Binary to hexadecimal is very easy because hexadecimal numbers are designed specifically so that each hex digit is exactly 4 bits (i.e. 16 different values). So if you had this binary number: binary: 100011011011110101000100001 You could put in commas every four places (starting on the left): binary: 100,0110,1101,1110,1010,0010,0001 Then you could write the hex values immediately below: binary: 0100,0110,1101,1110,1010,0010,0001 hex: 4 6 D E A 2 1 and the hex value would be 46DEA21.
It is E.
The value in hexadecimal of the decimal number 999910 is F41E6.
If you mean how do you convert 8 into binary numbers, here it is. 8 -- Eights place value
A
1101111010101101 in binary is equal to DEAD in hexadecimal.
Binary to hexadecimal is very easy because hexadecimal numbers are designed specifically so that each hex digit is exactly 4 bits (i.e. 16 different values). So if you had this binary number: binary: 100011011011110101000100001 You could put in commas every four places (starting on the left): binary: 100,0110,1101,1110,1010,0010,0001 Then you could write the hex values immediately below: binary: 0100,0110,1101,1110,1010,0010,0001 hex: 4 6 D E A 2 1 and the hex value would be 46DEA21.
It is E.
import java.util.Scanner; public class NumberSystem { public void displayConversion() { Scanner input = new Scanner(System.in); System.out.printf("%-20s%-20s%-20s%-20s\n", "Decimal", "Binary", "Octal", "Hexadecimal"); for ( int i = 1; i <= 256; i++ ) { String binary = Integer.toBinaryString(i); String octal = Integer.toOctalString(i); String hexadecimal = Integer.toHexString(i); System.out.format("%-20d%-20s%-20s%-20s\n", i, binary, octal, hexadecimal); } } // returns a string representation of the decimal number in binary public String toBinaryString( int dec ) { String binary = " "; while (dec >= 1 ) { int value = dec % 2; binary = value + binary; dec /= 2; } return binary; } //returns a string representation of the number in octal public String toOctalString( int dec ) { String octal = " "; while ( dec >= 1 ) { int value = dec % 8; octal = value + octal; dec /= 8; } return octal; } public String toHexString( int dec ) { String hexadecimal = " "; while ( dec >= 1 ) { int value = dec % 16; switch (value) { case 10: hexadecimal = "A" + hexadecimal; break; case 11: hexadecimal = "B" + hexadecimal; break; case 12: hexadecimal = "C" + hexadecimal; break; case 13: hexadecimal = "D" + hexadecimal; break; case 14: hexadecimal = "E" + hexadecimal; break; case 15: hexadecimal = "F" + hexadecimal; break; default: hexadecimal = value + hexadecimal; break; } dec /= 16; } return hexadecimal; } public static void main( String args[]) { NumberSystem apps = new NumberSystem(); apps.displayConversion(); } }
1100010000111010
1110 = E
import java.util.Scanner; public class NumberSystem { public void displayConversion() { Scanner input = new Scanner(System.in); System.out.printf("%-20s%-20s%-20s%-20s\n", "Decimal", "Binary", "Octal", "Hexadecimal"); for ( int i = 1; i <= 256; i++ ) { String binary = Integer.toBinaryString(i); String octal = Integer.toOctalString(i); String hexadecimal = Integer.toHexString(i); System.out.format("%-20d%-20s%-20s%-20s\n", i, binary, octal, hexadecimal); } } // returns a string representation of the decimal number in binary public String toBinaryString( int dec ) { String binary = " "; while (dec >= 1 ) { int value = dec % 2; binary = value + binary; dec /= 2; } return binary; } //returns a string representation of the number in octal public String toOctalString( int dec ) { String octal = " "; while ( dec >= 1 ) { int value = dec % 8; octal = value + octal; dec /= 8; } return octal; } public String toHexString( int dec ) { String hexadecimal = " "; while ( dec >= 1 ) { int value = dec % 16; switch (value) { case 10: hexadecimal = "A" + hexadecimal; break; case 11: hexadecimal = "B" + hexadecimal; break; case 12: hexadecimal = "C" + hexadecimal; break; case 13: hexadecimal = "D" + hexadecimal; break; case 14: hexadecimal = "E" + hexadecimal; break; case 15: hexadecimal = "F" + hexadecimal; break; default: hexadecimal = value + hexadecimal; break; } dec /= 16; } return hexadecimal; } public static void main( String args[]) { NumberSystem apps = new NumberSystem(); apps.displayConversion(); } }
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In binary: 11.00100100001111110110 In Decimal: 3.14159265358979323846 In hexadecimal: 3.243F6A8885A308D31319
16 is the 4th power of 2. So a hexadecimal number is converted to binary by replacing each hex digit by the 4-bit binary number having the same value. Conversely, in converting binary to hexadecimal, we group every 4 bits starting at the decimal (binary?) point and replace it with the equivalent hex digit. For example, the hexadecimal number 3F9 in binary is 1111111001, because 3 in binary is 11, F (decimal 15) is 1111, and 9 is 1001.
Hexadecimal is used whenever we want to notate a binary value because each hex digit maps directly with each 4 bit nybble (half a byte). A 64-bit binary value can therefore be reduced to a more concise 16-digit hex value.
It is only true in the sense that any numeric base, expressed in that base, is represented with the symbol "10". Confusing? Let's clarify that. Hexadecimal numbers use sixteen as the base. But how do you express the value sixteen in hexadecimal? Quite easy, it would be written as "10". The same is true in any other base. For example, in binary (base two), the value two is expressed as "10". In octal (base eight), the value eight is expressed as "10". In decimal (our familiar base ten), the value ten is expressed as "10". No matter what base you work in, the base itself will always be expressed as "10". That however is not the same thing as saying that hexadecimal numbers are based on the number ten. That is incorrect. Hexadecimal numbers use the base sixteen.