-0.016 738 891 601 562 496 530 06 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 06(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
-0.016 738 891 601 562 496 530 06(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:

|-0.016 738 891 601 562 496 530 06| = 0.016 738 891 601 562 496 530 06


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: 0.016 738 891 601 562 496 530 06.

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) 0.016 738 891 601 562 496 530 06 × 2 = 0 + 0.033 477 783 203 124 993 060 12;
  • 2) 0.033 477 783 203 124 993 060 12 × 2 = 0 + 0.066 955 566 406 249 986 120 24;
  • 3) 0.066 955 566 406 249 986 120 24 × 2 = 0 + 0.133 911 132 812 499 972 240 48;
  • 4) 0.133 911 132 812 499 972 240 48 × 2 = 0 + 0.267 822 265 624 999 944 480 96;
  • 5) 0.267 822 265 624 999 944 480 96 × 2 = 0 + 0.535 644 531 249 999 888 961 92;
  • 6) 0.535 644 531 249 999 888 961 92 × 2 = 1 + 0.071 289 062 499 999 777 923 84;
  • 7) 0.071 289 062 499 999 777 923 84 × 2 = 0 + 0.142 578 124 999 999 555 847 68;
  • 8) 0.142 578 124 999 999 555 847 68 × 2 = 0 + 0.285 156 249 999 999 111 695 36;
  • 9) 0.285 156 249 999 999 111 695 36 × 2 = 0 + 0.570 312 499 999 998 223 390 72;
  • 10) 0.570 312 499 999 998 223 390 72 × 2 = 1 + 0.140 624 999 999 996 446 781 44;
  • 11) 0.140 624 999 999 996 446 781 44 × 2 = 0 + 0.281 249 999 999 992 893 562 88;
  • 12) 0.281 249 999 999 992 893 562 88 × 2 = 0 + 0.562 499 999 999 985 787 125 76;
  • 13) 0.562 499 999 999 985 787 125 76 × 2 = 1 + 0.124 999 999 999 971 574 251 52;
  • 14) 0.124 999 999 999 971 574 251 52 × 2 = 0 + 0.249 999 999 999 943 148 503 04;
  • 15) 0.249 999 999 999 943 148 503 04 × 2 = 0 + 0.499 999 999 999 886 297 006 08;
  • 16) 0.499 999 999 999 886 297 006 08 × 2 = 0 + 0.999 999 999 999 772 594 012 16;
  • 17) 0.999 999 999 999 772 594 012 16 × 2 = 1 + 0.999 999 999 999 545 188 024 32;
  • 18) 0.999 999 999 999 545 188 024 32 × 2 = 1 + 0.999 999 999 999 090 376 048 64;
  • 19) 0.999 999 999 999 090 376 048 64 × 2 = 1 + 0.999 999 999 998 180 752 097 28;
  • 20) 0.999 999 999 998 180 752 097 28 × 2 = 1 + 0.999 999 999 996 361 504 194 56;
  • 21) 0.999 999 999 996 361 504 194 56 × 2 = 1 + 0.999 999 999 992 723 008 389 12;
  • 22) 0.999 999 999 992 723 008 389 12 × 2 = 1 + 0.999 999 999 985 446 016 778 24;
  • 23) 0.999 999 999 985 446 016 778 24 × 2 = 1 + 0.999 999 999 970 892 033 556 48;
  • 24) 0.999 999 999 970 892 033 556 48 × 2 = 1 + 0.999 999 999 941 784 067 112 96;
  • 25) 0.999 999 999 941 784 067 112 96 × 2 = 1 + 0.999 999 999 883 568 134 225 92;
  • 26) 0.999 999 999 883 568 134 225 92 × 2 = 1 + 0.999 999 999 767 136 268 451 84;
  • 27) 0.999 999 999 767 136 268 451 84 × 2 = 1 + 0.999 999 999 534 272 536 903 68;
  • 28) 0.999 999 999 534 272 536 903 68 × 2 = 1 + 0.999 999 999 068 545 073 807 36;
  • 29) 0.999 999 999 068 545 073 807 36 × 2 = 1 + 0.999 999 998 137 090 147 614 72;
  • 30) 0.999 999 998 137 090 147 614 72 × 2 = 1 + 0.999 999 996 274 180 295 229 44;
  • 31) 0.999 999 996 274 180 295 229 44 × 2 = 1 + 0.999 999 992 548 360 590 458 88;
  • 32) 0.999 999 992 548 360 590 458 88 × 2 = 1 + 0.999 999 985 096 721 180 917 76;
  • 33) 0.999 999 985 096 721 180 917 76 × 2 = 1 + 0.999 999 970 193 442 361 835 52;
  • 34) 0.999 999 970 193 442 361 835 52 × 2 = 1 + 0.999 999 940 386 884 723 671 04;
  • 35) 0.999 999 940 386 884 723 671 04 × 2 = 1 + 0.999 999 880 773 769 447 342 08;
  • 36) 0.999 999 880 773 769 447 342 08 × 2 = 1 + 0.999 999 761 547 538 894 684 16;
  • 37) 0.999 999 761 547 538 894 684 16 × 2 = 1 + 0.999 999 523 095 077 789 368 32;
  • 38) 0.999 999 523 095 077 789 368 32 × 2 = 1 + 0.999 999 046 190 155 578 736 64;
  • 39) 0.999 999 046 190 155 578 736 64 × 2 = 1 + 0.999 998 092 380 311 157 473 28;
  • 40) 0.999 998 092 380 311 157 473 28 × 2 = 1 + 0.999 996 184 760 622 314 946 56;
  • 41) 0.999 996 184 760 622 314 946 56 × 2 = 1 + 0.999 992 369 521 244 629 893 12;
  • 42) 0.999 992 369 521 244 629 893 12 × 2 = 1 + 0.999 984 739 042 489 259 786 24;
  • 43) 0.999 984 739 042 489 259 786 24 × 2 = 1 + 0.999 969 478 084 978 519 572 48;
  • 44) 0.999 969 478 084 978 519 572 48 × 2 = 1 + 0.999 938 956 169 957 039 144 96;
  • 45) 0.999 938 956 169 957 039 144 96 × 2 = 1 + 0.999 877 912 339 914 078 289 92;
  • 46) 0.999 877 912 339 914 078 289 92 × 2 = 1 + 0.999 755 824 679 828 156 579 84;
  • 47) 0.999 755 824 679 828 156 579 84 × 2 = 1 + 0.999 511 649 359 656 313 159 68;
  • 48) 0.999 511 649 359 656 313 159 68 × 2 = 1 + 0.999 023 298 719 312 626 319 36;
  • 49) 0.999 023 298 719 312 626 319 36 × 2 = 1 + 0.998 046 597 438 625 252 638 72;
  • 50) 0.998 046 597 438 625 252 638 72 × 2 = 1 + 0.996 093 194 877 250 505 277 44;
  • 51) 0.996 093 194 877 250 505 277 44 × 2 = 1 + 0.992 186 389 754 501 010 554 88;
  • 52) 0.992 186 389 754 501 010 554 88 × 2 = 1 + 0.984 372 779 509 002 021 109 76;
  • 53) 0.984 372 779 509 002 021 109 76 × 2 = 1 + 0.968 745 559 018 004 042 219 52;
  • 54) 0.968 745 559 018 004 042 219 52 × 2 = 1 + 0.937 491 118 036 008 084 439 04;
  • 55) 0.937 491 118 036 008 084 439 04 × 2 = 1 + 0.874 982 236 072 016 168 878 08;
  • 56) 0.874 982 236 072 016 168 878 08 × 2 = 1 + 0.749 964 472 144 032 337 756 16;
  • 57) 0.749 964 472 144 032 337 756 16 × 2 = 1 + 0.499 928 944 288 064 675 512 32;
  • 58) 0.499 928 944 288 064 675 512 32 × 2 = 0 + 0.999 857 888 576 129 351 024 64;

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:


0.016 738 891 601 562 496 530 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 530 06(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 06 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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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