Same Difference

When two things are the same, a lot of people have started saying, “Same difference.” In terms of pragmatics, presumably this arose from some arbitrary need to be contrary, whilst retaining the same meaning. We shan’t go into that — because pragmatics is a bit of a rabbit hole! — but instead, we’ll examine the semantics of the statement and mathematically prove that it is equivalent to “they’re the same”.

Please note that most of this will be somewhat verbose (even trivial) for mathematicians. This is for the benefit of linguists and other readers without a strong mathematical background.


The Same

Trivially, two things are the same if they are equal. That is, \(a\) and \(b\) are the same iff \(a=b\).

Note that equality is what’s known as an equivalence relation, in mathematics: A binary relation that is reflexive, symmetrical and transitive. We shan’t go into the details, as far as equality goes, as it has a very natural meaning.

Same Difference

Two things having the same difference is a little more tricky to analyse. We are basically saying that the difference between either item and some arbitrary item is the same. The trick is that, because our comparison item is arbitrary, the equal difference must hold for everything. Symbolically, we can write this as: \(a\) and \(b\) have the same difference iff \(\forall x\), \(d(a,x) = d(b,x)\).

So what is this \(d\) function? It is what’s known as a metric; a function which gives rigorous mathematical meaning to the notion of “difference”. Indeed, what we have here is a metric space, which is governed by three, axiomatic rules:

Metric Space

A set \(M\) and a function \(d:M^2\rightarrow \mathbb{R}\) is a metric space \((M,d)\) if, for any \(x,y,z\in M\), the following hold:

  1. Identity of Indiscernibles
    \(d(x,y)=0\Leftrightarrow x=y\)
  2. Symmetry
  3. Triangle Inequality
    \(d(x,z)\leq d(x,y)+d(y,z)\)

Note that a consequence of these three conditions is that the metric function is always non-negative. This corresponds with our intuition of how distance (or difference) is quantified. To put some further meat on these proverbial bones, imagine the distance between two points on a plane defined by the straight line between them:

  • Clearly, if the distance between two points is zero, those points must coincide.
  • Equally obvious is that the distance from A to B is the same as the distance from B to A, bearing in mind that we’re not constricted to road networks!
  • As for the triangle inequality, all this says is that the distance from A to B is shorter than the distance from A to B, via C; unless C happens to lie exactly between A and B, in which case the distance is the same. If you draw this out on a triangle — hence the name — it will become immediately obvious.

Thus this is a metric space; known as \((\mathbb{R}^2,d_{2})\), where \(d_2\) is called the Euclidean metric after the Ancient Greek mathematician who first codified geometry.

A question you might now be asking, however, is, “How does one quantify natural language in such a way that it can be made into a valid metric space?” Without this, the proof doesn’t hold and, unfortunately — to answer the question — I don’t believe it’s possible to rigorously define a metric over the space of natural language. That said, what I’ve tried to illustrate above is that our notion of difference is intuitively congruent with that of a metric. As such, we shall proceed with this one caveat.


If \(a\) and \(b\) have the same difference, then \(a\) and \(b\) are the same. That is, if \((M,d)\) is any metric space and \(a,b\in M\), then:

\(\forall x\in M, d(a,x)=d(b,x)\Leftrightarrow a=b\)


When we have a logical equivalence, we need to prove entailment in both directions. We start from right-to-left; that is, we assume that \(a=b\).

From the triangle inequality property of our metric, we know the following facts hold \(\forall x\in M\):

  • \(d(a,x)\leq d(a,b) + d(b,x)\)
  • \(d(b,x)\leq d(b,a) + d(a,x)\)

From the symmetry and identity properties of the metric, we also know that \(d(a,b)=d(b,a)=0\), because \(a=b\). We can thus remove these terms from the above equations, yielding:

  • \(d(a,x)\leq d(b,x)\)
  • \(d(b,x)\leq d(a,x)\)

That is, in short, \(d(a,x)\leq d(b,x)\leq d(a,x)\). However, for this inequality to be satisfied, we must have it that \(d(a,x)=d(b,x)\).

The other direction, where we assume \(d(a,x)=d(b,x), \forall x\in M\), is even easier to prove.

Again, we know from the triangle inequality that \(d(a,b)\leq d(a,x) + d(x,b)\). By symmetry, \(d(x,b)=d(b,x)\); as we know this is equal to \(d(a,x)\), we can substitute into our first inequality to get \(d(a,b)\leq 2d(a,x)\).

Now as this inequality has to apply for all \(x\in M\), we can replace the right hand side by a lower bound. This occurs when \(x=a\) and the metric yields 0 by the identity property. That is to say, \(d(a,b)\leq 2\min_{x\in M}d(a,x)=0\).

We know that the metric must be non-negative and, therefore, \(d(a,b)=0\Leftrightarrow a=b\), by the identity property.


— Xophmeister

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