Does $2^2=4$ imply $2=\pm \sqrt{4}$?

Question

I read the square root property from the book, College Algebra by Raymond A Barnett and Micheal R Ziegler that,

The square root property says,

If $A^2=C$ then $A=\pm \sqrt{C}$

I took the equality,

$2^2=4$

$\implies 2=\pm \sqrt{4}$

$\implies 2=\pm {2}$

but, How $2$ can be equal to $-2$?

That notation means: If $A^2 = C$ then $A = \sqrt{C}$ OR $A = -\sqrt{C}$. Not AND
$$ $$

And indeed, the statement

$$2 = 2 \ \ \ \ OR \ \ \ \ 2 = -2$$

is true. — Simon S, Apr 24 '15 at 19:10
@SimonS Maybe better as an answer since it certainly answers the question at hand :) — Daniel W. Farlow, Apr 24 '15 at 19:16

score 2 · Answer 1 · answered Apr 09 '17 at 20:22

Note that $\sqrt4=\pm2$ means that $\sqrt4 = +2 \;\lor -2$, not$\sqrt4 = +2 \land\; -2$. You must have misread the $\pm$ notation, which reads as "plus OR minus", not "plus and minus". However, $2 = \pm2$ is certainly true. Note that $\lor$ is the logical OR.

score 1 · Answer 2 · edited Apr 10 '17 at 00:35

1

$2$ must be equal to $2$ OR to $-2$. Just one of them, not both.

edited Apr 10 '17 at 00:35

Nosrati

29,995

answered Apr 09 '17 at 23:27

esmo

97

@Moo I feel like, being very low-hanging fruit, the answer should intuit the source of the confusion, which in my opinion it has not done. You are free to disagree. – The Count Apr 10 '17 at 01:15

farruhota · Answer 3 · 2017-09-04T13:33:25.937

1

Let's see how "If $A^2=C$ then $A=\pm \sqrt{C}$" is obtained:

$A^2=C \Rightarrow A^2-C=0 \Rightarrow (A-\sqrt{C})(A+\sqrt{C})=0 \Rightarrow$ $A-\sqrt{C}=0$ or $A+\sqrt{C}=0 \Rightarrow A_1=\sqrt{C}$ or $A_2=-\sqrt{C} \Rightarrow A=\pm \sqrt{C}$.

Now let's see what happens to "$2^2=4$":

$2^2=4 \Rightarrow 2^2-4=0 \Rightarrow (2-\sqrt{4})(2+\sqrt{4})=0 \Rightarrow$ $2-\sqrt{4}=0$ or $2+\sqrt{4}=0 \Rightarrow 2=\sqrt{4}$ or $2\ne-\sqrt{4} \Rightarrow 2=\sqrt{4}$.

edited Sep 04 '17 at 13:33

answered Apr 14 '17 at 08:50

farruhota

31,482

Funny. But $ (2-\sqrt{4})(2+\sqrt{4})=0 $ does not impy the $ 2 = -\sqrt{4} $. That only works for variables. – Peter Driscoll Sep 04 '17 at 13:20
@PeterDriscoll, yes, it does not, therefore $2 \ne -\sqrt{4}$. The point was to show this. – farruhota Sep 04 '17 at 13:36
Ah I see. $ 2+\sqrt{4} \vee 2=-\sqrt{4} $ is still true. It is true because the first part is true. :) It is funny how the mind works sometimes. – Peter Driscoll Sep 04 '17 at 13:39
@PeterDriscoll, yes, you are right. – farruhota Sep 04 '17 at 13:42

Does $2^2=4$ imply $2=\pm \sqrt{4}$?

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