Convert to binary........

Now, I believe you know how bases work-

suppose we are writing using base n. That means we have n digits that represent numbers:

0, 1, 2, ......n-1

for example, if n=2. we have 2 digits- 0 and 1.

Lets call the first digit the main digit (for example, the main digit in 11001010110 be 0, in 345 it will be 5 and so on)

If the digit d is K places to the left of the main digit, its value will be the base multiplied by itself K times, then multiplied by the digit. if the base is N it means the value will be d*N^K. It also means that if the digit is K places to the right of the main digit, its value will be d*N^-K=d/(N^K)

for example, if our base is 10, then the value of the marked digit in 0.007 will be 7*10^-3=7/(10³).

If our base is 2, it means the value of the marked digit in 0.01 will be 1*2^-2=1/(2²)

So, if we want to convert 3.141592 to base 2, we have to use those rules. 3.141592 (base 10)~~11.0011

Unfortunately for you, pi is irrational, meaning we cant write it with a finite number of digits (i know you didnt want me to do that but i just had to say that)

you wanted me to write 3.141592 in base 2, and im afraid to tell you we cant write this rational number with a finite number of digits either! but we CAN write it as a Repeating decimal. why? because if you dont want a fraction to be a repeating decimal (lets say the fraction is m₁/m₂) then after reducing the fraction you need all of m₂'s factors to be factors of the base N as well. It doesnt work with 141592/1000000, but we can write it as a repeating decimal.

Can you to that?

Convert to binary........

convert 3.141592 to base 2 with steps please, if possible.

Thank you for any help.

[pre-decimal point position]:

\(\begin{array}{|rcrcrr|} \hline && && & \text{Remainder} \\ \hline 3 &:& 2 &=& 1 & \color{red}{1} \\ 1 &:& 2 &=& 0 & \color{red}{1} \\ \hline \end{array}\)

decimals:

\(\begin{array}{|rcrcrr|} \hline && && & \text{Carry} \\ \hline 0.141592 &*& 2 &=& {\color{red}0}.283184 & \color{red}{0} \\ 0.283184 &*& 2 &=& {\color{red}0}.566368 & \color{red}{0} \\ 0.566368 &*& 2 &=& {\color{red}1}.132736 & \color{red}{1} \\ 0.132736 &*& 2 &=& {\color{red}0}.265472 & \color{red}{0} \\ 0.265472 &*& 2 &=& {\color{red}0}.530944 & \color{red}{0} \\ 0.530944 &*& 2 &=& {\color{red}1}.061888 & \color{red}{1} \\ 0.061888 &*& 2 &=& {\color{red}0}.123776 & \color{red}{0} \\ 0.123776 &*& 2 &=& {\color{red}0}.247552 & \color{red}{0} \\ 0.247552 &*& 2 &=& {\color{red}0}.495104 & \color{red}{0} \\ 0.495104 &*& 2 &=& {\color{red}0}.990208 & \color{red}{0} \\ 0.990208 &*& 2 &=& {\color{red}1}.980416 & \color{red}{1} \\ 0.980416 &*& 2 &=& {\color{red}1}.960832 & \color{red}{1} \\ 0.960832 &*& 2 &=& {\color{red}1}.921664 & \color{red}{1} \\ 0.921664 &*& 2 &=& {\color{red}1}.843328 & \color{red}{1} \\ 0.843328 &*& 2 &=& {\color{red}1}.686656 & \color{red}{1} \\ 0.686656 &*& 2 &=& {\color{red}1}.373312 & \color{red}{1} \\ 0.373312 &*& 2 &=& {\color{red}0}.746624 & \color{red}{0} \\ 0.746624 &*& 2 &=& {\color{red}1}.493248 & \color{red}{1} \\ 0.493248 &*& 2 &=& {\color{red}0}.986496 & \color{red}{0} \\ 0.986496 &*& 2 &=& {\color{red}1}.972992 & \color{red}{1} \\ \cdots \\ \hline \end{array}\)

\(3.141592_{10} = 11.00100100001111110101\ldots_2\)

Convert to binary........

convert 3.141592 to base 2 with steps please, if possible.

Thank you for any help.

[pre-decimal point position]:

\(\begin{array}{|rcrcrr|} \hline && && & \text{Remainder} \\ \hline 3 &:& 2 &=& \color{blue}{1} & \color{red}{1} \\ \color{blue}{1} &:& 2 &=& 0 & \color{red}{1} \\ \hline \end{array}\)

decimals:

\(\begin{array}{|rcrcrr|} \hline && && & \text{Carry} \\ \hline 0.141592 &*& 2 &=& {\color{red}0}.283184 & \color{red}{0} \\ 0.283184 &*& 2 &=& {\color{red}0}.566368 & \color{red}{0} \\ 0.566368 &*& 2 &=& {\color{red}1}.132736 & \color{red}{1} \\ 0.132736 &*& 2 &=& {\color{red}0}.265472 & \color{red}{0} \\ 0.265472 &*& 2 &=& {\color{red}0}.530944 & \color{red}{0} \\ 0.530944 &*& 2 &=& {\color{red}1}.061888 & \color{red}{1} \\ 0.061888 &*& 2 &=& {\color{red}0}.123776 & \color{red}{0} \\ 0.123776 &*& 2 &=& {\color{red}0}.247552 & \color{red}{0} \\ 0.247552 &*& 2 &=& {\color{red}0}.495104 & \color{red}{0} \\ 0.495104 &*& 2 &=& {\color{red}0}.990208 & \color{red}{0} \\ 0.990208 &*& 2 &=& {\color{red}1}.980416 & \color{red}{1} \\ 0.980416 &*& 2 &=& {\color{red}1}.960832 & \color{red}{1} \\ 0.960832 &*& 2 &=& {\color{red}1}.921664 & \color{red}{1} \\ 0.921664 &*& 2 &=& {\color{red}1}.843328 & \color{red}{1} \\ 0.843328 &*& 2 &=& {\color{red}1}.686656 & \color{red}{1} \\ 0.686656 &*& 2 &=& {\color{red}1}.373312 & \color{red}{1} \\ 0.373312 &*& 2 &=& {\color{red}0}.746624 & \color{red}{0} \\ 0.746624 &*& 2 &=& {\color{red}1}.493248 & \color{red}{1} \\ 0.493248 &*& 2 &=& {\color{red}0}.986496 & \color{red}{0} \\ 0.986496 &*& 2 &=& {\color{red}1}.972992 & \color{red}{1} \\ \cdots \\ \hline \end{array}\)

\(3.141592_{10} = 11.00100100001111110101\ldots_2\)

Thank you very much to both of you.

This is interesting. The incomprehensible slop at the top, well more irrational than Pi in decimal or binary form, has 2 points, but the concise answer by Heureka has zero. This is definitely an unbalanced set of solutions!

Thanks for the point, GingerAle.

This is _un^balanced! I placed this inequity of inequality before the Board of Directors of Lancelot Link’s A.P.E., to determine the optimum and relative value of Heureka’s solution.  After a debate, the BoD agreed that Heureka should receive bananas and peanuts and the first answer should get the peels and shells.  All the votes were unanimous, save one. Chip Loki and Chimp Stanley holds Lancelot’s two votes by proxy, while he’s on assignment in Tahiti. 

Chimp Perceval, Chancellor of the Exchequer, calculated the conversion of the bananas and peanuts into 17 points divided between the two posts by Heureka. A board member requested the Chancellor of the Exchequer to explain the math. 

One standard banana equals 1 point times the board’s vote counts (one for each regular member and two for Lancelot) of eight (8) yeas and one (1) nay equals eight points under the reinstated point system. Points to be applied to Heureka’s first post (#2).

Forty-seven (47) grams of shelled peanuts equals 1 point times the board’s vote counts (one for each regular member and two for Lancelot) of nine (9) yeas and zero (0) nays equals nine points under the reinstated point system. Points to be applied to Heureka’s second post (#3).

Additional comments and actions during the board meeting:

A gripe about the single point each member can authorize instead of the five under the old system.

A short debate preceded the vote on whether to give points to both the posts or only to the final post. The debate yielded a consensus: because this is a binary question, Heureka’s posting of the solution twice with the second post having two numbers more colorful, is both apposite and quite funny.

Most of the Board members threw banana peels and peanut shells at the member who voted ‘nay.’ The acting CEO ordered them swept up and tabled until the final call for a vote to offer them to the first poster responding to the question.

For the final order of business the acting CEO, Chimp Loki, ordered the Chancellor of the Exchequer to turn over the expropriated bananas to the company barkeeper for conversion to banana daiquiris to be served, with the expropriated peanuts, to all board members.   There were no objections or points of order. 

The deliberative assembly of the Board of Directors of Lancelot Link’s A.P.E. is closed.