If xy^2 = ab^2 : Does xy = ab?

[Femme_physics]

If

xy² = ab²

Does that mean automatically that xy = ab?

 

[Char. Limit]

Not necessarily. It's possible that xy = -ab.

 

[Femme_physics]

Got it :) Thanks

 

[I like Serena]

Let's try something.

$1 \cdot 2^2 = 4 \cdot 1^2$ is true isn't it?

But is: $1 \cdot 2 = 4 \cdot 1$ ?

 

[Char. Limit]

Oh, I was assuming the OP meant (xy)^2 = (ab)^2. If it's how you're suggesting, then (if x y^2 = a b^2, then xy = ab) only if y=b.

 

[I like Serena]

@CL: I suspect that it was what the OP meant too, but I felt the playful need to point out to the OP that round thingies should be used.

 

[Char. Limit]

Oh, of course. Round thingies should always be used. Or square thingies, they work too.

[xy]^2 = (ab)^2.

 

[NeroKid]

a^2-b^2 =0
<=> (a-b)(a+b)=0
<=> a=b or a=-b

 

[Mark44]

If

xy² = ab²

Does that mean automatically that xy = ab?

Femme_physics,
By chance, you don't mean (xy)² = (ab)², do you?

 

[Alex1812]

Assuming you mean (xy)^2 = (ab)^2, xy does not neccesarilly equal ab:

sqrt[(xy)^2] = sqrt[(ab)^2]
then
(+/-)xy = (+/-)ab
which is true for only two out of four cases:
+xy=+ab
-xy=-ab
But, you wind up with problems when -xy = +ab and +xy=-ab.

Hope that helps a little.

 

[Ouabache]

Assuming you mean (xy)^2 = (ab)^2, xy does not neccesarily equal ab:

sqrt[(xy)^2] = sqrt[(ab)^2]
then
(+/-)xy = (+/-)ab
which is true for only two out of four cases:
+xy=+ab
-xy=-ab
But, you wind up with problems when -xy = +ab and +xy=-ab.

I did not assume the OP meant {if $(xy)^2 = (ab)^2$, does $xy = ab$. I approached it as written {if $xy^2 = ab^2$ does $xy = ab$} which is false. Without proving it, by testing empirically, it becomes obvious it is false.

However if we assume it the first way, I don't see 4 cases, only 2, but that is sufficient to show the two equations are not equal (see below cases iii and iv, resolve to i and ii. Below my designation for case and equation numbering are interchangeable).

(i) xy = ab
(ii) xy= -ab
(iii) -xy= ab; multiply eq. by -1, you get xy= -ab (which is eq. ii)
(iv) -xy = -ab; multiply eq. by -1, you get xy = ab (which is eq. i )

 

[Alex1812]

You're right Ouabache, I didn't bother equating the pairs of equations to reduce it to two cases.

As written,
if (xy^2)=(ab^2)
then (x/a) = (b^2/y^2)

if xy=ab
then (x/a)=(b/y)

Then the original statement is true only iff (b/y)=(b^2/y^2)=((b/y)^2) which, as previously pointed out, is only true iff b=y.

And apologies for my ignorance of TeX.

 

[NeroKid]

if (xy^2)=(ab^2)
then (x/a) = (b^2/y^2)

if xy=ab
then (x/a)=(b/y)

Then the original statement is true only iff (b/y)=(b^2/y^2)=((b/y)^2) which, as previously pointed out, is only true iff b=y.
.

not true if y =0 and x = 0

 

[Alex1812]

That's true. Always good to keep in mind the impact of nul denominators. Anyway, I think we've disproven this more ways than Femme_Physics has time to read about lol