-0.000 035 666 967 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 967(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
-0.000 035 666 967(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:

|-0.000 035 666 967| = 0.000 035 666 967


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: 0.000 035 666 967.

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) 0.000 035 666 967 × 2 = 0 + 0.000 071 333 934;
  • 2) 0.000 071 333 934 × 2 = 0 + 0.000 142 667 868;
  • 3) 0.000 142 667 868 × 2 = 0 + 0.000 285 335 736;
  • 4) 0.000 285 335 736 × 2 = 0 + 0.000 570 671 472;
  • 5) 0.000 570 671 472 × 2 = 0 + 0.001 141 342 944;
  • 6) 0.001 141 342 944 × 2 = 0 + 0.002 282 685 888;
  • 7) 0.002 282 685 888 × 2 = 0 + 0.004 565 371 776;
  • 8) 0.004 565 371 776 × 2 = 0 + 0.009 130 743 552;
  • 9) 0.009 130 743 552 × 2 = 0 + 0.018 261 487 104;
  • 10) 0.018 261 487 104 × 2 = 0 + 0.036 522 974 208;
  • 11) 0.036 522 974 208 × 2 = 0 + 0.073 045 948 416;
  • 12) 0.073 045 948 416 × 2 = 0 + 0.146 091 896 832;
  • 13) 0.146 091 896 832 × 2 = 0 + 0.292 183 793 664;
  • 14) 0.292 183 793 664 × 2 = 0 + 0.584 367 587 328;
  • 15) 0.584 367 587 328 × 2 = 1 + 0.168 735 174 656;
  • 16) 0.168 735 174 656 × 2 = 0 + 0.337 470 349 312;
  • 17) 0.337 470 349 312 × 2 = 0 + 0.674 940 698 624;
  • 18) 0.674 940 698 624 × 2 = 1 + 0.349 881 397 248;
  • 19) 0.349 881 397 248 × 2 = 0 + 0.699 762 794 496;
  • 20) 0.699 762 794 496 × 2 = 1 + 0.399 525 588 992;
  • 21) 0.399 525 588 992 × 2 = 0 + 0.799 051 177 984;
  • 22) 0.799 051 177 984 × 2 = 1 + 0.598 102 355 968;
  • 23) 0.598 102 355 968 × 2 = 1 + 0.196 204 711 936;
  • 24) 0.196 204 711 936 × 2 = 0 + 0.392 409 423 872;
  • 25) 0.392 409 423 872 × 2 = 0 + 0.784 818 847 744;
  • 26) 0.784 818 847 744 × 2 = 1 + 0.569 637 695 488;
  • 27) 0.569 637 695 488 × 2 = 1 + 0.139 275 390 976;
  • 28) 0.139 275 390 976 × 2 = 0 + 0.278 550 781 952;
  • 29) 0.278 550 781 952 × 2 = 0 + 0.557 101 563 904;
  • 30) 0.557 101 563 904 × 2 = 1 + 0.114 203 127 808;
  • 31) 0.114 203 127 808 × 2 = 0 + 0.228 406 255 616;
  • 32) 0.228 406 255 616 × 2 = 0 + 0.456 812 511 232;
  • 33) 0.456 812 511 232 × 2 = 0 + 0.913 625 022 464;
  • 34) 0.913 625 022 464 × 2 = 1 + 0.827 250 044 928;
  • 35) 0.827 250 044 928 × 2 = 1 + 0.654 500 089 856;
  • 36) 0.654 500 089 856 × 2 = 1 + 0.309 000 179 712;
  • 37) 0.309 000 179 712 × 2 = 0 + 0.618 000 359 424;
  • 38) 0.618 000 359 424 × 2 = 1 + 0.236 000 718 848;
  • 39) 0.236 000 718 848 × 2 = 0 + 0.472 001 437 696;
  • 40) 0.472 001 437 696 × 2 = 0 + 0.944 002 875 392;
  • 41) 0.944 002 875 392 × 2 = 1 + 0.888 005 750 784;
  • 42) 0.888 005 750 784 × 2 = 1 + 0.776 011 501 568;
  • 43) 0.776 011 501 568 × 2 = 1 + 0.552 023 003 136;
  • 44) 0.552 023 003 136 × 2 = 1 + 0.104 046 006 272;
  • 45) 0.104 046 006 272 × 2 = 0 + 0.208 092 012 544;
  • 46) 0.208 092 012 544 × 2 = 0 + 0.416 184 025 088;
  • 47) 0.416 184 025 088 × 2 = 0 + 0.832 368 050 176;
  • 48) 0.832 368 050 176 × 2 = 1 + 0.664 736 100 352;
  • 49) 0.664 736 100 352 × 2 = 1 + 0.329 472 200 704;
  • 50) 0.329 472 200 704 × 2 = 0 + 0.658 944 401 408;
  • 51) 0.658 944 401 408 × 2 = 1 + 0.317 888 802 816;
  • 52) 0.317 888 802 816 × 2 = 0 + 0.635 777 605 632;
  • 53) 0.635 777 605 632 × 2 = 1 + 0.271 555 211 264;
  • 54) 0.271 555 211 264 × 2 = 0 + 0.543 110 422 528;
  • 55) 0.543 110 422 528 × 2 = 1 + 0.086 220 845 056;
  • 56) 0.086 220 845 056 × 2 = 0 + 0.172 441 690 112;
  • 57) 0.172 441 690 112 × 2 = 0 + 0.344 883 380 224;
  • 58) 0.344 883 380 224 × 2 = 0 + 0.689 766 760 448;
  • 59) 0.689 766 760 448 × 2 = 1 + 0.379 533 520 896;
  • 60) 0.379 533 520 896 × 2 = 0 + 0.759 067 041 792;
  • 61) 0.759 067 041 792 × 2 = 1 + 0.518 134 083 584;
  • 62) 0.518 134 083 584 × 2 = 1 + 0.036 268 167 168;
  • 63) 0.036 268 167 168 × 2 = 0 + 0.072 536 334 336;
  • 64) 0.072 536 334 336 × 2 = 0 + 0.145 072 668 672;
  • 65) 0.145 072 668 672 × 2 = 0 + 0.290 145 337 344;
  • 66) 0.290 145 337 344 × 2 = 0 + 0.580 290 674 688;
  • 67) 0.580 290 674 688 × 2 = 1 + 0.160 581 349 376;

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:


0.000 035 666 967(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 0100 1111 0001 1010 1010 0010 1100 001(2)

6. Positive number before normalization:

0.000 035 666 967(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 0100 1111 0001 1010 1010 0010 1100 001(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:


0.000 035 666 967(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 0100 1111 0001 1010 1010 0010 1100 001(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0111 0100 1111 0001 1010 1010 0010 1100 001(2) × 20 =

1.0010 1011 0011 0010 0011 1010 0111 1000 1101 0101 0001 0110 0001(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 0011 1010 0111 1000 1101 0101 0001 0110 0001


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 0011 1010 0111 1000 1101 0101 0001 0110 0001 =

0010 1011 0011 0010 0011 1010 0111 1000 1101 0101 0001 0110 0001


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 0011 1010 0111 1000 1101 0101 0001 0110 0001


Decimal number -0.000 035 666 967 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0011 1010 0111 1000 1101 0101 0001 0110 0001

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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