While solving exercise of Fuzzy Sets and Fuzzy Logic: Theory and Applications by George J Klir and Bo Yuan, I came across this question:-
Let $i$ be a t - norm such that
$$i(a, b + c) = i(a, b) + i(a, c)$$
for all $a, b, c \in [0, 1]$ and $b + c \leq 1$. Show that $i$ must be the algebraic product, i.e., $i(a, b) = ab$ for all $a, b \in [0, 1]$.
I have literally no insights about proving this. Help will be appreciated!
We will prove the statement for $i(a,b)$ where $b$ is rational Since the rationals are dense in $[0,1]$, you can use monotonicity of $t$-norms to generalize this to all $b\in [0,1]$.
To see what we will be doing, I will give an example for $b = 2/3$. In this case, we have $$ a = i(a, 1) = i\left(a, \frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right) =i\left(a, \frac{1}{3}\right) + i\left(a, \frac{1}{3}\right)+i\left(a, \frac{1}{3}\right) =3\cdot i\left(a, \frac{1}{3}\right) $$ Dividing either side by $3$, we see that $$ i\left(a, \frac{1}{3}\right) = \frac{a}{3} $$ But then, $$ i(a,b) = i\left(a, \frac{2}{3}\right) =i\left(a, \frac{1}{3} + \frac{1}{3}\right) =i\left(a, \frac{1}{3}\right)+i\left(a, \frac{1}{3}\right) =2\cdot i\left(a, \frac{1}{3}\right) $$ Since $i\left(a, \frac{1}{3}\right) = a/3$, the above is precisely $$ i(a,b) = 2\cdot \frac{a}{3} = a\cdot \frac{2}{3} = ab $$
We now prove the statement where $b$ is any rational in $[0,1]$. If $b=p/q$ for integers $p,q$ with $0\leq p\leq q$, $q\neq 0$. Then by the assumption $i(a, b + c) = i(a, b) + i(a, c)$ we have $$ a = i(a,1) = qi\left(a, \frac{1}{q}\right) \implies i\left(a, \frac{1}{q}\right)=\frac{a}{q} $$ Using again the assumption $i(a, b + c) = i(a, b) + i(a, c)$ we have $$ i(a,b) = i\left(1, \frac{p}{q}\right) = p\cdot i\left(1, \frac{1}{q}\right) =p\cdot\frac{a}{q} = ab $$
Now, suppose that $b$ is irrational. Then for all $n\in\mathbb{N}$, there exists rational number $b_1, b_2$ such that $$ b_1\leq b \leq b_2 $$ and $$ b_2-b < \frac{1}{n}, \quad b-b_1 < \frac{1}{n} $$ By monotonicity of $t$-norms, $$ i(a,b_1) \leq i(a,b) \leq i(a,b_2) $$ Since $b_1, b_2$ are rationals, the above is precisely $$ ab_1 \leq i(a,b) \leq ab_2 $$ Now, we may write $$ ab_1 = a(b_1+b-b) = ab - a(b-b_1) \quad\text{and}\quad ab_2 = ab + a(b_2-b) $$ So, we have $$ ab - a(b-b_1) \leq i(a,b) \leq ab + a(b_2-b) $$ Subtracting $ab$ on either side yields $$ - a(b-b_1) \leq i(a,b)-ab \leq a(b_2-b) $$ Therefore, $$ \lvert i(a,b)-ab\rvert \leq \max\{a(b-b_1), a(b_2-b)\} $$ Since $b_2-b < \frac{1}{n}$ and $b-b_1 < \frac{1}{n}$, the above is bounded by $$ \lvert i(a,b)-ab\rvert \leq \max\{a(b-b_1), a(b_2-b)\} \leq a\frac{1}{n} \leq \frac{1}{n} $$ where the last inequality holds since $0\leq a \leq 1$.
So, we conclude that $$ \lvert i(a,b)-ab\rvert \leq \frac{1}{n}\qquad\forall n\in\mathbb{N} $$ But this means that $i(a,b)=ab$ as desired.