Exponentation problems

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Exponentation problems

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i have a ${\sqrt{{\mathtt{a}}}}$ and the same as ${{\mathtt{a}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}$ ;

then it is equal to ${{\mathtt{a}}}^{\left({\mathtt{1}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)\right)}$ ;

according to the identities ( ${{\left({{\mathtt{a}}}^{{\mathtt{n}}}\right)}}^{{\mathtt{k}}} = {{\mathtt{a}}}^{\left({{\mathtt{n}}}^{{\mathtt{k}}}\right)} = {{\mathtt{a}}}^{\left({\mathtt{n}}{\mathtt{\,\times\,}}{\mathtt{k}}\right)}$ ) it is the same as ${{\mathtt{a}}}^{\left({{\mathtt{1}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)}$ ;

in this case it is the same as ${{\mathtt{a}}}^{{\sqrt{{\mathtt{1}}}}} = {{\mathtt{a}}}^{{\mathtt{1}}}$ ;

${{\mathtt{a}}}^{{\mathtt{1}}} = {\mathtt{a}}$ ;

SO: I have proven a false statement ${\mathtt{a}} = {\sqrt{{\mathtt{a}}}}$ ;

could you help me understand the problem?

juhaszbenedek08 Dec 28, 2014

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The product of $\sqrt{a}$ and $a^{\frac{1}{2}}$ is $a^{\frac{1}{2}}\times a^{\frac{1}{2}}=a^{\frac{1}{2}+\frac{1}{2}}=a^1=a$

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The general rule here is $a^n\times a^k=a^{n+k}$

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Alan Dec 28, 2014

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Alan · Answer 1 · 2014-12-28T07:03:09

The product of $\sqrt{a}$ and $a^{\frac{1}{2}}$ is $a^{\frac{1}{2}}\times a^{\frac{1}{2}}=a^{\frac{1}{2}+\frac{1}{2}}=a^1=a$

.

The general rule here is $a^n\times a^k=a^{n+k}$

.

Exponentation problems

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