We can, although not in very formal way, prove the property $((x)^b)^a$=$(x)^{ba}$ this way:
$((x)^a)^b=(x)^a*(x)^a*…*(x)^a$ which is a product of $b$ number of $(x)^a$ terms.
$\space\space\space\space\space\space\space\space\space\space\space\space=(x*x*…*x)*(x*x*…*x)*…*(x*x*…*x)$ where $(x*x*…*x)$ is a product of $a$ number of x terms. We can now conclude that this expansion equals $(x)^{ba}$ since multiplication is associative. But this is proof for $a,b\in\mathbb{N}$. How would you prove this formula for $a,b\in\mathbb{R}$ if it can be proved?
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ToTheSpace 2
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The problem is the base of the power that must be $>0$. Infact $((-2)^2)^{1/2}=[(-2)^{1/2}]^2$, where $(-2)^{1/2}$ is undefinite in $\mathbb R$. – Sebastiano Nov 12 '19 at 22:55
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Well your example shows that it cannot be proved for rational exponents : if it could, then $-2=2$ ! – Maxime Ramzi Nov 12 '19 at 22:56
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1Fractionary exponents are defined only positive numbers, and the usual formulæ for roots consesuently are valid only for positive numbers. – Bernard Nov 12 '19 at 22:58
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$f(x) = |x| := +\sqrt{ (x^{2})}$ – ZAF Nov 12 '19 at 23:00
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Cf. this question – J. W. Tanner Nov 12 '19 at 23:02
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Precise the base must be greater than or equal to zero. – Sebastiano Nov 12 '19 at 23:02
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Thank you all for your comments. I don't know why I haven't already stumbled upon this till now. I'll play with this more and see what's going on. As for now, thank you for pointing some things out. – ToTheSpace 2 Nov 12 '19 at 23:18
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You have stumbled upon the reason why the rule $$ (x^a)^b = x^{ab} $$ either requires $x>0$, or it requires $a, b$ to be integers (specifying what is valid at $x = 0$ requires additional care, as then it's not fractional exponents that are problematic, but negative exponents). This is the resolution to your issue: when you say $((-2)^{2})^{1/2} = (-2)^1$, that expression simply doesn't fulfill the requirements we set on well-behaved exponential expressions, and thus it's not a valid equality.
On the other hand, $((-2)^2)^{1/2} = 4^{1/2} = 2$ is entiely valid, because that's how parentheses work.
As for how to prove $(x^a)^b = x^{ab}$ for general real $a, b$ and $x> 0$, that depends entirely on how you define $x^a$ in the first place. If you're using continuity, then it follows directly from a standard continuity argument. If you have defined $x^a$ as $e^{a\ln x}$, then it follows from standard logarithm manipulations.
Arthur
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brackets need evaluated first. BEDMAS, remember? in this light, $((-2)^2)^\frac{1}{2}=(4)^\frac{1}{2}=2$
therefore,$((-2)^2)^\frac{1}{2}\ne(-2)^{2\cdot\frac{1}{2}}$
WindSoul
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