Maths

[[BTF]adam]

Math is codelike!

Stuck on a few math problems, wondering if anyone can help.

Firstly:

I need to work out what Q = 1.78 x 10-7(and after the 10 its alittle -7)

also I need to determine the base 10 number represented by 0101001011000110 S: (this is under the header of floating point numbers)

[QwazyWabbit]

Do you mean 1.78 x 10^-7 ? A raised little 7?
This would be read as one point seven eight times ten to the minus seventh.
Scientific notation.

Decimal notation would be 0.000000178. Or 1.78e-7 in Windows calculator.

0101001011000110  = 21190 decimal = 52C6 hex

[[BTF]adam]

yeah, a raised little -7, I thought it was to the power of -7 but wasn't sure.

Thanks for the help, appreciated.

[reaper]

"also I need to determine the base 10 number represented by 0101001011000110"

this is like the right most digit has a value, using base 10, of 2^0, and the left most 2^15.  you can remove the leading zeros because they don't mean anything.

so you have:
2,4,64,128,etc, and you can add them together to get the value in base 10.  there are quick ways to do this in your head, but there are also calculators

A group of 4 binary digits can be represented as one base16 (hex) digit, since the 1111 is the maximum value for one hex digit.

[QwazyWabbit]

The 0101001011000110 representation is 32 bits. The leading zero would be significant if it was signed binary. A negative integer would have the MSB set at 1 in 32-bit signed binary representation. One does not casually toss out leading zeros in binary representations.

[reaper]

That's 32 bits?  If you count them it's not close to that.

In IPv6, the whole representation so it's human readable tosses out leading 0's.

If you declare a signed integer value and alter it with bitwise operations, that's one thing.  But above is not anything near 32 bits, or something doing bitwise operations on the integer variable.

[QwazyWabbit]

A typo. It's 16 bits. I had a plane to catch and I was late.

The whole topic has nothing to do with IPv6 and its representation of addresses.

The point is that a cluster of bits with a leading zero to make it an even power of two number of bits is perfectly acceptable.

The point is also that without more knowledge about the type of the data you cannot assume leading zeros are insignificant.

I would also point out that adam's original post stated "this is under the header of floating point numbers" and if those 16 bits are the mantissa of a floating point number then the zero is VERY significant.

[[BTF]adam]

Yeah it's a 16 bit (two bytes). How did you get that base 10 number from 0101001011000110? trying to understand how it works as well as complete the tasks you see, wouldnt mind knowing :p - if it helps, it says the first 12 bits are used for the mantissa, and the remaining 4 for the exponent.

I also have another question underneath which I truly do not know where to begin with, all this mantissa and exponent malarky is confusing.

Write -0.3125 as a floating point number? i'm bad enough with bits/bytes/binary etc but when there are fractions or negatives in there it screws me riiight over.

[[BTF]adam]

Do you mean 1.78 x 10^-7 ? A raised little 7?
This would be read as one point seven eight times ten to the minus seventh.
Scientific notation.

Decimal notation would be 0.000000178. Or 1.78e-7 in Windows calculator.

also - is this the actual answer to 1.78 x 10 (little raised -7)? reason being is I need to do "PQ" and that was Q, P I think I worked out myself:

4.32 x 10(raised)3

 = 4320

so I would be adding 4320 to 0.000000178 ? :O

[reaper]

Just converting that number from base2 to base 10, just goes:
furthest bit on the right is 2^1, then in order from the left 2^2,2^3,2^4,etc and then you add the values

So for instance:

1111 would be 15

becuase 1+2+4+8=-15

But if the binary number is referenced as a floating point value you might want to check:
http://en.wikipedia.org/wiki/Floating_point

[[BTF]adam]

it is a floating point number yes.

[reaper]

with a 16 bit binary number it is probably meant to do just that, convert it from binary to base10

21190


Just converting that number from base2 to base 10, just goes:
furthest bit on the right is 2^1, then in order from the left 2^2,2^3,2^4,etc and then you add the values

If the question is really talking about the the bits of a floating point value, and what the value of the number actually is.  It is a 16 bit value, so from the wikipedia article you can see:
"
[edit] Internal representation
Floating-point numbers are typically packed into a computer datum as the sign bit, the exponent field, and the significand (mantissa), from left to right. For the IEEE 754 binary formats they are apportioned as follows:

Type Sign Exponent Exponent bias significand total
Half (IEEE 754-2008) 1 5 15 10 16
"

So the value of the number is dependent on where the decimal point is, which in floating point terms, means you need to work out the above. Please read the wikipedia article and maybe ask if you are just supposed to convert the base2 to base10, or it is a 16 bit floating point value as described in the wikipedia article.  couldn't hurt to read it either....

 

[haunted]

Adam, the carrot key (^) denotes the 'raise' you're talking about. As in x squared = x^2. It's pretty much always like that over computer and on most calculators that I've used, and you will have to type it that way in different kinds of software if you take 3 dimensional calculus as well as multivariate. Just pointing that out for ya!

[[BTF]adam]

Thanks both!

I did read it reaper, but all the stuff about IEEE and whatnot seemed pretty irrelevant to this minor task of a few sums I have to try and complete ^^ I was lost at like the second paragraph.

[reaper]

You declare a floating point variable in your C source code.

float blah;

When you compile then run the program, 4 bytes of memory is allocated to store this variable.

If you took at a look at the contents of the memory assigned for that floating point variable, you might see something like:
01010101010101010101110101011101 , with 0x denoting where the memory is located.

With declaring a floating point variable, you cannot just take the binary number and convert it to base 10, and say..this is my number, 60,200 (just made that up).  Because to do the floating point calculations, the processor and code work together, and there is meaning to the bits.

As described by wikipedia here is the meanings, a portion of the description
"
Floating-point numbers are typically packed into a computer datum as the sign bit, the exponent field, and the significand (mantissa), from left to right. For the IEEE 754 binary formats they are apportioned as follows:
"

It also goes on to say the format of a 16bit floating point number, which is what you were posed with.

However, you might just be asked to convert it to binary from base10, but that doesn't seem to make sense if it's under the category of floating point numbers.  So you are going to need to figure out which bits are the sign bit, exponent field, ect, and calculate the value of the number.  Floating point shifts the decimal around, and the values changes where the decimal point in.  You need to factor in the meaning of the special bits, and figure out what the number is..  You should probably read the article