#12 Python Tutorial for Beginners | Number System Conversion in Python

All I agree, but why hi Aliens? 😂

if you want to convert from one system to another. you can use function for ex: hex(int('1010',2)) in the output: oxa

That's a wonderful session regarding number system conversion to one from another 👌
a = 15
b = 12
x = (a // 4 + b ** 3) < 2000 and (b % 4 != 0)
here b ** 3 = 1728 and a // 4 = 3. So, (a // 4 + b ** 3) = 1731 which is < 2000 --> True
b % 4 = 0 remainder so, b % 4 != 0 is False.
so output of x is False

decimal to binary conversion -->
31 - 0b11111, 52 - 0b110100, 65 - 0b1000001
binary to decimal conversion -->
0b110011010 - 410

It gives the output as False because it satisfies the first condition (1731<2000) and for the second condition it fails (12%4! =0).
For the 'and' operator both conditions should be true. Therefore, the answer is False. @Telusko

The answer is false

Please anna repu 3 ki ibm exam vundhi coding help cheyyandi anna

Hw answer is false because 1731<2000(true) but 12%3 is not 0(false) hence it is false and no op due to no print function

The ans for that question that u have given is "False"becoz it is (and) operator before the and operator it is True value and after (and) the value is "False" sir

The answer is False because the final answer would contain True and False, according to the truth table of and. Both have to be True for the result to be True. some common terms to note when solving are as follows: //-removes float, **-means raised to the power of a number, % remainder of a division, !-not. With these you should understand how to solve the question.

Really fun in learning, engineering didn't teach how to use this

Bro this de

output will be False as right condition of 'and' operator is false

Ok i did my code for today (18)

sir its using ip conversion formula to convert in binary form 00000000 . means 128 64 32 16 8 4 2 0 it is easy to understand

Hometask done...bin(31)=0b11111, bin(52)=0b110100, bin(65)0b1000001, the output of x is false as 1731<2000 is True but 0!=0 is False, so as per Truth Table 'and ' operation refers if one case case is true and other is false then ultimate output is False...Thanks again Mr. Reddy for teaching these concepts so well..

Output:False

ans to quiz is false because the second condition is not satisfying

true

45

False

done

Thank you very much

it is true

110011010...ans 256..
M i ri8

false
a//4 + b**3 = True
but
b%4 =0 hence 0!=0 is False

Sir can u tell the output at last question

False. Cause the result of the first parenthesis is 1731. And that is not less than 2000. The second parenthesis gives True though

Its possible to convert floating value to binary?? for example 23.5 to binary???

you really didn't focus on the octal or the hexadecimal in this specific lesson as much as you focused on the binary to decimal (not saying that is bad tho as I did learn something regarding conversions)...

Can anyone suggest me any videos to understand this topic deeply because I'm from non computer background it's difficult to understand for me

x=1731 is <2000 and thus True
b%4 =0 and thus b%4 !=0 is False
since & operator is used both conditions have to be true for the final statement to be true. but 2nd part is False. so the output will be False

for 110011010 it said 410 why is that?

I think r palakadam raadhu kadha

X=false

Wow. I love your teaching. I tested my self on a lot of binary equations both ways and was getting all of them correct. I’m also a total beginner. Thank you.

Hi Navin, I am a serious fan of you and your videos and the way you teach and finally smartly say 'bye bye'. It's like I am stick with you like glue. Can you please help me to be a python Master.

While changing from Binary to decimal getting traceback errors. I am putting Ob11111 and expecting result as 31 but getting Traceback error, name error: name 'Ob111111'is not defined. Please advise me

X
False

For 31-'0b1111'
For 52-'0b1001000'
For 65- 'ob1000001'


For 110011010 -820.

Manually calculated if any mistakes iam sorry sir.

Answer is False (1728<2000) and False

false

x=(a//4+b**3)<2000 and (b%4!=0)
here first we need to finish the operations mentioned in the brackets
a//4 is 3 (actual value is 3.75 but this symbol "//' takes only int values)
b**3 is b raised to 3 so 1728
x=(a//4+b**3) output is 1728+3= 1731


b%4 the remainder is 0
(b%4!=0) is false i.e 0 is equal to 0
1728 is less than 2000 but the statement as false so it doesn't satisfy both conditions
1728< 2000 and false
output of x= false

Thanks I'm 1in the class 👍🏻👍🏻

Output : False

Answer for homework question is false

False

int,float,complex,bool

Answer: False
First of all - (//:print the quotient as integer, ** - Power, % - Modulus(Get the remainder), != not equal to)
(a // 4 + b**3) < 2000 and (b%4 !=0)
= (15 // 4 + 12**3) < 2000 and (12%4 !=0)
According to bodmas = 3 + 1728 < 2000 and 0 !=0
Hence, = 1731 < 2000 and 0!= 0
The first condition is true but the second is false, hence the solution will come as 'False'
[Do not include brackets in solution]

A = 15
B = 12
X = (a // 4 + b ** 3) < 2000 and (b % 4 != 0)
Ans : False
Because
( a // 4 + b ** 3) = 1731
(b % 4 != 0) = [0 = 0] this statement is wrong.....
So x = False
......................