$(-1)^3$ has different results when evaluated as $(-1)times(-1)times(-1) = -1$ vs $((-1)^2)^{3/2} = 1$. Which...
$begingroup$
I know that
$$(-1)^3=(-1)times(-1)times(-1)=-1 tag{1}$$
but also
$$(-1)^3=((-1)^2)^{3/2}=1^{3/2}=1 tag{2}$$
So which gives the correct value of $(-1)^3$?
arithmetic exponentiation fractions
edited Jan 13 at 8:06
Blue
49.1k870156
asked Jan 13 at 7:49
senxsenx
163
$endgroup$
add a comment |
$begingroup$
I know that
$$(-1)^3=(-1)times(-1)times(-1)=-1 tag{1}$$
but also
$$(-1)^3=((-1)^2)^{3/2}=1^{3/2}=1 tag{2}$$
So which gives the correct value of $(-1)^3$?
arithmetic exponentiation fractions
edited Jan 13 at 8:06
Blue
49.1k870156
asked Jan 13 at 7:49
senxsenx
163
$endgroup$
add a comment |
$begingroup$
I know that
$$(-1)^3=(-1)times(-1)times(-1)=-1 tag{1}$$
but also
$$(-1)^3=((-1)^2)^{3/2}=1^{3/2}=1 tag{2}$$
So which gives the correct value of $(-1)^3$?
arithmetic exponentiation fractions
edited Jan 13 at 8:06
Blue
49.1k870156
asked Jan 13 at 7:49
senxsenx
163
$endgroup$
I know that
$$(-1)^3=(-1)times(-1)times(-1)=-1 tag{1}$$
but also
$$(-1)^3=((-1)^2)^{3/2}=1^{3/2}=1 tag{2}$$
So which gives the correct value of $(-1)^3$?
arithmetic exponentiation fractions
arithmetic exponentiation fractions
edited Jan 13 at 8:06
Blue
49.1k870156
asked Jan 13 at 7:49
senxsenx
163
edited Jan 13 at 8:06
Blue
49.1k870156
asked Jan 13 at 7:49
senxsenx
163
edited Jan 13 at 8:06
Blue
49.1k870156
edited Jan 13 at 8:06
Blue
49.1k870156
edited Jan 13 at 8:06
Blue
49.1k870156
49.1k870156
asked Jan 13 at 7:49
senxsenx
163
asked Jan 13 at 7:49
senxsenx
163
asked Jan 13 at 7:49
senxsenx
163
163
add a comment |
add a comment |
1 Answer
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$begingroup$
The first equality is correct. The second one isn't: it assumes that $a^{bc}=(a^b)^c$. This is true indeed if $a>0$ (and $b,cinmathbb R$), but $-1<0$.
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
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$begingroup$
The first equality is correct. The second one isn't: it assumes that $a^{bc}=(a^b)^c$. This is true indeed if $a>0$ (and $b,cinmathbb R$), but $-1<0$.
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
The first equality is correct. The second one isn't: it assumes that $a^{bc}=(a^b)^c$. This is true indeed if $a>0$ (and $b,cinmathbb R$), but $-1<0$.
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
The first equality is correct. The second one isn't: it assumes that $a^{bc}=(a^b)^c$. This is true indeed if $a>0$ (and $b,cinmathbb R$), but $-1<0$.
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
The first equality is correct. The second one isn't: it assumes that $a^{bc}=(a^b)^c$. This is true indeed if $a>0$ (and $b,cinmathbb R$), but $-1<0$.
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 13 at 7:55
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
add a comment |
add a comment |
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