Conversion from binary to decimal

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frctl

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I am reading my notes trying to understand, is someone able to provide me with further explanation for the result and the whole process taking place? conversion.png
 
frctl said:
I am reading my notes trying to understand, is someone able to provide me with further explanation for the result and the whole process taking place? View attachment 33978
Click to expand...
What, specifically, is giving you troubles here? The concept of writing a multi-digit number is that something like xyz represents
[imath]x \times b^2 + y \times b^1 + z \times b^0[/imath], where b is the base you are working in. (A similar rule holds for decimal places.)

That's pretty much it.

-Dan
 
I envision x = 1, y = 0, and z = 1 so [math]1×2^2+0×2^1+1×2_0[/math] Does this make sense?
@Subhotosh Khan

Shouldn't the answer be 8 in this case?
I don't know how the result 5 is obtained.
 
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frctl said:
I am reading my notes trying to understand, is someone able to provide me with further explanation for the result and the whole process taking place? View attachment 33978
Click to expand...
This exercise can at best be the most frustrating for learners. But slow down and take one step at a time.
First step is to convince oneself that each number has its own digit set.
Base two has two digits: [imath]\mathscr{B}_{2}=\{0,1\}|[/imath] Base 10 has ten digits: [imath]\mathscr{B}_{10}=\{0,1,2,3\cdots 8,9\}|[/imath]
NOW Base sixteen has sixteen digits: [imath]\mathscr{B}_{16}=\{0,1,2,3\cdots 8,9,A,B,C,D,E,F\}|[/imath]
In the last line of wgat you posted that is where the [imath]10[/imath] comes from.
Thus [imath]C140BF_{16}=12\times 16^5+1\times 16^4+4\times 16^3+0\times 16^2+11\times 16^1+15\times 16^0[/imath]
Please post some of your work so that it can be checked.

[imath][/imath][imath][/imath][imath][/imath][imath][/imath]
 
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frctl said:
I envision x = 1, y = 0, and z = 1 so [math]1×2^2+0×2^1+1×2_0[/math] Does this make sense?
@Subhotosh Khan

Shouldn't the answer be 8 in this case?
I don't know how the result 5 is obtained.
Click to expand...

[math]1×2^2+0×2^1+1×2^0 = 5[/math]
Please show us what you think is the value is for each of these...
[math]1×2^2=?\\ 0×2^1=?\\ 1×2^0=?[/math]
 
frctl said:
I envision x = 1, y = 0, and z = 1 so [math]1×2^2+0×2^1+1×2_0[/math] Does this make sense?
@Subhotosh Khan

Shouldn't the answer be 8 in this case?
I don't know how the result 5 is obtained.
Click to expand...
The answer would be 8 if 0x21=2 and if 1x20 = 2.
The real question is whether or not the above equalities are correct.
 
Cubist said:
[math]1×2^2+0×2^1+1×2^0 = 5[/math]
Please show us what you think is the value is for each of these...
[math]1×2^2=?\\ 0×2^1=?\\ 1×2^0=?[/math]
Click to expand...
By the way, I wasn't trying to embarrass you by asking that. I just wanted to see which part of the calculation you don't understand so that someone here could help you out by explaining it :)
 
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