-0.000 006 491 636 509 588 76 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 006 491 636 509 588 76(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
-0.000 006 491 636 509 588 76(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:

|-0.000 006 491 636 509 588 76| = 0.000 006 491 636 509 588 76


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: 0.000 006 491 636 509 588 76.

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) 0.000 006 491 636 509 588 76 × 2 = 0 + 0.000 012 983 273 019 177 52;
  • 2) 0.000 012 983 273 019 177 52 × 2 = 0 + 0.000 025 966 546 038 355 04;
  • 3) 0.000 025 966 546 038 355 04 × 2 = 0 + 0.000 051 933 092 076 710 08;
  • 4) 0.000 051 933 092 076 710 08 × 2 = 0 + 0.000 103 866 184 153 420 16;
  • 5) 0.000 103 866 184 153 420 16 × 2 = 0 + 0.000 207 732 368 306 840 32;
  • 6) 0.000 207 732 368 306 840 32 × 2 = 0 + 0.000 415 464 736 613 680 64;
  • 7) 0.000 415 464 736 613 680 64 × 2 = 0 + 0.000 830 929 473 227 361 28;
  • 8) 0.000 830 929 473 227 361 28 × 2 = 0 + 0.001 661 858 946 454 722 56;
  • 9) 0.001 661 858 946 454 722 56 × 2 = 0 + 0.003 323 717 892 909 445 12;
  • 10) 0.003 323 717 892 909 445 12 × 2 = 0 + 0.006 647 435 785 818 890 24;
  • 11) 0.006 647 435 785 818 890 24 × 2 = 0 + 0.013 294 871 571 637 780 48;
  • 12) 0.013 294 871 571 637 780 48 × 2 = 0 + 0.026 589 743 143 275 560 96;
  • 13) 0.026 589 743 143 275 560 96 × 2 = 0 + 0.053 179 486 286 551 121 92;
  • 14) 0.053 179 486 286 551 121 92 × 2 = 0 + 0.106 358 972 573 102 243 84;
  • 15) 0.106 358 972 573 102 243 84 × 2 = 0 + 0.212 717 945 146 204 487 68;
  • 16) 0.212 717 945 146 204 487 68 × 2 = 0 + 0.425 435 890 292 408 975 36;
  • 17) 0.425 435 890 292 408 975 36 × 2 = 0 + 0.850 871 780 584 817 950 72;
  • 18) 0.850 871 780 584 817 950 72 × 2 = 1 + 0.701 743 561 169 635 901 44;
  • 19) 0.701 743 561 169 635 901 44 × 2 = 1 + 0.403 487 122 339 271 802 88;
  • 20) 0.403 487 122 339 271 802 88 × 2 = 0 + 0.806 974 244 678 543 605 76;
  • 21) 0.806 974 244 678 543 605 76 × 2 = 1 + 0.613 948 489 357 087 211 52;
  • 22) 0.613 948 489 357 087 211 52 × 2 = 1 + 0.227 896 978 714 174 423 04;
  • 23) 0.227 896 978 714 174 423 04 × 2 = 0 + 0.455 793 957 428 348 846 08;
  • 24) 0.455 793 957 428 348 846 08 × 2 = 0 + 0.911 587 914 856 697 692 16;
  • 25) 0.911 587 914 856 697 692 16 × 2 = 1 + 0.823 175 829 713 395 384 32;
  • 26) 0.823 175 829 713 395 384 32 × 2 = 1 + 0.646 351 659 426 790 768 64;
  • 27) 0.646 351 659 426 790 768 64 × 2 = 1 + 0.292 703 318 853 581 537 28;
  • 28) 0.292 703 318 853 581 537 28 × 2 = 0 + 0.585 406 637 707 163 074 56;
  • 29) 0.585 406 637 707 163 074 56 × 2 = 1 + 0.170 813 275 414 326 149 12;
  • 30) 0.170 813 275 414 326 149 12 × 2 = 0 + 0.341 626 550 828 652 298 24;
  • 31) 0.341 626 550 828 652 298 24 × 2 = 0 + 0.683 253 101 657 304 596 48;
  • 32) 0.683 253 101 657 304 596 48 × 2 = 1 + 0.366 506 203 314 609 192 96;
  • 33) 0.366 506 203 314 609 192 96 × 2 = 0 + 0.733 012 406 629 218 385 92;
  • 34) 0.733 012 406 629 218 385 92 × 2 = 1 + 0.466 024 813 258 436 771 84;
  • 35) 0.466 024 813 258 436 771 84 × 2 = 0 + 0.932 049 626 516 873 543 68;
  • 36) 0.932 049 626 516 873 543 68 × 2 = 1 + 0.864 099 253 033 747 087 36;
  • 37) 0.864 099 253 033 747 087 36 × 2 = 1 + 0.728 198 506 067 494 174 72;
  • 38) 0.728 198 506 067 494 174 72 × 2 = 1 + 0.456 397 012 134 988 349 44;
  • 39) 0.456 397 012 134 988 349 44 × 2 = 0 + 0.912 794 024 269 976 698 88;
  • 40) 0.912 794 024 269 976 698 88 × 2 = 1 + 0.825 588 048 539 953 397 76;
  • 41) 0.825 588 048 539 953 397 76 × 2 = 1 + 0.651 176 097 079 906 795 52;
  • 42) 0.651 176 097 079 906 795 52 × 2 = 1 + 0.302 352 194 159 813 591 04;
  • 43) 0.302 352 194 159 813 591 04 × 2 = 0 + 0.604 704 388 319 627 182 08;
  • 44) 0.604 704 388 319 627 182 08 × 2 = 1 + 0.209 408 776 639 254 364 16;
  • 45) 0.209 408 776 639 254 364 16 × 2 = 0 + 0.418 817 553 278 508 728 32;
  • 46) 0.418 817 553 278 508 728 32 × 2 = 0 + 0.837 635 106 557 017 456 64;
  • 47) 0.837 635 106 557 017 456 64 × 2 = 1 + 0.675 270 213 114 034 913 28;
  • 48) 0.675 270 213 114 034 913 28 × 2 = 1 + 0.350 540 426 228 069 826 56;
  • 49) 0.350 540 426 228 069 826 56 × 2 = 0 + 0.701 080 852 456 139 653 12;
  • 50) 0.701 080 852 456 139 653 12 × 2 = 1 + 0.402 161 704 912 279 306 24;
  • 51) 0.402 161 704 912 279 306 24 × 2 = 0 + 0.804 323 409 824 558 612 48;
  • 52) 0.804 323 409 824 558 612 48 × 2 = 1 + 0.608 646 819 649 117 224 96;
  • 53) 0.608 646 819 649 117 224 96 × 2 = 1 + 0.217 293 639 298 234 449 92;
  • 54) 0.217 293 639 298 234 449 92 × 2 = 0 + 0.434 587 278 596 468 899 84;
  • 55) 0.434 587 278 596 468 899 84 × 2 = 0 + 0.869 174 557 192 937 799 68;
  • 56) 0.869 174 557 192 937 799 68 × 2 = 1 + 0.738 349 114 385 875 599 36;
  • 57) 0.738 349 114 385 875 599 36 × 2 = 1 + 0.476 698 228 771 751 198 72;
  • 58) 0.476 698 228 771 751 198 72 × 2 = 0 + 0.953 396 457 543 502 397 44;
  • 59) 0.953 396 457 543 502 397 44 × 2 = 1 + 0.906 792 915 087 004 794 88;
  • 60) 0.906 792 915 087 004 794 88 × 2 = 1 + 0.813 585 830 174 009 589 76;
  • 61) 0.813 585 830 174 009 589 76 × 2 = 1 + 0.627 171 660 348 019 179 52;
  • 62) 0.627 171 660 348 019 179 52 × 2 = 1 + 0.254 343 320 696 038 359 04;
  • 63) 0.254 343 320 696 038 359 04 × 2 = 0 + 0.508 686 641 392 076 718 08;
  • 64) 0.508 686 641 392 076 718 08 × 2 = 1 + 0.017 373 282 784 153 436 16;
  • 65) 0.017 373 282 784 153 436 16 × 2 = 0 + 0.034 746 565 568 306 872 32;
  • 66) 0.034 746 565 568 306 872 32 × 2 = 0 + 0.069 493 131 136 613 744 64;
  • 67) 0.069 493 131 136 613 744 64 × 2 = 0 + 0.138 986 262 273 227 489 28;
  • 68) 0.138 986 262 273 227 489 28 × 2 = 0 + 0.277 972 524 546 454 978 56;
  • 69) 0.277 972 524 546 454 978 56 × 2 = 0 + 0.555 945 049 092 909 957 12;
  • 70) 0.555 945 049 092 909 957 12 × 2 = 1 + 0.111 890 098 185 819 914 24;

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:


0.000 006 491 636 509 588 76(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1011 1101 0000 01(2)

6. Positive number before normalization:

0.000 006 491 636 509 588 76(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1011 1101 0000 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 18 positions to the right, so that only one non zero digit remains to the left of it:


0.000 006 491 636 509 588 76(10) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1011 1101 0000 01(2) =

0.0000 0000 0000 0000 0110 1100 1110 1001 0101 1101 1101 0011 0101 1001 1011 1101 0000 01(2) × 20 =

1.1011 0011 1010 0101 0111 0111 0100 1101 0110 0110 1111 0100 0001(2) × 2-18


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): -18

Mantissa (not normalized):
1.1011 0011 1010 0101 0111 0111 0100 1101 0110 0110 1111 0100 0001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-18 + 2(11-1) - 1 =

(-18 + 1 023)(10) =

1 005(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;
  • 1 005 ÷ 2 = 502 + 1;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1005(10) =

011 1110 1101(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. 1011 0011 1010 0101 0111 0111 0100 1101 0110 0110 1111 0100 0001 =

1011 0011 1010 0101 0111 0111 0100 1101 0110 0110 1111 0100 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 1110 1101

Mantissa (52 bits) =
1011 0011 1010 0101 0111 0111 0100 1101 0110 0110 1111 0100 0001


Decimal number -0.000 006 491 636 509 588 76 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1101 - 1011 0011 1010 0101 0111 0111 0100 1101 0110 0110 1111 0100 0001

» Convert decimal -0.000 006 491 636 509 589 22 to 64 bit double precision IEEE 754 binary floating point representation standard

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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