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Is there someone show me Why this is not true ?

$$-1=(-1)^1=(-1)^\frac{2}{2}=({(-1)}^{2})^{1/2}=\sqrt{1}=1$$

then :$$-1=1$$

Thank you for any help

Asaf Karagila
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zeraoulia rafik
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1 Answers1

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You are not allowed to split indices like that if you have a negative number. $x=a^{bc} \not \implies x=(a^b)^c $ if a is negative. So the step $(-1)^1=((-1)^2)^{(1/2)}$ is wrong.

RowanS
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  • indices for what ? pleas look well the given step – zeraoulia rafik Jun 26 '15 at 23:03
  • ok, i see it's not duplicate, in this case discussed about power not product under sqrt root – zeraoulia rafik Jun 26 '15 at 23:10
  • and how to convice someone that : 21/2 \neq1/22 where a is negative ? – zeraoulia rafik Jun 26 '15 at 23:14
  • @user51189 You got to be very careful with the standard rules of exponents when the base is negative. In essence, the standard rules of exponents on negative basis do not always apply the way we do on positive basis. Your example is one of them, but there are many more. Most people only prove the rules according to High School Standard. A more appropriate approach is to use logarithms (the natural logarithm in particular) and e-powers to prove all the rules of exponents. In the real world, the natural log of a negative cannot be taken, so negative basis are then omitted. – imranfat Jun 27 '15 at 00:30
  • @user51189 There are numerous discussions of this issue on this site (have a look, for instance, at the "Related" links in the right-hand margin of this "page" and the related links you can follow still further from those). In brief, roots of negative real numbers are not unique and careless handling of them can lead to various sorts of false "equations". [For a start, this question is similar to yours: http://math.stackexchange.com/questions/472227/math-fallacy-problem-1-13-16-2-sqrt-16-1?rq=1 . ] – colormegone Jun 27 '15 at 01:25