Approximate $\displaystyle \sqrt{2}$
The Babylonian Method works well here. It makes use of the bisection method and the fact that if you divide a number by its square root, you get the square root.
You need to make an initial guess.
Since $\displaystyle \sqrt{1} < \sqrt{2} < \sqrt{4}$ that means $\displaystyle 1 < \sqrt{2} < 2$.
Average these endpoints to get $\displaystyle 1.5$ is a good starting guess.
$\displaystyle \frac{2}{1.5} = 1.33333333$.
So that means $\displaystyle 1.33333333 < \sqrt{2} < 1.5$.
Average these out and you get a new guess.
Follow this process until you reach your desired level of accuracy.
razemsoft21, it helps if you tell us the full details in the first post. Saying Approximate $\displaystyle \sqrt2$ isn't enough.
Also, there are many methods of solving squareroots. If you don't know how to solve it, then maybe asking for someone to tell you a method of solving squareroots.
Doing research would help too: Computing Square roots - Wikipedia. Here, it shows you many methods of how to solve a square root of a number.
Hello, razemsoft21!
Where did this problem come from?
It's unlikely that you would be assigned this problem
, , without being taught any approximation methods.
Approximate $\displaystyle \sqrt{2}$
The very worst you could do is guess-and-adjust.
Since $\displaystyle \sqrt{2}$ is between 1 and 2 . . .
$\displaystyle \begin{array}{ccccccc}
\text{Try 1.5:} & 1.5^2 &=& 2.25 & \text{too big} \\
\text{Try 1.4:} & 1.5^2 &=& 2.25 & \text{too small} \\
\text{Try 1.45:} & 1.45^2 &=& 2.1025 & \text{too big} \\
\text{Try 1.42:} & 1.42^2 &=& 2.0164 & \text{too big} \\
\text{Try 1.42:} & 1.41^2 &=& 1.9881 & \text{too small}
\end{array}$
We have the two decimal places.
Which is the "better" answer: 1.41 or 1.42 ?
1.41 gives an error of: .$\displaystyle 2 - 1.9881 \:=\:0.0119$
1.42 gives an error of: .$\displaystyle 2.0164 - 2 \:=\:0.0164$
Since 1.41 has the smaller error: .$\displaystyle \sqrt{2}\:\approx\:1.41$
refer to this page Square root of 2 - Wikipedia, the free encyclopedia
. . . An elementary approximation of $\displaystyle \sqrt{N}$
Let $\displaystyle a_1$ be your first approximation to $\displaystyle \sqrt{N}.$
Substitute into: .$\displaystyle \dfrac{N+a_1^2}{2a_1} \;=\;a_2$ .[1]
This produces $\displaystyle a_2$, a better approximation.
For more accuracy, substitute $\displaystyle a_2$ into [1] . . . and so on.
Example: .$\displaystyle \sqrt{2},\;\;N = 2$
Let $\displaystyle a_1 = 1.5$
We have: .$\displaystyle a_2 \:=\:\dfrac{2 + 1.5^2}{2(1.5)} \;=\;1.41666\hdots $
. . Check: .$\displaystyle (1.41666\hdots)^2 \:=\:2.006944443$
Then: .$\displaystyle a_3 \:=\;\dfrac{2 + (1.4166\hdots)^2}{2(1.41666\hdots)} \:=\:1.414215686$
. . Check: .$\displaystyle 1.414256862^2 \:=\:2.000007749$
Then: .$\displaystyle a_4 \:=\:\dfrac{2 + 1.41425686^2}{2(1.414256862)} \:=\:1.414213562 $
. . Check: .$\displaystyle 1.414213562^2 \:=\:1.999999999$
Therefore: .$\displaystyle \sqrt{2} \;\approx\;1.424213562$ .to 9 decimal places.
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Why does [1] provide better and better approximations?
We want an approximation of $\displaystyle \sqrt{N}.$
Our first guess is $\displaystyle a_1.$
We have factored $\displaystyle N$ into two factors: .$\displaystyle a_1\,\text{ and }\,\dfrac{N}{a_1} $
If we are extremely lucky, the two factors will be equal.
. . (Think about it!)
Since it is unlikely that the two factors are equal,
. . a better approximation is the average of the two factors.
. . (The number exactly halfway between them.)
So we use: .$\displaystyle a_2 \;=\;\dfrac{a_1 + \frac{N}{a_1}}{2}
$
. . which simplfies to: .$\displaystyle a_2 \;=\;\dfrac{N+a_1^2}{2a_1}$ . . . . . ta-DAA!
we must appreciate the works of Robert Nemiroff he calculated to 1 million digits just check this out!
The Square Root of Two to 1 Million Digits