64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -160.207 049 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -160.207 049(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-160.207 049| = 160.207 049

2. First, convert to binary (in base 2) the integer part: 160.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 160 ÷ 2 = 80 + 0;
  • 80 ÷ 2 = 40 + 0;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


160(10) =


1010 0000(2)


4. Convert to binary (base 2) the fractional part: 0.207 049.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.207 049 × 2 = 0 + 0.414 098;
  • 2) 0.414 098 × 2 = 0 + 0.828 196;
  • 3) 0.828 196 × 2 = 1 + 0.656 392;
  • 4) 0.656 392 × 2 = 1 + 0.312 784;
  • 5) 0.312 784 × 2 = 0 + 0.625 568;
  • 6) 0.625 568 × 2 = 1 + 0.251 136;
  • 7) 0.251 136 × 2 = 0 + 0.502 272;
  • 8) 0.502 272 × 2 = 1 + 0.004 544;
  • 9) 0.004 544 × 2 = 0 + 0.009 088;
  • 10) 0.009 088 × 2 = 0 + 0.018 176;
  • 11) 0.018 176 × 2 = 0 + 0.036 352;
  • 12) 0.036 352 × 2 = 0 + 0.072 704;
  • 13) 0.072 704 × 2 = 0 + 0.145 408;
  • 14) 0.145 408 × 2 = 0 + 0.290 816;
  • 15) 0.290 816 × 2 = 0 + 0.581 632;
  • 16) 0.581 632 × 2 = 1 + 0.163 264;
  • 17) 0.163 264 × 2 = 0 + 0.326 528;
  • 18) 0.326 528 × 2 = 0 + 0.653 056;
  • 19) 0.653 056 × 2 = 1 + 0.306 112;
  • 20) 0.306 112 × 2 = 0 + 0.612 224;
  • 21) 0.612 224 × 2 = 1 + 0.224 448;
  • 22) 0.224 448 × 2 = 0 + 0.448 896;
  • 23) 0.448 896 × 2 = 0 + 0.897 792;
  • 24) 0.897 792 × 2 = 1 + 0.795 584;
  • 25) 0.795 584 × 2 = 1 + 0.591 168;
  • 26) 0.591 168 × 2 = 1 + 0.182 336;
  • 27) 0.182 336 × 2 = 0 + 0.364 672;
  • 28) 0.364 672 × 2 = 0 + 0.729 344;
  • 29) 0.729 344 × 2 = 1 + 0.458 688;
  • 30) 0.458 688 × 2 = 0 + 0.917 376;
  • 31) 0.917 376 × 2 = 1 + 0.834 752;
  • 32) 0.834 752 × 2 = 1 + 0.669 504;
  • 33) 0.669 504 × 2 = 1 + 0.339 008;
  • 34) 0.339 008 × 2 = 0 + 0.678 016;
  • 35) 0.678 016 × 2 = 1 + 0.356 032;
  • 36) 0.356 032 × 2 = 0 + 0.712 064;
  • 37) 0.712 064 × 2 = 1 + 0.424 128;
  • 38) 0.424 128 × 2 = 0 + 0.848 256;
  • 39) 0.848 256 × 2 = 1 + 0.696 512;
  • 40) 0.696 512 × 2 = 1 + 0.393 024;
  • 41) 0.393 024 × 2 = 0 + 0.786 048;
  • 42) 0.786 048 × 2 = 1 + 0.572 096;
  • 43) 0.572 096 × 2 = 1 + 0.144 192;
  • 44) 0.144 192 × 2 = 0 + 0.288 384;
  • 45) 0.288 384 × 2 = 0 + 0.576 768;
  • 46) 0.576 768 × 2 = 1 + 0.153 536;
  • 47) 0.153 536 × 2 = 0 + 0.307 072;
  • 48) 0.307 072 × 2 = 0 + 0.614 144;
  • 49) 0.614 144 × 2 = 1 + 0.228 288;
  • 50) 0.228 288 × 2 = 0 + 0.456 576;
  • 51) 0.456 576 × 2 = 0 + 0.913 152;
  • 52) 0.913 152 × 2 = 1 + 0.826 304;
  • 53) 0.826 304 × 2 = 1 + 0.652 608;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.207 049(10) =


0.0011 0101 0000 0001 0010 1001 1100 1011 1010 1011 0110 0100 1001 1(2)


6. Positive number before normalization:

160.207 049(10) =


1010 0000.0011 0101 0000 0001 0010 1001 1100 1011 1010 1011 0110 0100 1001 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


160.207 049(10) =


1010 0000.0011 0101 0000 0001 0010 1001 1100 1011 1010 1011 0110 0100 1001 1(2) =


1010 0000.0011 0101 0000 0001 0010 1001 1100 1011 1010 1011 0110 0100 1001 1(2) × 20 =


1.0100 0000 0110 1010 0000 0010 0101 0011 1001 0111 0101 0110 1100 1001 0011(2) × 27


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)


Exponent (unadjusted): 7


Mantissa (not normalized):
1.0100 0000 0110 1010 0000 0010 0101 0011 1001 0111 0101 0110 1100 1001 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


7 + 2(11-1) - 1 =


(7 + 1 023)(10) =


1 030(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1030(10) =


100 0000 0110(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0100 0000 0110 1010 0000 0010 0101 0011 1001 0111 0101 0110 1100 1001 0011 =


0100 0000 0110 1010 0000 0010 0101 0011 1001 0111 0101 0110 1100


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
100 0000 0110


Mantissa (52 bits) =
0100 0000 0110 1010 0000 0010 0101 0011 1001 0111 0101 0110 1100


The base ten decimal number -160.207 049 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 0110 - 0100 0000 0110 1010 0000 0010 0101 0011 1001 0111 0101 0110 1100

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