2

I'm having trouble seeing the error in the following "proof": $$ (-4)=(-4)^1=(-4)^\frac{2}{2}=[(-4)^2]^\frac{1}{2}=[16]^\frac{1}{2}=4$$ therefore $(-4)=4$.

Obviously this is incorrect, but I'm not seeing where the error is occuring. I appreciate any help. Thanks.

martini
  • 86,011
Jeffrey
  • 79
  • This mistake comes from using the rules of exponentiation as they are developed on positive real base numbers and believing that they behave nicely when you use them on negative numbers. This is not the case, as you've just discovered. – Arthur Mar 15 '16 at 09:04

3 Answers3

3

Your mistake starts here: $$(-4)=(-4)^1=(-4)^\frac{2}{2}\color{red}{=[(-4)^2]^\frac{1}{2}}=[16]^\frac{1}{2}=4$$

because $\sqrt{a^2}=\mid a\mid$

3SAT
  • 7,627
0

$x^{\frac{1}{2}}$ gives two values, both additively inverse to each other, on all nonzero real numbers $x$. You can think of an equation like $2^\frac{1}{2} = \pm \sqrt{2}$ as saying that the square root of $2$ is "$\sqrt{2}$ or $-\sqrt{2}$."

mad_algebraist
  • 870
  • 4
    Do you think $e^{1/2}$ has two values? That would be inconsistent with a lot of written mathematics. – hmakholm left over Monica Mar 15 '16 at 09:03
  • Doesn't it? A square root contains both positive and negative values, as $(-a)^2 = a^2$ – frog1944 Mar 15 '16 at 09:05
  • 3
    @frog1944: The exponential function is a function, giving one output for each one input, and writing this function as $e^x$ has hysterically large precedent in mathematics in general. – hmakholm left over Monica Mar 15 '16 at 09:07
  • 3
    On the other hand, insisting that $x^{1/2}$ has to mean any $a$ that solves $a^2=x$ seems to be the invention of the answerer here, and is directly in conflict with how mathematics uses expressions like $e^{1/2}$. – hmakholm left over Monica Mar 15 '16 at 09:08
  • I see what you're saying @HenningMakholm but when you look at wolfram alpha (http://www.wolframalpha.com/input/?i=e%5E(1%2F2)), it states that the is 2 answers to $e^{1/2}$ – frog1944 Mar 15 '16 at 09:10
  • 1
    @frog1944: Wolfram Alpha is a machine that tries to figure out the right mathematical meaning of the user's ASCII representation of mathematical notation. Sometimes it gets it wrong. – hmakholm left over Monica Mar 15 '16 at 09:44
  • The function $x \mapsto x^{\frac{1}{2}}$ is one of the simplest examples of a function on the complex numbers with multiple branches. You are right, however, in pointing out that the exponential function is single-valued. – mad_algebraist Mar 15 '16 at 10:45
0

The rule $a^{bc}=(a^b)^c$ is not true for arbitrary $a$, $b$, and $c$. The most important cases where it works are

  • When $b$ and $c$ are both integers.
  • When $a$ is a nonnegative real, and $b$ and $c$ are reals.

Here you're applying it with $a=-4$ and $c=\frac12$, which does not fall into any of those cases. Therefore you don't have any guarantee that the rule holds in this case, and indeed it doesn't.

hmakholm left over Monica
  • 291,611