# Approximate Sqrt(2)

• Sep 4th 2010, 08:05 PM
razemsoft21
Approximate Sqrt(2)
Approximate $\displaystyle \sqrt{2}$
• Sep 4th 2010, 08:12 PM
Educated
$\displaystyle \sqrt2\approx1.414$
• Sep 4th 2010, 08:23 PM
razemsoft21
Approximate $\displaystyle \sqrt{2}$ to #.## and show how please.
• Sep 4th 2010, 08:33 PM
Prove It
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.

• Sep 4th 2010, 09:22 PM
mr fantastic
Quote:

Originally Posted by razemsoft21
Approximate $\displaystyle \sqrt{2}$

It helps if you post all of the information relevant to the question. Approximate using what method? To how many decimal places? etc.
• Sep 4th 2010, 11:29 PM
razemsoft21
Quote:

Originally Posted by mr fantastic
It helps if you post all of the information relevant to the question. Approximate using what method? To how many decimal places? etc.

Approximate using what method? ----> If I know the method, no need to ask the question !
To how many decimal places? ----> already mentioned in post #3 (#.## means to 2 decimal places.

I think the method of Prove It works
• Sep 5th 2010, 12:43 AM
Educated
Quote:

Originally Posted by razemsoft21
Approximate using what method? ----> If I know the method, no need to ask the question !
To how many decimal places? ----> already mentioned in post #3 (#.## means to 2 decimal places.

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.
• Sep 5th 2010, 04:17 AM
Soroban
Hello, razemsoft21!

Where did this problem come from?
It's unlikely that you would be assigned this problem
, , without being taught any approximation methods.

Quote:

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: .$\displaystyle \sqrt{2}\:\approx\:1.41$
• Sep 5th 2010, 04:42 AM
grgrsanjay
• Sep 5th 2010, 04:46 AM
Soroban
. . . 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.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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.

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!
• Sep 5th 2010, 05:05 AM
ggn
this is more accurate

1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799
• Sep 5th 2010, 05:09 AM
ggn
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
• Sep 5th 2010, 10:40 AM
razemsoft21
Thank you for help and especial THANKS for:
Prove It , Educated , Soroban
• Sep 5th 2010, 04:36 PM
HallsofIvy
Quote:

Originally Posted by razemsoft21
Approximate $\displaystyle \sqrt{2}$ to #.## and show how please.

Here's a perfectly valid method: a calculator (for example, the one included with "Windows") gives $\displaystyle \sqrt{2}$ as 1.4142135623730950488016887242097. To two decimal places, that is 1.41.