-0.000 035 666 978 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 978(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
-0.000 035 666 978(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:

|-0.000 035 666 978| = 0.000 035 666 978


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: 0.000 035 666 978.

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) 0.000 035 666 978 × 2 = 0 + 0.000 071 333 956;
  • 2) 0.000 071 333 956 × 2 = 0 + 0.000 142 667 912;
  • 3) 0.000 142 667 912 × 2 = 0 + 0.000 285 335 824;
  • 4) 0.000 285 335 824 × 2 = 0 + 0.000 570 671 648;
  • 5) 0.000 570 671 648 × 2 = 0 + 0.001 141 343 296;
  • 6) 0.001 141 343 296 × 2 = 0 + 0.002 282 686 592;
  • 7) 0.002 282 686 592 × 2 = 0 + 0.004 565 373 184;
  • 8) 0.004 565 373 184 × 2 = 0 + 0.009 130 746 368;
  • 9) 0.009 130 746 368 × 2 = 0 + 0.018 261 492 736;
  • 10) 0.018 261 492 736 × 2 = 0 + 0.036 522 985 472;
  • 11) 0.036 522 985 472 × 2 = 0 + 0.073 045 970 944;
  • 12) 0.073 045 970 944 × 2 = 0 + 0.146 091 941 888;
  • 13) 0.146 091 941 888 × 2 = 0 + 0.292 183 883 776;
  • 14) 0.292 183 883 776 × 2 = 0 + 0.584 367 767 552;
  • 15) 0.584 367 767 552 × 2 = 1 + 0.168 735 535 104;
  • 16) 0.168 735 535 104 × 2 = 0 + 0.337 471 070 208;
  • 17) 0.337 471 070 208 × 2 = 0 + 0.674 942 140 416;
  • 18) 0.674 942 140 416 × 2 = 1 + 0.349 884 280 832;
  • 19) 0.349 884 280 832 × 2 = 0 + 0.699 768 561 664;
  • 20) 0.699 768 561 664 × 2 = 1 + 0.399 537 123 328;
  • 21) 0.399 537 123 328 × 2 = 0 + 0.799 074 246 656;
  • 22) 0.799 074 246 656 × 2 = 1 + 0.598 148 493 312;
  • 23) 0.598 148 493 312 × 2 = 1 + 0.196 296 986 624;
  • 24) 0.196 296 986 624 × 2 = 0 + 0.392 593 973 248;
  • 25) 0.392 593 973 248 × 2 = 0 + 0.785 187 946 496;
  • 26) 0.785 187 946 496 × 2 = 1 + 0.570 375 892 992;
  • 27) 0.570 375 892 992 × 2 = 1 + 0.140 751 785 984;
  • 28) 0.140 751 785 984 × 2 = 0 + 0.281 503 571 968;
  • 29) 0.281 503 571 968 × 2 = 0 + 0.563 007 143 936;
  • 30) 0.563 007 143 936 × 2 = 1 + 0.126 014 287 872;
  • 31) 0.126 014 287 872 × 2 = 0 + 0.252 028 575 744;
  • 32) 0.252 028 575 744 × 2 = 0 + 0.504 057 151 488;
  • 33) 0.504 057 151 488 × 2 = 1 + 0.008 114 302 976;
  • 34) 0.008 114 302 976 × 2 = 0 + 0.016 228 605 952;
  • 35) 0.016 228 605 952 × 2 = 0 + 0.032 457 211 904;
  • 36) 0.032 457 211 904 × 2 = 0 + 0.064 914 423 808;
  • 37) 0.064 914 423 808 × 2 = 0 + 0.129 828 847 616;
  • 38) 0.129 828 847 616 × 2 = 0 + 0.259 657 695 232;
  • 39) 0.259 657 695 232 × 2 = 0 + 0.519 315 390 464;
  • 40) 0.519 315 390 464 × 2 = 1 + 0.038 630 780 928;
  • 41) 0.038 630 780 928 × 2 = 0 + 0.077 261 561 856;
  • 42) 0.077 261 561 856 × 2 = 0 + 0.154 523 123 712;
  • 43) 0.154 523 123 712 × 2 = 0 + 0.309 046 247 424;
  • 44) 0.309 046 247 424 × 2 = 0 + 0.618 092 494 848;
  • 45) 0.618 092 494 848 × 2 = 1 + 0.236 184 989 696;
  • 46) 0.236 184 989 696 × 2 = 0 + 0.472 369 979 392;
  • 47) 0.472 369 979 392 × 2 = 0 + 0.944 739 958 784;
  • 48) 0.944 739 958 784 × 2 = 1 + 0.889 479 917 568;
  • 49) 0.889 479 917 568 × 2 = 1 + 0.778 959 835 136;
  • 50) 0.778 959 835 136 × 2 = 1 + 0.557 919 670 272;
  • 51) 0.557 919 670 272 × 2 = 1 + 0.115 839 340 544;
  • 52) 0.115 839 340 544 × 2 = 0 + 0.231 678 681 088;
  • 53) 0.231 678 681 088 × 2 = 0 + 0.463 357 362 176;
  • 54) 0.463 357 362 176 × 2 = 0 + 0.926 714 724 352;
  • 55) 0.926 714 724 352 × 2 = 1 + 0.853 429 448 704;
  • 56) 0.853 429 448 704 × 2 = 1 + 0.706 858 897 408;
  • 57) 0.706 858 897 408 × 2 = 1 + 0.413 717 794 816;
  • 58) 0.413 717 794 816 × 2 = 0 + 0.827 435 589 632;
  • 59) 0.827 435 589 632 × 2 = 1 + 0.654 871 179 264;
  • 60) 0.654 871 179 264 × 2 = 1 + 0.309 742 358 528;
  • 61) 0.309 742 358 528 × 2 = 0 + 0.619 484 717 056;
  • 62) 0.619 484 717 056 × 2 = 1 + 0.238 969 434 112;
  • 63) 0.238 969 434 112 × 2 = 0 + 0.477 938 868 224;
  • 64) 0.477 938 868 224 × 2 = 0 + 0.955 877 736 448;
  • 65) 0.955 877 736 448 × 2 = 1 + 0.911 755 472 896;
  • 66) 0.911 755 472 896 × 2 = 1 + 0.823 510 945 792;
  • 67) 0.823 510 945 792 × 2 = 1 + 0.647 021 891 584;

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:


0.000 035 666 978(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1000 0001 0000 1001 1110 0011 1011 0100 111(2)

6. Positive number before normalization:

0.000 035 666 978(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1000 0001 0000 1001 1110 0011 1011 0100 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 978(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1000 0001 0000 1001 1110 0011 1011 0100 111(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 1000 0001 0000 1001 1110 0011 1011 0100 111(2) × 20 =

1.0010 1011 0011 0010 0100 0000 1000 0100 1111 0001 1101 1010 0111(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0010 0100 0000 1000 0100 1111 0001 1101 1010 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 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) =

1008(10) =

011 1111 0000(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. 0010 1011 0011 0010 0100 0000 1000 0100 1111 0001 1101 1010 0111 =

0010 1011 0011 0010 0100 0000 1000 0100 1111 0001 1101 1010 0111


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 0000

Mantissa (52 bits) =
0010 1011 0011 0010 0100 0000 1000 0100 1111 0001 1101 1010 0111


Decimal number -0.000 035 666 978 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0100 0000 1000 0100 1111 0001 1101 1010 0111

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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