Operators A & B Commute: Explain Why or Why Not?

[terp.asessed]

Homework Statement

Determine whether or not the following pairs of operators commute...and there was one I could not solve...according to the back of the textbook, I do understand 14.c does NOT commute, but I don't understand...

(14)c.
A = SQR
B = SQRT

Homework Equations

ABf(x) - BAf(x) = 0

The Attempt at a Solution

ABf(x) = A[f(x)]^1/2 = f(x)
BAf(x) = Bf²(x) = f(x)...so I thought they DO commute, but the textbook says NO! Could someone explain? Thanks!

 

[Mark44]

Homework Statement

Determine whether or not the following pairs of operators commute...and there was one I could not solve...according to the back of the textbook, I do understand 14.c does NOT commute, but I don't understand...

(14)c.
A = SQR
B = SQRT

Homework Equations

ABf(x) - BAf(x) = 0

The Attempt at a Solution

ABf(x) = A[f(x)]^1/2 = f(x)
BAf(x) = Bf²(x) = f(x)...so I thought they DO commute, but the textbook says NO! Could someone explain? Thanks!

I think you're getting lost in the symbolism. The question is asking whether
$(\sqrt{f(x)})^2 = \sqrt{(f(x))^2}$
Can you always take the square root of something?

 

[terp.asessed]

Wait, you mean,

$(\sqrt{f(x)})^2 = \sqrt{(f(x))^2}$
Can you always take the square root of something?

are different? I'm sorry, but I'm confused...

 

[Mark44]

Well, if they're different, then the two operations aren't commutative. If you believe they are different, why are they different?

I'll ask again, can you always take the square root of something?

 

[terp.asessed]

Come to think of it, one can't square root of the function that is negative...so NOT always...

 

[Mark44]

Come to think of it, one can't square root of the function that is negative...so NOT always...

Right. That's why the two operations don't commute. Good!

 

[terp.asessed]

GOTCHA----thank you!