-0.000 035 666 835 439 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 835 439(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
-0.000 035 666 835 439(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:

|-0.000 035 666 835 439| = 0.000 035 666 835 439


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: 0.000 035 666 835 439.

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) 0.000 035 666 835 439 × 2 = 0 + 0.000 071 333 670 878;
  • 2) 0.000 071 333 670 878 × 2 = 0 + 0.000 142 667 341 756;
  • 3) 0.000 142 667 341 756 × 2 = 0 + 0.000 285 334 683 512;
  • 4) 0.000 285 334 683 512 × 2 = 0 + 0.000 570 669 367 024;
  • 5) 0.000 570 669 367 024 × 2 = 0 + 0.001 141 338 734 048;
  • 6) 0.001 141 338 734 048 × 2 = 0 + 0.002 282 677 468 096;
  • 7) 0.002 282 677 468 096 × 2 = 0 + 0.004 565 354 936 192;
  • 8) 0.004 565 354 936 192 × 2 = 0 + 0.009 130 709 872 384;
  • 9) 0.009 130 709 872 384 × 2 = 0 + 0.018 261 419 744 768;
  • 10) 0.018 261 419 744 768 × 2 = 0 + 0.036 522 839 489 536;
  • 11) 0.036 522 839 489 536 × 2 = 0 + 0.073 045 678 979 072;
  • 12) 0.073 045 678 979 072 × 2 = 0 + 0.146 091 357 958 144;
  • 13) 0.146 091 357 958 144 × 2 = 0 + 0.292 182 715 916 288;
  • 14) 0.292 182 715 916 288 × 2 = 0 + 0.584 365 431 832 576;
  • 15) 0.584 365 431 832 576 × 2 = 1 + 0.168 730 863 665 152;
  • 16) 0.168 730 863 665 152 × 2 = 0 + 0.337 461 727 330 304;
  • 17) 0.337 461 727 330 304 × 2 = 0 + 0.674 923 454 660 608;
  • 18) 0.674 923 454 660 608 × 2 = 1 + 0.349 846 909 321 216;
  • 19) 0.349 846 909 321 216 × 2 = 0 + 0.699 693 818 642 432;
  • 20) 0.699 693 818 642 432 × 2 = 1 + 0.399 387 637 284 864;
  • 21) 0.399 387 637 284 864 × 2 = 0 + 0.798 775 274 569 728;
  • 22) 0.798 775 274 569 728 × 2 = 1 + 0.597 550 549 139 456;
  • 23) 0.597 550 549 139 456 × 2 = 1 + 0.195 101 098 278 912;
  • 24) 0.195 101 098 278 912 × 2 = 0 + 0.390 202 196 557 824;
  • 25) 0.390 202 196 557 824 × 2 = 0 + 0.780 404 393 115 648;
  • 26) 0.780 404 393 115 648 × 2 = 1 + 0.560 808 786 231 296;
  • 27) 0.560 808 786 231 296 × 2 = 1 + 0.121 617 572 462 592;
  • 28) 0.121 617 572 462 592 × 2 = 0 + 0.243 235 144 925 184;
  • 29) 0.243 235 144 925 184 × 2 = 0 + 0.486 470 289 850 368;
  • 30) 0.486 470 289 850 368 × 2 = 0 + 0.972 940 579 700 736;
  • 31) 0.972 940 579 700 736 × 2 = 1 + 0.945 881 159 401 472;
  • 32) 0.945 881 159 401 472 × 2 = 1 + 0.891 762 318 802 944;
  • 33) 0.891 762 318 802 944 × 2 = 1 + 0.783 524 637 605 888;
  • 34) 0.783 524 637 605 888 × 2 = 1 + 0.567 049 275 211 776;
  • 35) 0.567 049 275 211 776 × 2 = 1 + 0.134 098 550 423 552;
  • 36) 0.134 098 550 423 552 × 2 = 0 + 0.268 197 100 847 104;
  • 37) 0.268 197 100 847 104 × 2 = 0 + 0.536 394 201 694 208;
  • 38) 0.536 394 201 694 208 × 2 = 1 + 0.072 788 403 388 416;
  • 39) 0.072 788 403 388 416 × 2 = 0 + 0.145 576 806 776 832;
  • 40) 0.145 576 806 776 832 × 2 = 0 + 0.291 153 613 553 664;
  • 41) 0.291 153 613 553 664 × 2 = 0 + 0.582 307 227 107 328;
  • 42) 0.582 307 227 107 328 × 2 = 1 + 0.164 614 454 214 656;
  • 43) 0.164 614 454 214 656 × 2 = 0 + 0.329 228 908 429 312;
  • 44) 0.329 228 908 429 312 × 2 = 0 + 0.658 457 816 858 624;
  • 45) 0.658 457 816 858 624 × 2 = 1 + 0.316 915 633 717 248;
  • 46) 0.316 915 633 717 248 × 2 = 0 + 0.633 831 267 434 496;
  • 47) 0.633 831 267 434 496 × 2 = 1 + 0.267 662 534 868 992;
  • 48) 0.267 662 534 868 992 × 2 = 0 + 0.535 325 069 737 984;
  • 49) 0.535 325 069 737 984 × 2 = 1 + 0.070 650 139 475 968;
  • 50) 0.070 650 139 475 968 × 2 = 0 + 0.141 300 278 951 936;
  • 51) 0.141 300 278 951 936 × 2 = 0 + 0.282 600 557 903 872;
  • 52) 0.282 600 557 903 872 × 2 = 0 + 0.565 201 115 807 744;
  • 53) 0.565 201 115 807 744 × 2 = 1 + 0.130 402 231 615 488;
  • 54) 0.130 402 231 615 488 × 2 = 0 + 0.260 804 463 230 976;
  • 55) 0.260 804 463 230 976 × 2 = 0 + 0.521 608 926 461 952;
  • 56) 0.521 608 926 461 952 × 2 = 1 + 0.043 217 852 923 904;
  • 57) 0.043 217 852 923 904 × 2 = 0 + 0.086 435 705 847 808;
  • 58) 0.086 435 705 847 808 × 2 = 0 + 0.172 871 411 695 616;
  • 59) 0.172 871 411 695 616 × 2 = 0 + 0.345 742 823 391 232;
  • 60) 0.345 742 823 391 232 × 2 = 0 + 0.691 485 646 782 464;
  • 61) 0.691 485 646 782 464 × 2 = 1 + 0.382 971 293 564 928;
  • 62) 0.382 971 293 564 928 × 2 = 0 + 0.765 942 587 129 856;
  • 63) 0.765 942 587 129 856 × 2 = 1 + 0.531 885 174 259 712;
  • 64) 0.531 885 174 259 712 × 2 = 1 + 0.063 770 348 519 424;
  • 65) 0.063 770 348 519 424 × 2 = 0 + 0.127 540 697 038 848;
  • 66) 0.127 540 697 038 848 × 2 = 0 + 0.255 081 394 077 696;
  • 67) 0.255 081 394 077 696 × 2 = 0 + 0.510 162 788 155 392;

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:


0.000 035 666 835 439(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1010 1000 1001 0000 1011 000(2)

6. Positive number before normalization:

0.000 035 666 835 439(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1010 1000 1001 0000 1011 000(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:


0.000 035 666 835 439(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1010 1000 1001 0000 1011 000(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0100 1010 1000 1001 0000 1011 000(2) × 20 =

1.0010 1011 0011 0001 1111 0010 0010 0101 0100 0100 1000 0101 1000(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 0001 1111 0010 0010 0101 0100 0100 1000 0101 1000


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 0001 1111 0010 0010 0101 0100 0100 1000 0101 1000 =

0010 1011 0011 0001 1111 0010 0010 0101 0100 0100 1000 0101 1000


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 0001 1111 0010 0010 0101 0100 0100 1000 0101 1000


Decimal number -0.000 035 666 835 439 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0010 0010 0101 0100 0100 1000 0101 1000

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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