# Is equality of two fractions ( like $2/10$ and $1/5$) really equality or equivalence?

In propositional logic, one could not correctly say that : $$(A \& B) = (B \& A)$$.

The reason is syntactic: the first conjunct of $$(A \& B)$$ is $$A$$, while the first conjunct of $$(B \& A)$$ is $$B$$. So the two formulas are not identical, they are not the same formula.

The only thing one can say is that the two formulas are equivalent.

My question is: does this syntactic argument also hold for fractions?

Remark. I write this question after having watched a video by Herbert Gross where he expresses his reluctancy to call two fractions like $$1/5$$ and $$2/10$$" equal". According to Gross, they would be better called "equivalent" inasmuch as they " name the same number"

Remark. I do not ask whether the equivalence relation between fractions is the same as the logical equivalence relation " formula X is true in exactly the same interpretations as Y ". My question is not :

does " 1/5 = 2/10 mean 1/5 <=> 2/10" ?

I simply ask whether the equal sign between fractions should be read as some sort of arithmetical equivalence ( not a logical one of course).

• The construction of $\Bbb Q$ as the fraction field of $\Bbb Z$ does exactly this. It consists of equivalence classes. Apr 27, 2019 at 10:47
• they are equal in value but equivalent in notation. $\frac{1}{5}$ does not look like $\frac{2}{10}$ at all. and as Burde said, it's construction. Apr 27, 2019 at 10:48
• To reference Morris Kline's Mathematics for the Nonmathematician, (page 62) he describes that in order to add the fractions $2/3$ and $7/5$, one must "express each fraction in an equivalent form such that the denominators are now alike..." Apr 27, 2019 at 10:50
• They are two names of the same rational number. We use them in an expession like $\dfrac {2}{10} \times 5 = \dfrac 1 5 \times 5$ according to the axiom fro equality : $x=y \to (f(x)=f(y))$. Apr 27, 2019 at 11:03
• There is some confucion here between logical symbols and set-theoretical/arithmetical ones. In logic language, "logical equivalence", i.e. bi-conditional, is a symbols expressing a relation between formulas, while equlity is a relation between terms. Thus $1+1=2$ and not $1+1 \leftrightarrow 2$. Apr 27, 2019 at 11:12

Whether $$2/10$$ and $$1/5$$ are equivalent or equal depends on how you define the meaning of the formal expression "$$a/b$$".

If $$a/b$$ is just a convenient way to write the ordered pair $$(a,b)$$ of integers when you are discussing the rational numbers, then those two fractions are equivalent - they define the same rational number.

If $$a/b$$ is just a way to write the rational number that solves the equation $$bx=a$$ then those two fractions are equal.

In an application they may not even be equivalent. Kids are taugh to model "$$1/5$$" as "cut a pie in $$5$$ parts and take $$1$$ of them". That is not the same physical operation as "cut a pie in $$10$$ parts and take $$2$$". That lack of equivalence is even clearer for the commutativity of multiplication: two kids with three cookies each is not the same as three kids with two cookies each even though the number of cookies is the same.

Construction of the field of rational numbers $$\mathbb{Q}$$, from the integers $$\mathbb{Z}$$, may help. In this construction, we deal with equivalent classes $$[(a,b)]$$ for $$b\ne 0$$, defined by $$[(a,b)]=\{(x,y)\in\mathbb{Z}\times(\mathbb{Z}-\{0\}): ay=bx\}.$$

Therefore, for example, the fraction $$\frac{1}{5}$$, is equal to the class $$[(1,5)]$$ which also contains infinitely many elements like $$(2,10),(3,15),\ldots$$ and all of this elements are a representation of the same class, namely $$\frac{1}{5}$$.

• Yes, what's important to note is that although the couples $(1,5)$ and $(2,10)$ are merely equivalent, the fractions $\frac{1}{5}$ and $\frac{2}{10}$, which are equivalence classes, are literally equal Apr 27, 2019 at 11:42

$$\dfrac2{10}$$ and $$\dfrac15$$ denote the same rational number. As rationals, they are equal. As fractions, up to your taste.