Mathematics Asked by Physor on February 12, 2021
The equation $x^2 = 9$ has the solution set {-3,3}, that is $(x = -3 vee x = 3)$. So I would understand that
$$
x^2 = 9 iff (x = -3 vee x = 3)
$$
Now "usually" (perhaps) one writes in course of solving an equation
$$ dotsquad x^2 = 9 Rightarrow x = 3 $$
but I think that is wrong, because a set implies a "larger" or equal set, So the correct one should read
$$ x^2 = 9 Leftarrow x = 3 $$
So which one is correct ?
$x^2 = 9 iff x^2 - 9 = 0$ is clear. Of course $x^2 -9 = (x-3)(x+3)$, and in all fields (like $Bbb R$ or $Bbb C$) we have that $$ab= 0 iff (a = 0) lor (b=0)tag{1}$$ for all $a,b$ in that field.
So $$x^2 = 9 iff (x-3)(x+3)= 0 iff (x-3 = 0) lor (x+3=0) iff x=3 lor x= -3$$
So every step is reversible by standard facts about inverses and $(1)$.
Answered by Henno Brandsma on February 12, 2021
To begin with, $x^2=9 Leftarrow x=3$ is certainly correct.
Regarding the other way around, $x^2=9 Rightarrow x=3$ it depends. If $x$ earlier has been restricted to be non-negative, e.g. if $x in mathbb{N}$, then the implication is correct. If $x$ can be negative, the implication is not correct, since also $x=-3$ is a possibility.
Answered by md2perpe on February 12, 2021
$$x^2=9Rightarrow x=3$$ is wrong.
Try $x=-3.$ $$x=3Rightarrow x^2=9$$ is true of course.
Also $x^2=9$ is equivalent to $x=3$ or $x=-3$ because $x^2-9=(x-3)(x+3).$
Answered by Michael Rozenberg on February 12, 2021
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