Problem:
Find the points of trisection of the line segment AB, where $A\equiv(4,0), B\equiv(0,3)$
Process 1:
Let the P & Q be the points of trisection. Let $x_1$ & $y_1$ be the abscissa and ordinate of P, and let $x_2$ & $y_2$ be the abscissa and ordinate of Q. Now,
$$(x_1,y_1)=\left(\frac{(2)(4)+(1)(0)}{3},\frac{(2)(0)+(1)(3)}{3}\right)$$
$$=\left(\frac{8}{3}, 1\right)$$
$$...$$
Process 2:
Let the P & Q be the points of trisection. Let $x_1$ & $y_1$ be the abscissa and ordinate of P, and let $x_2$ & $y_2$ be the abscissa and ordinate of Q. Now,
$$(x_1,y_1)\equiv\left(\frac{(2)(4)+(1)(0)}{3},\frac{(2)(0)+(1)(3)}{3}\right)$$
$$\equiv\left(\frac{8}{3}, 1\right)\tag{1}$$
$$...$$
Question:
- Should I use the $=$ sign or the $\equiv$ sign?
- If process 2 is correct, then in line $(1)$ should I use $=$ or $\equiv$ before $\left(\frac{8}{3}, 1\right)$? Currently, there is an $\equiv$ before $\left(\frac{8}{3}, 1\right)$ in line $(1)$.
Comments:
The usage of $=$ in process 1 seems completely fine to me. However, in my book, it has been written like process 2. So, am I correct to write in process 1?
The symbol $≡$ has multiple meanings in mathematics, to indicate a definition, a mathematical or logical equivalence between two statements, an equivalence relation (like in modular arithmetic, and congruence of geometric figures), an identity (an equation like $x^2-y^2≡(x+y)(x-y)$ that is true for every allowable combination of variable values), etc.
In your quoted example, there is no compelling reason to switch $=$ and $≡.$ The latter is just confusing here, unless the author has prefaced it with a note on their particular usage of $≡.$