continuous at x=0 and that, if we define F(x) by
F(x) = (1/x)*x when x is in R-{0}
F(x) = 1 when x=0

then we have 'extended' f to the continuous function F and that F(x) = g(x) for all x.

This is not a trivial problem. Indeed the study of the ratio 0/0 led to the formalization of the differential Calculus and the resulting theory is a template for epistemology.

Good reply, Ivo. :)

Though I'd like to say that through my own informal analysis the multiplicative inverse of 0 is any given Real.

Consider if you take a circle and separate it into it's two constituent pieces(cosine and sine) and then set the two as a ratio of each other we form the tangent function (sine / cosine). Observing the convergence of the opposite to the adjacent over a single period (0 to pi) we see the tangent function goes from 0 to +complex infinity on the left of (pi / 2) to -negative complex infinity to 0 on the right. What this shows is that dividing by 0 or approaching nothingness pulls out all parts of all numbers over a single phase (which makes sense when viewed under the lens of Lim x->inf 1/x = 0 and heck it even makes sense in terms of the Heisenberg-uncertainty principle). No number is duplicated in the y (other than 0) along the single phase from 0 to pi. Yet amazingly all Reals are generated as these two waveforms converge towards and away from each other.

Maybe one day I'll have the insight to formalize why I see this to be the case.

nice point...

Let f(x)=x*(1/x).
f(x)=1 for x not=0.

f(x) undefined for x=0.

This reminds of a lecture I had when I was taking calc I. The instructor described three types of discontinuities at a point, jump (a finite difference between the left and right limits), point (a hole in the graph of the eqn), and asymptotic. "Yes", he said, "you're gonna have to know the difference between an asymptote and a hole in the graph."

Hi Philidor,
always your posts are quite interesting. Some captious question, isn't? I have read two good replies (mathman and Ivoyster).
Let me tell you what I think about your question. But, hey, I'm not mathematician, just a fan!
Well,
(1) x.1/x = 1
is an axiom in the multiplicative group (X-{0},.) with identity 1, whereas
(2) 1=1
is a particular assertion on the refexive property of equality(a binary predicate in Logics) x=x.
So, in my opinion, (1) and (2) are not equivalent.
perucho.

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