Your code has a lot of issues:

1. Your indentation is off. Indentation is very important in Python since that is how it knows what goes with what.
2. You need to useisinstance to see if an object is a float. I assume this is what you are trying to do witha == float. But, that doesn't make sense because, in Python 3.x.,input always returns a string object. So,a is a string. However, iffloat is actually a variable, then you should change its name. Naming a variablefloat is a bad practice since it overrides the built-in.
3. You are missing a colon at the end of eachelse.
4. You are missing a closing parenthesis on the first line.
5. Thestr in the first line is unnecessary (not an error, but I just thought I'd mention it).

However, instead of fixing all this, I'm going to introduce you to thebin built-in:

>>> n = 127>>> bin(n)>>> # The "0b" at the start means "binary".'0b1111111'>>> # This gets rid of the "0b">>> bin(n)[2:]'1111111'>>>

It was built explicitly to do what you are trying to do.

Also, here are some references on Python you might enjoy:

http://www.tutorialspoint.com/python/python_overview.htm

http://wiki.python.org/moin/BeginnersGuide/Programmers

You can just use thebin function:

>>> bin(100)'0b1100100'

Ignore the0b infront of the string. You can always get the raw binary numbers using usingbin(your_numer)[2:].

Also, you can get this using theformat function:

>>> format(100, 'b')'1100100'

If you need to print it in binary you can just do: print(bin(a))

This is what i made

while True:    print("FIND OUT WHAT BINARY THIS IS")    space = " "    num1 = int(input())    while num1 > 0:        if num1 % 2 == 0:          space = space + "0"        else:          space = space + "1"        num1 = int(num1 / 2)    else:     space = space[::-1]     print(space)

• python
• binary
• integer