-0.016 738 891 601 562 496 544 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 544 4(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
-0.016 738 891 601 562 496 544 4(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:

|-0.016 738 891 601 562 496 544 4| = 0.016 738 891 601 562 496 544 4


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: 0.016 738 891 601 562 496 544 4.

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) 0.016 738 891 601 562 496 544 4 × 2 = 0 + 0.033 477 783 203 124 993 088 8;
  • 2) 0.033 477 783 203 124 993 088 8 × 2 = 0 + 0.066 955 566 406 249 986 177 6;
  • 3) 0.066 955 566 406 249 986 177 6 × 2 = 0 + 0.133 911 132 812 499 972 355 2;
  • 4) 0.133 911 132 812 499 972 355 2 × 2 = 0 + 0.267 822 265 624 999 944 710 4;
  • 5) 0.267 822 265 624 999 944 710 4 × 2 = 0 + 0.535 644 531 249 999 889 420 8;
  • 6) 0.535 644 531 249 999 889 420 8 × 2 = 1 + 0.071 289 062 499 999 778 841 6;
  • 7) 0.071 289 062 499 999 778 841 6 × 2 = 0 + 0.142 578 124 999 999 557 683 2;
  • 8) 0.142 578 124 999 999 557 683 2 × 2 = 0 + 0.285 156 249 999 999 115 366 4;
  • 9) 0.285 156 249 999 999 115 366 4 × 2 = 0 + 0.570 312 499 999 998 230 732 8;
  • 10) 0.570 312 499 999 998 230 732 8 × 2 = 1 + 0.140 624 999 999 996 461 465 6;
  • 11) 0.140 624 999 999 996 461 465 6 × 2 = 0 + 0.281 249 999 999 992 922 931 2;
  • 12) 0.281 249 999 999 992 922 931 2 × 2 = 0 + 0.562 499 999 999 985 845 862 4;
  • 13) 0.562 499 999 999 985 845 862 4 × 2 = 1 + 0.124 999 999 999 971 691 724 8;
  • 14) 0.124 999 999 999 971 691 724 8 × 2 = 0 + 0.249 999 999 999 943 383 449 6;
  • 15) 0.249 999 999 999 943 383 449 6 × 2 = 0 + 0.499 999 999 999 886 766 899 2;
  • 16) 0.499 999 999 999 886 766 899 2 × 2 = 0 + 0.999 999 999 999 773 533 798 4;
  • 17) 0.999 999 999 999 773 533 798 4 × 2 = 1 + 0.999 999 999 999 547 067 596 8;
  • 18) 0.999 999 999 999 547 067 596 8 × 2 = 1 + 0.999 999 999 999 094 135 193 6;
  • 19) 0.999 999 999 999 094 135 193 6 × 2 = 1 + 0.999 999 999 998 188 270 387 2;
  • 20) 0.999 999 999 998 188 270 387 2 × 2 = 1 + 0.999 999 999 996 376 540 774 4;
  • 21) 0.999 999 999 996 376 540 774 4 × 2 = 1 + 0.999 999 999 992 753 081 548 8;
  • 22) 0.999 999 999 992 753 081 548 8 × 2 = 1 + 0.999 999 999 985 506 163 097 6;
  • 23) 0.999 999 999 985 506 163 097 6 × 2 = 1 + 0.999 999 999 971 012 326 195 2;
  • 24) 0.999 999 999 971 012 326 195 2 × 2 = 1 + 0.999 999 999 942 024 652 390 4;
  • 25) 0.999 999 999 942 024 652 390 4 × 2 = 1 + 0.999 999 999 884 049 304 780 8;
  • 26) 0.999 999 999 884 049 304 780 8 × 2 = 1 + 0.999 999 999 768 098 609 561 6;
  • 27) 0.999 999 999 768 098 609 561 6 × 2 = 1 + 0.999 999 999 536 197 219 123 2;
  • 28) 0.999 999 999 536 197 219 123 2 × 2 = 1 + 0.999 999 999 072 394 438 246 4;
  • 29) 0.999 999 999 072 394 438 246 4 × 2 = 1 + 0.999 999 998 144 788 876 492 8;
  • 30) 0.999 999 998 144 788 876 492 8 × 2 = 1 + 0.999 999 996 289 577 752 985 6;
  • 31) 0.999 999 996 289 577 752 985 6 × 2 = 1 + 0.999 999 992 579 155 505 971 2;
  • 32) 0.999 999 992 579 155 505 971 2 × 2 = 1 + 0.999 999 985 158 311 011 942 4;
  • 33) 0.999 999 985 158 311 011 942 4 × 2 = 1 + 0.999 999 970 316 622 023 884 8;
  • 34) 0.999 999 970 316 622 023 884 8 × 2 = 1 + 0.999 999 940 633 244 047 769 6;
  • 35) 0.999 999 940 633 244 047 769 6 × 2 = 1 + 0.999 999 881 266 488 095 539 2;
  • 36) 0.999 999 881 266 488 095 539 2 × 2 = 1 + 0.999 999 762 532 976 191 078 4;
  • 37) 0.999 999 762 532 976 191 078 4 × 2 = 1 + 0.999 999 525 065 952 382 156 8;
  • 38) 0.999 999 525 065 952 382 156 8 × 2 = 1 + 0.999 999 050 131 904 764 313 6;
  • 39) 0.999 999 050 131 904 764 313 6 × 2 = 1 + 0.999 998 100 263 809 528 627 2;
  • 40) 0.999 998 100 263 809 528 627 2 × 2 = 1 + 0.999 996 200 527 619 057 254 4;
  • 41) 0.999 996 200 527 619 057 254 4 × 2 = 1 + 0.999 992 401 055 238 114 508 8;
  • 42) 0.999 992 401 055 238 114 508 8 × 2 = 1 + 0.999 984 802 110 476 229 017 6;
  • 43) 0.999 984 802 110 476 229 017 6 × 2 = 1 + 0.999 969 604 220 952 458 035 2;
  • 44) 0.999 969 604 220 952 458 035 2 × 2 = 1 + 0.999 939 208 441 904 916 070 4;
  • 45) 0.999 939 208 441 904 916 070 4 × 2 = 1 + 0.999 878 416 883 809 832 140 8;
  • 46) 0.999 878 416 883 809 832 140 8 × 2 = 1 + 0.999 756 833 767 619 664 281 6;
  • 47) 0.999 756 833 767 619 664 281 6 × 2 = 1 + 0.999 513 667 535 239 328 563 2;
  • 48) 0.999 513 667 535 239 328 563 2 × 2 = 1 + 0.999 027 335 070 478 657 126 4;
  • 49) 0.999 027 335 070 478 657 126 4 × 2 = 1 + 0.998 054 670 140 957 314 252 8;
  • 50) 0.998 054 670 140 957 314 252 8 × 2 = 1 + 0.996 109 340 281 914 628 505 6;
  • 51) 0.996 109 340 281 914 628 505 6 × 2 = 1 + 0.992 218 680 563 829 257 011 2;
  • 52) 0.992 218 680 563 829 257 011 2 × 2 = 1 + 0.984 437 361 127 658 514 022 4;
  • 53) 0.984 437 361 127 658 514 022 4 × 2 = 1 + 0.968 874 722 255 317 028 044 8;
  • 54) 0.968 874 722 255 317 028 044 8 × 2 = 1 + 0.937 749 444 510 634 056 089 6;
  • 55) 0.937 749 444 510 634 056 089 6 × 2 = 1 + 0.875 498 889 021 268 112 179 2;
  • 56) 0.875 498 889 021 268 112 179 2 × 2 = 1 + 0.750 997 778 042 536 224 358 4;
  • 57) 0.750 997 778 042 536 224 358 4 × 2 = 1 + 0.501 995 556 085 072 448 716 8;
  • 58) 0.501 995 556 085 072 448 716 8 × 2 = 1 + 0.003 991 112 170 144 897 433 6;

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:


0.016 738 891 601 562 496 544 4(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 544 4(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 544 4(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 496 544 4 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 1111

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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