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You learn in algebra that $$\sqrt{ab}=\sqrt{a} \sqrt{b}$$ and that $$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$ You also learn to never make the fatal mistake of thinking $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$

However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.

$$\sqrt{a+b}=\text{?}$$

Michael Hardy
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Tdonut
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  • Note that the first two identities only hold when $a, b \geq 0$. As for your question, it depends on what one means by "rule", but the short answer is that there is no comparably simple expression. Since $\sqrt{\cdot}$ is concave, one can write down inequalities satisfies by $\sqrt{a + b}$. – Travis Willse Jun 19 '15 at 02:00
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    I suppose it depends on what you mean by a "rule". If you mean "some formula that lets you express $\sqrt{a+b}$ as some sort of combination involving $\sqrt a$ and $\sqrt b$", then no, there is no rule. – mweiss Jun 19 '15 at 02:01
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    @mweiss Of course, one can write $\sqrt{a + b} = \sqrt{(\sqrt{a})^2 + (\sqrt{b})^2}$, but this is hardly illuminating and really just emphasizes your point that the answer depends on what 'rule' means. – Travis Willse Jun 19 '15 at 02:04
  • @Travis : $;;;$ More generally, they hold when $: \operatorname{Im}\hspace{.02 in}(a) = 0 \leq \operatorname{Re}(a) :$ or $: \operatorname{Im}\hspace{.02 in}(b) = 0 \leq \operatorname{Re}(b) ;$. $\hspace{.88 in}$ –  Jun 19 '15 at 02:19
  • I suppose you could use imaginary numbers if you wanted... – Kbot Jun 19 '15 at 02:44
  • The nice rules that you know involve multiplication only, since the relation of square root to multiplication is so direct. But the fact of the matter is that multiplication and addition don’t play together very well, being connected only via distributivity. So any formula for $\sqrt{a+b,}$, if it exists, will be very ugly, as you see from the fine response of @lhf. – Lubin Jun 19 '15 at 02:55

3 Answers3

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There is no finite rule but there is the binomial series: $$ (1 + x)^\alpha = \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k $$ where $$ {\alpha \choose k} = \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!} $$ This series converges for $|x|<1$. Taking $\alpha=1/2$, we get $$ (1+x)^{1/2} = 1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5 x^4}{128}+\frac{7 x^5}{256}+\cdots $$

We can apply this to $\sqrt{a+b}$ as follows. Assume wlog that $a>b$. Then $$ \sqrt{a+b}=(a+b)^{1/2} = \sqrt{a}(1+x)^{1/2} $$ with $x=b/a<1$. We then have $$ \sqrt{a+b} = \sqrt{a}\left(1+\frac{b}{2a}-\frac{b^2}{8a^2}+\frac{b^3}{16a^3}-\frac{5 b^4}{128a^4}+\frac{7 b^5}{256a^5}+\cdots\right) $$

lhf
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Provided that $a>0$ and $a+b>0$, we have $$\sqrt{a+b}=\sqrt{a}\cdot\sqrt{1+\frac{b}{a}}$$

vadim123
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  • It's just like rewriting $a$ as $1\cdot a$ – Vim Jun 19 '15 at 10:07
  • This answer kind of reminds me of your answer here http://math.stackexchange.com/questions/454333/can-the-golden-ratio-accurately-be-expressed-in-terms-of-e-and-pi/454344#454344 I am a fan of your work! – graydad Jun 24 '15 at 01:35
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The general answer is no. However, you may encounter some special occasions like: $$\sqrt{3+2\sqrt 2}=1+\sqrt 2$$ in which $\sqrt{3+2\sqrt 2}$ happens to be $(1+\sqrt 2)^2$.

Vim
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