## Newtons Square root method

1. The problem statement, all variables and given/known data

Let e be the number close to sqrt(a) by Newtons Method (That is picking a number, diving a by it, and taking their average, divide a by average, get a number, find their average, so on). Using |e<sqrt(a)+e|
prove that if |a/e-e|<1/10
then |sqrt(a)-e|<1/10

Note that e is using the Newtons method a few times, not necessarily infinity, for any number of times. Also this is about positive integers, and 0 only, root and a.

2. Relevant equations

3. The attempt at a solution
So were trying to prove the second one smaller then first (I think), that is:
|sqrt(a)-e|<|a/e-e|
sqrt(a)-e<a/e-e (since both are positive, as using the given inequality subtract e from both sides) so sqrt(a)<a/e
e*sqrt(a)<a,
but e is not necessarily smaller then sqrt a, what am I missing?
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 e =>0 and e0 right? if e == sqr(a) then you'd have e*e = a and hence a < a which is wrong
 Sorry, i had a typo, its a +, not a -,| e< sqrt(a)+e|

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## Newtons Square root method

 Quote by Bonaparte Sorry, i had a typo, its a +, not a -,| e< sqrt(a)+e|
Has that resolved your problem or are you still stuck?
 Recognitions: Gold Member Science Advisor Staff Emeritus To find $\sqrt{a}$, we choose some starting value, e, and calculate a/e. There are three possibilities: 1) $e= \sqrt{a}$. Then $e^2= a$ so that $e= a/e$. We get the same number again and so know that we are done. 2) $e< \sqrt{a}$. Then multiplying both sides by $\sqrt{a}$, $e\sqrt{a}< a$ and $\sqrt{a}< a/e$. That is, $e< \sqrt{a}< a/e$. The square root of a is somewhere e and a/e and the midpoint, (e+ a/e)/2, is as good a place to look as any. 3) $e> \sqrt{a}$. Then multiplying both sides by $\sqrt{a}$, $e\sqrt{a}> a$ and $\sqrt{a}> a/e$. That is, $e> \sqrt{a}> a/e$. The square root of a is somewhere e and a/e and the midpoint, (e+ a/e)/2, is as good a place to look as any.