How Can I Easily Tackle Binary and Hexadecimal Problems?

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Discussion Overview

The discussion focuses on methods for converting between binary, decimal, and hexadecimal number systems. Participants share their techniques and approaches for tackling these conversions, exploring both theoretical and practical aspects.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant outlines their methods for converting binary to decimal, decimal to binary, binary to hexadecimal, hexadecimal to binary, and hexadecimal to decimal, expressing uncertainty about the correctness of their calculations.
  • Another participant suggests using repeated division by 2 as a straightforward method for converting decimal to binary, providing a step-by-step example with the number 79.
  • This second participant also explains how to convert binary to hexadecimal by grouping binary digits into sets of four and translating them into hexadecimal values.
  • A third participant reiterates the second participant's points, emphasizing the ease of converting between binary and hexadecimal, and provides an example with a more complex hexadecimal number.
  • There is a challenge regarding a potential error in the addition of hexadecimal values, with one participant questioning the accuracy of the conversion results presented by another.
  • A later reply critiques the phrasing of "halving by 2," suggesting that it is redundant, while also affirming the method described.

Areas of Agreement / Disagreement

Participants express differing views on the clarity and correctness of the methods presented, with some agreeing on the techniques while others raise questions or corrections regarding specific calculations. The discussion remains unresolved regarding the accuracy of certain conversions.

Contextual Notes

Some methods rely on specific assumptions about the understanding of binary and hexadecimal systems, and there are unresolved questions about the correctness of certain calculations presented by participants.

Amathproblem22
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I don't need help with a problem just help with easy ways of tackling the problems. Below are my current methods. (Hopefully, all is correct I was in a rush).

Binary to Decimal:
1286432168421
11110001

I already know $$128+64+32= 224$$ so then $$ 16+1=17$$, meaning $$224+17=241$$

Decimal to Binary:
$$79$$
1286432168421
01001111
So I do this by basically figuring what adds up to 79 or by figuring if the number can fit into any (hopefully that makes sense).

Binary to Hexadecimal:
84218421
10100101

I break it up into two sets of 8-4-2-1 and add up the 1's below it accordingly. $$10=A$$ in the first column, and $$5$$ in the scond column so $$A5$$

Hexadecimal to Binary:
$$B4$$

$$B=11,4=4$$ so I but ones in the columns above so it adds up to the desired numbers leaving me 10110100.

Hexadecimal to Decimal
Using B4 from above = 11, then times that by 16 which in this case is easy and = 176 then adding 4 = 180
 
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The simplest way to change decimal to binary repeated division by 2. 79/2= 39 with remainder 1. 39/2= 19 with remainder 1. 19/2= 9 with remainder 1. 9/2= 4 with remainder 1. 4/2= 2 with remainder 0, 2/2= 1 with remainder 0 1/2= 0 with remainder . So 79= 100111. Notice that "1"s and "0"s are the reverse of those calculations.

Binary to hexadecimal is easy because 16= 2^4 = 10000 in binary. Divide the binary number into groups of 4 digits: 10100101= 1010 0101. Yes, 1010 is 8+2= 10 which is represented by "A" in hexadecimal and 0101 is 4+ 2= 6. 10100101 in binary is A6 in hexadecimal.

To go from hexadecimal to binary is also easy- convert each hexadecimal "digit" to binary.

B= 8+ 3= 1011 and 4 is 0100 so B4 is 10110100.
For something more complicated, E423BF: E= 8+ 4+ 2= 1110. 4= 0100, 2= 0010, 3= 0011, B= 1011, and F is 1111 so E423BF is 111001000010001110111111 in base 2.
 
Country Boy said:
The simplest way to change decimal to binary repeated division by 2. 79/2= 39 with remainder 1. 39/2= 19 with remainder 1. 19/2= 9 with remainder 1. 9/2= 4 with remainder 1. 4/2= 2 with remainder 0, 2/2= 1 with remainder 0 1/2= 0 with remainder . So 79= 100111. Notice that "1"s and "0"s are the reverse of those calculations.

Binary to hexadecimal is easy because 16= 2^4 = 10000 in binary. Divide the binary number into groups of 4 digits: 10100101= 1010 0101. Yes, 1010 is 8+2= 10 which is represented by "A" in hexadecimal and 0101 is 4+ 2= 6. 10100101 in binary is A6 in hexadecimal.

To go from hexadecimal to binary is also easy- convert each hexadecimal "digit" to binary.

B= 8+ 3= 1011 and 4 is 0100 so B4 is 10110100.
For something more complicated, E423BF: E= 8+ 4+ 2= 1110. 4= 0100, 2= 0010, 3= 0011, B= 1011, and F is 1111 so E423BF is 111001000010001110111111 in base 2.
This is an easier version of what you were saying I guess halving by two and put a 1 if the number above is odd and a 0 if the number is 0.
01249193979
01001111

Assuming you added A6 wrong? and dropped a 1 off 100111? Or was I wrong lol
 
Except that I would never say "halving by 2" (what else can you halve by?) that is essentially what I said.
 

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