I am trying to learn the fundamentals behind the conversion of binary to decimal/Hex/Oct. For now I am focusing on decimal to binary conversion, without the use of the engineering functions dec2bin. I haven't found much online regarding this.
Thus far I have figured out how to convert a cell with a positive decimal number inside of it into 8 boxes (8 bits) making up 0's and 1's.
The way I have done this is using the MOD and INT functions.
First bit with formula
2nd Bit with formula
In the above two images I show how I manually calculate the process of converting from decimal to binary. The question is, this is the long way of doing it and I would like help to make it so that it is simply two boxes; one with a decimal and the other with the binary, so that when I enter the decimal it is instantly calculated into the binary beside it, as demonstrated in picture 3.
decimal to binary short form
If anyone could help me understand how to make my formula work using just one cell instead of 8 separate and different formulaes, it would be appreciated.
Use CONCAT as an Array formula:
=CONCAT(INT(MOD(A3/2^(8-ROW($1:$8)),2)))
Being an array formula it must be confirmed with Ctrl-Shift-Enter instead of Enter when exiting edit mode.
CONCAT was introduced with Office 365 Excel.
If one does not have Concat, this can be done with TEXT and SUMPRODUCT:
=TEXT(SUMPRODUCT(INT(MOD(A3/2^(8-ROW($1:$8)),2))*(10^(8-ROW($1:$8)))),"00000000")
Related
I have a list of fractions in Excel which I want to format as fractions, including integers. However, by default Excel formats integers as integers which is understandable.
Is there any way to force Excel to format, say, 4/4 as 4/4 instead of 1?
I need it to be stores as values and not as text, so '4/4 wont work. As I need to average a bunch of values from it afterwards.
Apparently I'm the first person ever to take issue with this, because google provides absolutely no help whatsoever :o
Yes, use a custom number format:
?/4
I don't believe what you are trying to accomplish is doable outright, as fractions are really division problems. However, with some formula trickery, you may be able to get something that will work for you.
If you place '7/8 in cell A1 and then use the following formula in cell B1
=DECIMAL(MID(A1,1,FIND("/",A1,1)-1),10)/DECIMAL(MID(A1,FIND("/",A1,1)+1,LEN(A1)),10)
The cell will display the decimal value of the "fraction", in this case 0.875 allowing you to change the denominator at will and still perform math functions.
This works because the formula slices up the "fraction" stored as text and converts it to a number and performs the math.
A colleague of mine sent me their Excel sheet and asked me to take a look at it. The issue is that with a very specific number (56136.598), Excel is automatically extrapolating that number out to 10 decimal places completely regardless of the formatting options.
The cell displays the number to the correct 3 decimal places, but if you look at the number in the formula bar it displays all 10 decimal places. It even changes the number to 10 decimal places if I write the formula =round(56136.598,3) to =round(56136.5979999999,3).
Unfortunately, given the industry I am in, I need some explanation as to why this very specific number induces this change. It's not enough to just use a round or trunc function to lop it off at 3 decimal places, the fact that this number and this cell have a different set up then the rest of the parallel cell calculations is drawing some criticism. Has anyone ran into this before? I have tried it in Excel 2010 and 2019 and in new worksheets, same issue. It seems that excel refuses to accept the number at 3 decimal places and forcing an expansion to 10 decimal places on its own.
This is a normal behavior. See the image below where I just entered 56136,598 into the cell.
This happens due to the fact that Excel is a numeric calculation program and not an algebraic one. So it is a problem of precision. Also see Numeric precision in Microsoft Excel.
Excels results are not absolute but very close to correct. The difference between these to numbers is almost 0 (the difference is 0,0000000001).
And this is actually how most common calculators will act too (you just don't see that). It is just the nature of how calculators (and computers) work.
So there is nothing to worry about.
More about this: Understanding Floating Point Precision, aka “Why does Excel Give Me Seemingly Wrong Answers?”
I am very new to Excel and I have a problem with a simple multiplication (I know, it is depressing but I'm stuck).
I have to multiply the numeric content of 2 cells (these value are calculated using 2 different formulas).
The problem is that it seems that these 2 cells contain numeric values having different format and I obtain a strange result.
Infact I have:
1) The K3 cell containing this value: 0,0783272400
2) The K6 cell containing this value: 728.454911165
In another cell I simply do:
=K3*K6
but now I am obtaining this nonsense value: 57.057.862.655,9996000000
I think that the problem could be related to the fact that the first one use the , do divide integer section and decimal section, and in the other one I am using . to divider the integer section and decimal section.
How can I correctly handle this situation?
Format both values as Currency in Excel and forget about the issue.
You are getting it, because the floating point values are not represented differently in many programming languages. In Excel probably the best way to make sure you do not give strange values is to format as Currency.
Or in VBA to use the CDec and to convert to decimal.
Is floating point math broken?
Excel is treating 0,0783272400 as something less than one tenth and 728.454911165 as getting on for one thousand billion. The result is formatted with . for thousands separator and , for decimal separator - and is not nonsensical (though the choice of formatting is).
I have a excel sheet with a few formulas like this:
A1,A2,A3= 0.13,1.25,2.21
A4: =(A1*A2) =0.16 ( 2 decimal points)
A5: =(A2*A3) =2.76 ( 2 decimal points)
A6: =SUM(A4;A5) =2.93 ( 2 decimal points )
And i want to show 0.16+2.76=2.92
well, there's my problem in bold. i want to add the values from the cells, not the formuls result. How can i do that ? Thank you
Presumably you're working with money which is why you need this.
One way to resolve this is to use =ROUND(A1*A2, 2) etc. and base your subsequent calculations from that.
Do be aware though that you will still occasionally get spurious results due to Excel using a 64 bit IEEE754 floating point double to represent numbers. (Although it does have some extremely clever circumvention techniques - see how it evaluates 1/3 + 1/3 + 1/3 - it will not resolve every possible oddity). If you're building an accounting-style sheet you are best off working in pence, and dividing the final result.
Round the values before you sum, ie:
=ROUND(A1*A2,2)
=ROUND(A2*A3,2)
You could wrap your formulas with the ROUND function:
=ROUND(A1*A2,2)
This will give you 0.16 as opposed to 0.163. Do this for each of your calculations and you'll only be calculating everything to two decimal places. Although I'm not sure why you'd want to do that.
I have put two date time values in cells A1 and B1. The date time is precisely 41703.0416666667. One is ouput from an SQL database the other manually written.
The result of =if(A1=B1,1,0) is 1.
The result of =MATCH(B1,A1,0) is #N/A.
Does anyone have any theories as to why this may be happening?
Probably an issue with converting decimal to binary. See related answer:
Simple HLOOKUP Failing with Excel 2010
This appears to be a limitation of storing floating point numbers in binary - as described here: http://support.microsoft.com/kb/214118
Many combinations of arithmetic operations on floating-point numbers
in Microsoft Excel and Microsoft Works may produce results that appear
to be incorrect by very small amounts. For example, the equation
=1*(.5-.4-.1) may be evaluated to the quantity (-2.78E-17), or -0.0000000000000000278 instead of 0.
This problem is not unique to excel either but rather a result of:
IEEE 754 specifies that numbers be stored in binary format to reduce
storage requirements and allow the built-in binary arithmetic
instructions that are available on all microprocessors to process the
data in a relatively rapid fashion. However, some numbers that are
simple, nonrepeating decimal numbers are converted into repeating
binary numbers that cannot be stored with perfect accuracy.
The issue is with the floating-point calculation (see http://support.microsoft.com/kb/78113), one possible workaround is to work with the round() function. In your case, rounding to 10 or 12 decimals would probably be enough to address the issue.