-0.016 738 891 601 562 496 556 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 556 8(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
-0.016 738 891 601 562 496 556 8(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:

|-0.016 738 891 601 562 496 556 8| = 0.016 738 891 601 562 496 556 8


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: 0.016 738 891 601 562 496 556 8.

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) 0.016 738 891 601 562 496 556 8 × 2 = 0 + 0.033 477 783 203 124 993 113 6;
  • 2) 0.033 477 783 203 124 993 113 6 × 2 = 0 + 0.066 955 566 406 249 986 227 2;
  • 3) 0.066 955 566 406 249 986 227 2 × 2 = 0 + 0.133 911 132 812 499 972 454 4;
  • 4) 0.133 911 132 812 499 972 454 4 × 2 = 0 + 0.267 822 265 624 999 944 908 8;
  • 5) 0.267 822 265 624 999 944 908 8 × 2 = 0 + 0.535 644 531 249 999 889 817 6;
  • 6) 0.535 644 531 249 999 889 817 6 × 2 = 1 + 0.071 289 062 499 999 779 635 2;
  • 7) 0.071 289 062 499 999 779 635 2 × 2 = 0 + 0.142 578 124 999 999 559 270 4;
  • 8) 0.142 578 124 999 999 559 270 4 × 2 = 0 + 0.285 156 249 999 999 118 540 8;
  • 9) 0.285 156 249 999 999 118 540 8 × 2 = 0 + 0.570 312 499 999 998 237 081 6;
  • 10) 0.570 312 499 999 998 237 081 6 × 2 = 1 + 0.140 624 999 999 996 474 163 2;
  • 11) 0.140 624 999 999 996 474 163 2 × 2 = 0 + 0.281 249 999 999 992 948 326 4;
  • 12) 0.281 249 999 999 992 948 326 4 × 2 = 0 + 0.562 499 999 999 985 896 652 8;
  • 13) 0.562 499 999 999 985 896 652 8 × 2 = 1 + 0.124 999 999 999 971 793 305 6;
  • 14) 0.124 999 999 999 971 793 305 6 × 2 = 0 + 0.249 999 999 999 943 586 611 2;
  • 15) 0.249 999 999 999 943 586 611 2 × 2 = 0 + 0.499 999 999 999 887 173 222 4;
  • 16) 0.499 999 999 999 887 173 222 4 × 2 = 0 + 0.999 999 999 999 774 346 444 8;
  • 17) 0.999 999 999 999 774 346 444 8 × 2 = 1 + 0.999 999 999 999 548 692 889 6;
  • 18) 0.999 999 999 999 548 692 889 6 × 2 = 1 + 0.999 999 999 999 097 385 779 2;
  • 19) 0.999 999 999 999 097 385 779 2 × 2 = 1 + 0.999 999 999 998 194 771 558 4;
  • 20) 0.999 999 999 998 194 771 558 4 × 2 = 1 + 0.999 999 999 996 389 543 116 8;
  • 21) 0.999 999 999 996 389 543 116 8 × 2 = 1 + 0.999 999 999 992 779 086 233 6;
  • 22) 0.999 999 999 992 779 086 233 6 × 2 = 1 + 0.999 999 999 985 558 172 467 2;
  • 23) 0.999 999 999 985 558 172 467 2 × 2 = 1 + 0.999 999 999 971 116 344 934 4;
  • 24) 0.999 999 999 971 116 344 934 4 × 2 = 1 + 0.999 999 999 942 232 689 868 8;
  • 25) 0.999 999 999 942 232 689 868 8 × 2 = 1 + 0.999 999 999 884 465 379 737 6;
  • 26) 0.999 999 999 884 465 379 737 6 × 2 = 1 + 0.999 999 999 768 930 759 475 2;
  • 27) 0.999 999 999 768 930 759 475 2 × 2 = 1 + 0.999 999 999 537 861 518 950 4;
  • 28) 0.999 999 999 537 861 518 950 4 × 2 = 1 + 0.999 999 999 075 723 037 900 8;
  • 29) 0.999 999 999 075 723 037 900 8 × 2 = 1 + 0.999 999 998 151 446 075 801 6;
  • 30) 0.999 999 998 151 446 075 801 6 × 2 = 1 + 0.999 999 996 302 892 151 603 2;
  • 31) 0.999 999 996 302 892 151 603 2 × 2 = 1 + 0.999 999 992 605 784 303 206 4;
  • 32) 0.999 999 992 605 784 303 206 4 × 2 = 1 + 0.999 999 985 211 568 606 412 8;
  • 33) 0.999 999 985 211 568 606 412 8 × 2 = 1 + 0.999 999 970 423 137 212 825 6;
  • 34) 0.999 999 970 423 137 212 825 6 × 2 = 1 + 0.999 999 940 846 274 425 651 2;
  • 35) 0.999 999 940 846 274 425 651 2 × 2 = 1 + 0.999 999 881 692 548 851 302 4;
  • 36) 0.999 999 881 692 548 851 302 4 × 2 = 1 + 0.999 999 763 385 097 702 604 8;
  • 37) 0.999 999 763 385 097 702 604 8 × 2 = 1 + 0.999 999 526 770 195 405 209 6;
  • 38) 0.999 999 526 770 195 405 209 6 × 2 = 1 + 0.999 999 053 540 390 810 419 2;
  • 39) 0.999 999 053 540 390 810 419 2 × 2 = 1 + 0.999 998 107 080 781 620 838 4;
  • 40) 0.999 998 107 080 781 620 838 4 × 2 = 1 + 0.999 996 214 161 563 241 676 8;
  • 41) 0.999 996 214 161 563 241 676 8 × 2 = 1 + 0.999 992 428 323 126 483 353 6;
  • 42) 0.999 992 428 323 126 483 353 6 × 2 = 1 + 0.999 984 856 646 252 966 707 2;
  • 43) 0.999 984 856 646 252 966 707 2 × 2 = 1 + 0.999 969 713 292 505 933 414 4;
  • 44) 0.999 969 713 292 505 933 414 4 × 2 = 1 + 0.999 939 426 585 011 866 828 8;
  • 45) 0.999 939 426 585 011 866 828 8 × 2 = 1 + 0.999 878 853 170 023 733 657 6;
  • 46) 0.999 878 853 170 023 733 657 6 × 2 = 1 + 0.999 757 706 340 047 467 315 2;
  • 47) 0.999 757 706 340 047 467 315 2 × 2 = 1 + 0.999 515 412 680 094 934 630 4;
  • 48) 0.999 515 412 680 094 934 630 4 × 2 = 1 + 0.999 030 825 360 189 869 260 8;
  • 49) 0.999 030 825 360 189 869 260 8 × 2 = 1 + 0.998 061 650 720 379 738 521 6;
  • 50) 0.998 061 650 720 379 738 521 6 × 2 = 1 + 0.996 123 301 440 759 477 043 2;
  • 51) 0.996 123 301 440 759 477 043 2 × 2 = 1 + 0.992 246 602 881 518 954 086 4;
  • 52) 0.992 246 602 881 518 954 086 4 × 2 = 1 + 0.984 493 205 763 037 908 172 8;
  • 53) 0.984 493 205 763 037 908 172 8 × 2 = 1 + 0.968 986 411 526 075 816 345 6;
  • 54) 0.968 986 411 526 075 816 345 6 × 2 = 1 + 0.937 972 823 052 151 632 691 2;
  • 55) 0.937 972 823 052 151 632 691 2 × 2 = 1 + 0.875 945 646 104 303 265 382 4;
  • 56) 0.875 945 646 104 303 265 382 4 × 2 = 1 + 0.751 891 292 208 606 530 764 8;
  • 57) 0.751 891 292 208 606 530 764 8 × 2 = 1 + 0.503 782 584 417 213 061 529 6;
  • 58) 0.503 782 584 417 213 061 529 6 × 2 = 1 + 0.007 565 168 834 426 123 059 2;

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:


0.016 738 891 601 562 496 556 8(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 556 8(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:


0.016 738 891 601 562 496 556 8(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 -0.016 738 891 601 562 496 556 8 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