0

Let $\mathrel{R}$ by a relation defined on the set $\mathbb{R}$

$a\mathrel{R}b \iff$ a is a solution to the equation $X²-2bX+b²=0$

How do I prove that it's a transitive relation?

Here's what I tried :

Let a, b, c in R such that $a\mathrel{R}b$ and $b\mathrel{R}c$

$a\mathrel{R}b \iff$ $a²-2ba+b²=0$

$b\mathrel{R}c \iff$ $b²-2cb+c²=0$

This is where I am stuck, I don't know how to advance any further to prove that $a\mathrel{R}c$

Cheeze
  • 69
  • 4
    Have you thought of factoring $x^2-2bx+b^2$? You should be able to find that your relation can be written in a much simpler way without having to mention polynomials or anything of the sort at all.... "$a\mathcal{R}b\iff a$___$b$____" Fill in the blanks. Use this way of writing the relation to completely trivialize this problem. – JMoravitz Feb 22 '21 at 13:59

1 Answers1

1

Hint: Factor $x^2-2bx+b^2$

$(x-b)^2$

Recognize that if $a$ is a solution to $x^2-2bx+b^2=0$ then

$a=b$

Your relation can then be rewritten as

$a\mathcal{R} b\iff a=b$

JMoravitz
  • 79,518