-0.000 062 259 888 373 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 062 259 888 373(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 062 259 888 373(10) to 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:

|-0.000 062 259 888 373| = 0.000 062 259 888 373


2. 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;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 062 259 888 373.

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) 0.000 062 259 888 373 × 2 = 0 + 0.000 124 519 776 746;
  • 2) 0.000 124 519 776 746 × 2 = 0 + 0.000 249 039 553 492;
  • 3) 0.000 249 039 553 492 × 2 = 0 + 0.000 498 079 106 984;
  • 4) 0.000 498 079 106 984 × 2 = 0 + 0.000 996 158 213 968;
  • 5) 0.000 996 158 213 968 × 2 = 0 + 0.001 992 316 427 936;
  • 6) 0.001 992 316 427 936 × 2 = 0 + 0.003 984 632 855 872;
  • 7) 0.003 984 632 855 872 × 2 = 0 + 0.007 969 265 711 744;
  • 8) 0.007 969 265 711 744 × 2 = 0 + 0.015 938 531 423 488;
  • 9) 0.015 938 531 423 488 × 2 = 0 + 0.031 877 062 846 976;
  • 10) 0.031 877 062 846 976 × 2 = 0 + 0.063 754 125 693 952;
  • 11) 0.063 754 125 693 952 × 2 = 0 + 0.127 508 251 387 904;
  • 12) 0.127 508 251 387 904 × 2 = 0 + 0.255 016 502 775 808;
  • 13) 0.255 016 502 775 808 × 2 = 0 + 0.510 033 005 551 616;
  • 14) 0.510 033 005 551 616 × 2 = 1 + 0.020 066 011 103 232;
  • 15) 0.020 066 011 103 232 × 2 = 0 + 0.040 132 022 206 464;
  • 16) 0.040 132 022 206 464 × 2 = 0 + 0.080 264 044 412 928;
  • 17) 0.080 264 044 412 928 × 2 = 0 + 0.160 528 088 825 856;
  • 18) 0.160 528 088 825 856 × 2 = 0 + 0.321 056 177 651 712;
  • 19) 0.321 056 177 651 712 × 2 = 0 + 0.642 112 355 303 424;
  • 20) 0.642 112 355 303 424 × 2 = 1 + 0.284 224 710 606 848;
  • 21) 0.284 224 710 606 848 × 2 = 0 + 0.568 449 421 213 696;
  • 22) 0.568 449 421 213 696 × 2 = 1 + 0.136 898 842 427 392;
  • 23) 0.136 898 842 427 392 × 2 = 0 + 0.273 797 684 854 784;
  • 24) 0.273 797 684 854 784 × 2 = 0 + 0.547 595 369 709 568;
  • 25) 0.547 595 369 709 568 × 2 = 1 + 0.095 190 739 419 136;
  • 26) 0.095 190 739 419 136 × 2 = 0 + 0.190 381 478 838 272;
  • 27) 0.190 381 478 838 272 × 2 = 0 + 0.380 762 957 676 544;
  • 28) 0.380 762 957 676 544 × 2 = 0 + 0.761 525 915 353 088;
  • 29) 0.761 525 915 353 088 × 2 = 1 + 0.523 051 830 706 176;
  • 30) 0.523 051 830 706 176 × 2 = 1 + 0.046 103 661 412 352;
  • 31) 0.046 103 661 412 352 × 2 = 0 + 0.092 207 322 824 704;
  • 32) 0.092 207 322 824 704 × 2 = 0 + 0.184 414 645 649 408;
  • 33) 0.184 414 645 649 408 × 2 = 0 + 0.368 829 291 298 816;
  • 34) 0.368 829 291 298 816 × 2 = 0 + 0.737 658 582 597 632;
  • 35) 0.737 658 582 597 632 × 2 = 1 + 0.475 317 165 195 264;
  • 36) 0.475 317 165 195 264 × 2 = 0 + 0.950 634 330 390 528;
  • 37) 0.950 634 330 390 528 × 2 = 1 + 0.901 268 660 781 056;
  • 38) 0.901 268 660 781 056 × 2 = 1 + 0.802 537 321 562 112;
  • 39) 0.802 537 321 562 112 × 2 = 1 + 0.605 074 643 124 224;
  • 40) 0.605 074 643 124 224 × 2 = 1 + 0.210 149 286 248 448;
  • 41) 0.210 149 286 248 448 × 2 = 0 + 0.420 298 572 496 896;
  • 42) 0.420 298 572 496 896 × 2 = 0 + 0.840 597 144 993 792;
  • 43) 0.840 597 144 993 792 × 2 = 1 + 0.681 194 289 987 584;
  • 44) 0.681 194 289 987 584 × 2 = 1 + 0.362 388 579 975 168;
  • 45) 0.362 388 579 975 168 × 2 = 0 + 0.724 777 159 950 336;
  • 46) 0.724 777 159 950 336 × 2 = 1 + 0.449 554 319 900 672;
  • 47) 0.449 554 319 900 672 × 2 = 0 + 0.899 108 639 801 344;
  • 48) 0.899 108 639 801 344 × 2 = 1 + 0.798 217 279 602 688;
  • 49) 0.798 217 279 602 688 × 2 = 1 + 0.596 434 559 205 376;
  • 50) 0.596 434 559 205 376 × 2 = 1 + 0.192 869 118 410 752;
  • 51) 0.192 869 118 410 752 × 2 = 0 + 0.385 738 236 821 504;
  • 52) 0.385 738 236 821 504 × 2 = 0 + 0.771 476 473 643 008;
  • 53) 0.771 476 473 643 008 × 2 = 1 + 0.542 952 947 286 016;
  • 54) 0.542 952 947 286 016 × 2 = 1 + 0.085 905 894 572 032;
  • 55) 0.085 905 894 572 032 × 2 = 0 + 0.171 811 789 144 064;
  • 56) 0.171 811 789 144 064 × 2 = 0 + 0.343 623 578 288 128;
  • 57) 0.343 623 578 288 128 × 2 = 0 + 0.687 247 156 576 256;
  • 58) 0.687 247 156 576 256 × 2 = 1 + 0.374 494 313 152 512;
  • 59) 0.374 494 313 152 512 × 2 = 0 + 0.748 988 626 305 024;
  • 60) 0.748 988 626 305 024 × 2 = 1 + 0.497 977 252 610 048;
  • 61) 0.497 977 252 610 048 × 2 = 0 + 0.995 954 505 220 096;
  • 62) 0.995 954 505 220 096 × 2 = 1 + 0.991 909 010 440 192;
  • 63) 0.991 909 010 440 192 × 2 = 1 + 0.983 818 020 880 384;
  • 64) 0.983 818 020 880 384 × 2 = 1 + 0.967 636 041 760 768;
  • 65) 0.967 636 041 760 768 × 2 = 1 + 0.935 272 083 521 536;
  • 66) 0.935 272 083 521 536 × 2 = 1 + 0.870 544 167 043 072;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 062 259 888 373(10) =

0.0000 0000 0000 0100 0001 0100 1000 1100 0010 1111 0011 0101 1100 1100 0101 0111 11(2)

6. Positive number before normalization:

0.000 062 259 888 373(10) =

0.0000 0000 0000 0100 0001 0100 1000 1100 0010 1111 0011 0101 1100 1100 0101 0111 11(2)

7. 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:


0.000 062 259 888 373(10) =

0.0000 0000 0000 0100 0001 0100 1000 1100 0010 1111 0011 0101 1100 1100 0101 0111 11(2) =

0.0000 0000 0000 0100 0001 0100 1000 1100 0010 1111 0011 0101 1100 1100 0101 0111 11(2) × 20 =

1.0000 0101 0010 0011 0000 1011 1100 1101 0111 0011 0001 0101 1111(2) × 2-14


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): -14

Mantissa (not normalized):
1.0000 0101 0010 0011 0000 1011 1100 1101 0111 0011 0001 0101 1111


9. 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)


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;
  • 1 009 ÷ 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;

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) =

1009(10) =

011 1111 0001(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0000 0101 0010 0011 0000 1011 1100 1101 0111 0011 0001 0101 1111 =

0000 0101 0010 0011 0000 1011 1100 1101 0111 0011 0001 0101 1111


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) =
011 1111 0001

Mantissa (52 bits) =
0000 0101 0010 0011 0000 1011 1100 1101 0111 0011 0001 0101 1111


Decimal number -0.000 062 259 888 373 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 0000 0101 0010 0011 0000 1011 1100 1101 0111 0011 0001 0101 1111

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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