MHB Symmetry in the algebraic expressions

AI Thread Summary
The expression $(x+a+b)(x+b+c)(x+c+a)$ exhibits symmetry in the variables a, b, and c, meaning that switching any two of these variables does not change the overall expression. This is referred to as "cyclic symmetry," where the expression remains invariant under cyclic permutations of the variables. However, it is also described as "fully symmetric," which is a stronger condition; in this case, swapping any pair of variables results in the same expression. An example of a non-fully symmetric expression is $(a - b)(b - c)(c - a)$, which changes sign when two variables are swapped. Understanding these symmetries is crucial for analyzing algebraic expressions and their properties.
NotaMathPerson
Messages
82
Reaction score
0
Hello!

I read about the symmetry of the following product

$(x+a+b)(x+b+c)(x+c+a)$ my book says that a, b and c occur symmetrically. Why is that?
 
Mathematics news on Phys.org
NotaMathPerson said:
Hello!

I read about the symmetry of the following product

$(x+a+b)(x+b+c)(x+c+a)$ my book says that a, b and c occur symmetrically. Why is that?

This is also called "cyclic symmetry" and it simply means that if you switch the two variables in any of the 3 pairs in $(a,b,c)$, you still have the same expression. Try it to verify. :)
 
Actually, it's "fully symmetric" which is *different* than cyclic symmetry. To see the difference, consider:

$(a - b)(b - c)(c - a)$

If we send $a \to b \to c \to a$, we get:

$(b - c)(c - a)(a - b)$, which is the same expression (even though the factors are in a different order).

If we just switch $a$ and $b$, we get

$(b - a)(a - c)(c - b) = -(a - b)(b - c)(c - a)$, which is *not* the original expression, but its negative.

That is-if we can swap any pair and get the original expression, it is fully symmetric (permutations are generated by pair-swaps), full symmetry implies cyclic symmetry, but the reverse is not so.

(Note that $(a - b)(b - c)(c - a)$ also remains unchanged under the transformation $a \to c \to b \to a$ which gives:

$(c - a)(a - b)(b - c)$).

The transformation (substitution) $a \to b \to c \to a$ is called a 3-cycle, as it changes 3 things to something else, and repeating it three times "completes the cycle" and leaves you where you were.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top