64bit IEEE 754: Decimal -> Double Precision Floating Point Binary: 0.000 079 762 373 5 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.000 079 762 373 5(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. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

2. 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.


0(10) =


0(2)


3. Convert to binary (base 2) the fractional part: 0.000 079 762 373 5.

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) 0.000 079 762 373 5 × 2 = 0 + 0.000 159 524 747;
  • 2) 0.000 159 524 747 × 2 = 0 + 0.000 319 049 494;
  • 3) 0.000 319 049 494 × 2 = 0 + 0.000 638 098 988;
  • 4) 0.000 638 098 988 × 2 = 0 + 0.001 276 197 976;
  • 5) 0.001 276 197 976 × 2 = 0 + 0.002 552 395 952;
  • 6) 0.002 552 395 952 × 2 = 0 + 0.005 104 791 904;
  • 7) 0.005 104 791 904 × 2 = 0 + 0.010 209 583 808;
  • 8) 0.010 209 583 808 × 2 = 0 + 0.020 419 167 616;
  • 9) 0.020 419 167 616 × 2 = 0 + 0.040 838 335 232;
  • 10) 0.040 838 335 232 × 2 = 0 + 0.081 676 670 464;
  • 11) 0.081 676 670 464 × 2 = 0 + 0.163 353 340 928;
  • 12) 0.163 353 340 928 × 2 = 0 + 0.326 706 681 856;
  • 13) 0.326 706 681 856 × 2 = 0 + 0.653 413 363 712;
  • 14) 0.653 413 363 712 × 2 = 1 + 0.306 826 727 424;
  • 15) 0.306 826 727 424 × 2 = 0 + 0.613 653 454 848;
  • 16) 0.613 653 454 848 × 2 = 1 + 0.227 306 909 696;
  • 17) 0.227 306 909 696 × 2 = 0 + 0.454 613 819 392;
  • 18) 0.454 613 819 392 × 2 = 0 + 0.909 227 638 784;
  • 19) 0.909 227 638 784 × 2 = 1 + 0.818 455 277 568;
  • 20) 0.818 455 277 568 × 2 = 1 + 0.636 910 555 136;
  • 21) 0.636 910 555 136 × 2 = 1 + 0.273 821 110 272;
  • 22) 0.273 821 110 272 × 2 = 0 + 0.547 642 220 544;
  • 23) 0.547 642 220 544 × 2 = 1 + 0.095 284 441 088;
  • 24) 0.095 284 441 088 × 2 = 0 + 0.190 568 882 176;
  • 25) 0.190 568 882 176 × 2 = 0 + 0.381 137 764 352;
  • 26) 0.381 137 764 352 × 2 = 0 + 0.762 275 528 704;
  • 27) 0.762 275 528 704 × 2 = 1 + 0.524 551 057 408;
  • 28) 0.524 551 057 408 × 2 = 1 + 0.049 102 114 816;
  • 29) 0.049 102 114 816 × 2 = 0 + 0.098 204 229 632;
  • 30) 0.098 204 229 632 × 2 = 0 + 0.196 408 459 264;
  • 31) 0.196 408 459 264 × 2 = 0 + 0.392 816 918 528;
  • 32) 0.392 816 918 528 × 2 = 0 + 0.785 633 837 056;
  • 33) 0.785 633 837 056 × 2 = 1 + 0.571 267 674 112;
  • 34) 0.571 267 674 112 × 2 = 1 + 0.142 535 348 224;
  • 35) 0.142 535 348 224 × 2 = 0 + 0.285 070 696 448;
  • 36) 0.285 070 696 448 × 2 = 0 + 0.570 141 392 896;
  • 37) 0.570 141 392 896 × 2 = 1 + 0.140 282 785 792;
  • 38) 0.140 282 785 792 × 2 = 0 + 0.280 565 571 584;
  • 39) 0.280 565 571 584 × 2 = 0 + 0.561 131 143 168;
  • 40) 0.561 131 143 168 × 2 = 1 + 0.122 262 286 336;
  • 41) 0.122 262 286 336 × 2 = 0 + 0.244 524 572 672;
  • 42) 0.244 524 572 672 × 2 = 0 + 0.489 049 145 344;
  • 43) 0.489 049 145 344 × 2 = 0 + 0.978 098 290 688;
  • 44) 0.978 098 290 688 × 2 = 1 + 0.956 196 581 376;
  • 45) 0.956 196 581 376 × 2 = 1 + 0.912 393 162 752;
  • 46) 0.912 393 162 752 × 2 = 1 + 0.824 786 325 504;
  • 47) 0.824 786 325 504 × 2 = 1 + 0.649 572 651 008;
  • 48) 0.649 572 651 008 × 2 = 1 + 0.299 145 302 016;
  • 49) 0.299 145 302 016 × 2 = 0 + 0.598 290 604 032;
  • 50) 0.598 290 604 032 × 2 = 1 + 0.196 581 208 064;
  • 51) 0.196 581 208 064 × 2 = 0 + 0.393 162 416 128;
  • 52) 0.393 162 416 128 × 2 = 0 + 0.786 324 832 256;
  • 53) 0.786 324 832 256 × 2 = 1 + 0.572 649 664 512;
  • 54) 0.572 649 664 512 × 2 = 1 + 0.145 299 329 024;
  • 55) 0.145 299 329 024 × 2 = 0 + 0.290 598 658 048;
  • 56) 0.290 598 658 048 × 2 = 0 + 0.581 197 316 096;
  • 57) 0.581 197 316 096 × 2 = 1 + 0.162 394 632 192;
  • 58) 0.162 394 632 192 × 2 = 0 + 0.324 789 264 384;
  • 59) 0.324 789 264 384 × 2 = 0 + 0.649 578 528 768;
  • 60) 0.649 578 528 768 × 2 = 1 + 0.299 157 057 536;
  • 61) 0.299 157 057 536 × 2 = 0 + 0.598 314 115 072;
  • 62) 0.598 314 115 072 × 2 = 1 + 0.196 628 230 144;
  • 63) 0.196 628 230 144 × 2 = 0 + 0.393 256 460 288;
  • 64) 0.393 256 460 288 × 2 = 0 + 0.786 512 920 576;
  • 65) 0.786 512 920 576 × 2 = 1 + 0.573 025 841 152;
  • 66) 0.573 025 841 152 × 2 = 1 + 0.146 051 682 304;

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...)


4. 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:


0.000 079 762 373 5(10) =


0.0000 0000 0000 0101 0011 1010 0011 0000 1100 1001 0001 1111 0100 1100 1001 0100 11(2)


5. Positive number before normalization:

0.000 079 762 373 5(10) =


0.0000 0000 0000 0101 0011 1010 0011 0000 1100 1001 0001 1111 0100 1100 1001 0100 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 079 762 373 5(10) =


0.0000 0000 0000 0101 0011 1010 0011 0000 1100 1001 0001 1111 0100 1100 1001 0100 11(2) =


0.0000 0000 0000 0101 0011 1010 0011 0000 1100 1001 0001 1111 0100 1100 1001 0100 11(2) × 20 =


1.0100 1110 1000 1100 0011 0010 0100 0111 1101 0011 0010 0101 0011(2) × 2-14


7. 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 0 (a positive number)


Exponent (unadjusted): -14


Mantissa (not normalized):
1.0100 1110 1000 1100 0011 0010 0100 0111 1101 0011 0010 0101 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


-14 + 2(11-1) - 1 =


(-14 + 1 023)(10) =


1 009(10)


9. 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;
  • 1 009 ÷ 2 = 504 + 1;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


1009(10) =


011 1111 0001(2)


11. 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, only if necessary (not the case here).


Mantissa (normalized) =


1. 0100 1110 1000 1100 0011 0010 0100 0111 1101 0011 0010 0101 0011 =


0100 1110 1000 1100 0011 0010 0100 0111 1101 0011 0010 0101 0011


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

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
011 1111 0001


Mantissa (52 bits) =
0100 1110 1000 1100 0011 0010 0100 0111 1101 0011 0010 0101 0011


The base ten decimal number 0.000 079 762 373 5 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 0001 - 0100 1110 1000 1100 0011 0010 0100 0111 1101 0011 0010 0101 0011

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 1

      47
    • 1

      46
    • 1

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 1

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 1

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 1

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 0

      5
    • 1

      4
    • 0

      3
    • 0

      2
    • 1

      1
    • 1

      0

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 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): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

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

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =


    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100