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$\sqrt{4}=2$

But is it same as writing... $4^{1/2}=2$?

Basically I do not understand why $\sqrt{}$-sign equals $1/2$?

Cornman
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Saksham Sharma
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    For a speculative argument, square √4=2 and 4^1/2=2 and see what you get. If you mean a formal proof, post the definitions you use for $,\sqrt{,\cdot,},$ and $,(,\cdot,)^{1/2},$. – dxiv Aug 11 '18 at 05:00
  • Does √ halves the number of times the number is multiplied by itself? – Saksham Sharma Aug 11 '18 at 05:02
  • Sorry, but that doesn't make math sense. Replace $4$ in your question with $2$ (or $\pi$ for that matter), and it should become more obvious. What's under the $\sqrt{\cdot}$ is not necessarily an even integer power of something simple, so how would you halve the number of times the number is multiplied by itself? – dxiv Aug 11 '18 at 05:07

2 Answers2

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Basically it is just a definition of the $\sqrt{}$-sign. Note, that the $\sqrt{}$-sign is one of the few examples (maybe the only one?) in mathematics, where you write "nothing" (since $\sqrt{x}=\sqrt[2]{x}$) and actually mean $2$.

So yes. It is $\sqrt{4}=4^{1/2}=2$

To make it more clear, we have $\sqrt[n]{x}=x^{1/n}$, where $\sqrt[n]{x}$ notes the $n-th$ root of $x$. For example $\sqrt[3]{8}=2$, because $2\cdot 2\cdot 2=2^3=8$.

Cornman
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  • I am actually confused. Does √ halves the number of times the number is multiplied by itself? – Saksham Sharma Aug 11 '18 at 05:00
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    You missphrase it. $\sqrt{x}$ "asks" for the number which you have to multipy with itself to get $x$. So for $\sqrt{16}$ for example, we search for the number which we have to take to the power of $2$, to get 16. Which is $4$. – Cornman Aug 11 '18 at 05:02
  • Then does $x^1/2$ mean what is the number which when multiplied by itself two times gives x? – Saksham Sharma Aug 11 '18 at 05:05
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    @SakshamSharma Yes, that's correct. – Toby Mak Aug 11 '18 at 05:06
  • And is '2' in $√x$ always understood? Like $√x$ same as $2√x$? – Saksham Sharma Aug 11 '18 at 05:08
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    @Saksham Sharma No. You missphrase it again. $x^{1/2}$ is the number, which you have to multiply with itself to get $x$. :) It is just the sign, which askes the question. And for that you can either write $\sqrt{}$, $\sqrt[2]{}$ or ^(1/2) – Cornman Aug 11 '18 at 05:08
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    @SakshamSharma For mathjax, if you want to write more in the power, you have to use brackets: x^{1/2} is $x^{1/2}$. If you want to write the sqrt-sign with an power, you have to go like this: \sqrt[n]{x} to get $\sqrt[n]{x}$ – Cornman Aug 11 '18 at 05:10
  • @Cornman x^1/2 is the number, which you have to multiply with itself to get x That should be "the *positive* number". Because of course $,(-2)(-2)=4,$ but $,-2 \ne \sqrt{4},$. – dxiv Aug 11 '18 at 05:18
  • @dxiv Yes. Thanks for pointing that out. Actually I didnt write it, because I did not want to confuse Saksham Sharma more. – Cornman Aug 11 '18 at 05:20
  • @Cornman Not spelling it out might confuse them even worse ;-) – dxiv Aug 11 '18 at 05:22
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Unfortunately, I think, roots were studied before powers with arbitrary exponents were dared explored, so that the (ugly, in my opinion) symbol $\sqrt{(\cdot)}$ came much earlier than the other notation $(\cdot)^{1/2},$ which is a more natural notation as it falls out of the extension of exponentiation to rational numbers.

Specifically, a motivation for why $(\cdot)^{1/2}$ is considered to be the same as the operator $\sqrt{(\cdot)}$ is this. The (positive) square root $x$ of a positive number $r$ is the number (why this $x$ is unique is proved in calculus) satisfying the relation $x^2=r.$ Once we grasp this characteristic property of square roots, it becomes easy to see that if we want to extend exponentiation to rational exponents so that the usual laws (especially in this case that $(a^m)^n=(a^n)^m=a^{mn}$ and $a^1=a$) are preserved, then we must have $$(x^2)^{1/2}=r^{1/2},$$ whose left hand side becomes $x$ under our assumptions, so that we have the square root $x$ of $r$ given by $r^{1/2}.$

Allawonder
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