-0.016 738 891 601 562 496 524 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 524 2(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
-0.016 738 891 601 562 496 524 2(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:

|-0.016 738 891 601 562 496 524 2| = 0.016 738 891 601 562 496 524 2


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: 0.016 738 891 601 562 496 524 2.

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) 0.016 738 891 601 562 496 524 2 × 2 = 0 + 0.033 477 783 203 124 993 048 4;
  • 2) 0.033 477 783 203 124 993 048 4 × 2 = 0 + 0.066 955 566 406 249 986 096 8;
  • 3) 0.066 955 566 406 249 986 096 8 × 2 = 0 + 0.133 911 132 812 499 972 193 6;
  • 4) 0.133 911 132 812 499 972 193 6 × 2 = 0 + 0.267 822 265 624 999 944 387 2;
  • 5) 0.267 822 265 624 999 944 387 2 × 2 = 0 + 0.535 644 531 249 999 888 774 4;
  • 6) 0.535 644 531 249 999 888 774 4 × 2 = 1 + 0.071 289 062 499 999 777 548 8;
  • 7) 0.071 289 062 499 999 777 548 8 × 2 = 0 + 0.142 578 124 999 999 555 097 6;
  • 8) 0.142 578 124 999 999 555 097 6 × 2 = 0 + 0.285 156 249 999 999 110 195 2;
  • 9) 0.285 156 249 999 999 110 195 2 × 2 = 0 + 0.570 312 499 999 998 220 390 4;
  • 10) 0.570 312 499 999 998 220 390 4 × 2 = 1 + 0.140 624 999 999 996 440 780 8;
  • 11) 0.140 624 999 999 996 440 780 8 × 2 = 0 + 0.281 249 999 999 992 881 561 6;
  • 12) 0.281 249 999 999 992 881 561 6 × 2 = 0 + 0.562 499 999 999 985 763 123 2;
  • 13) 0.562 499 999 999 985 763 123 2 × 2 = 1 + 0.124 999 999 999 971 526 246 4;
  • 14) 0.124 999 999 999 971 526 246 4 × 2 = 0 + 0.249 999 999 999 943 052 492 8;
  • 15) 0.249 999 999 999 943 052 492 8 × 2 = 0 + 0.499 999 999 999 886 104 985 6;
  • 16) 0.499 999 999 999 886 104 985 6 × 2 = 0 + 0.999 999 999 999 772 209 971 2;
  • 17) 0.999 999 999 999 772 209 971 2 × 2 = 1 + 0.999 999 999 999 544 419 942 4;
  • 18) 0.999 999 999 999 544 419 942 4 × 2 = 1 + 0.999 999 999 999 088 839 884 8;
  • 19) 0.999 999 999 999 088 839 884 8 × 2 = 1 + 0.999 999 999 998 177 679 769 6;
  • 20) 0.999 999 999 998 177 679 769 6 × 2 = 1 + 0.999 999 999 996 355 359 539 2;
  • 21) 0.999 999 999 996 355 359 539 2 × 2 = 1 + 0.999 999 999 992 710 719 078 4;
  • 22) 0.999 999 999 992 710 719 078 4 × 2 = 1 + 0.999 999 999 985 421 438 156 8;
  • 23) 0.999 999 999 985 421 438 156 8 × 2 = 1 + 0.999 999 999 970 842 876 313 6;
  • 24) 0.999 999 999 970 842 876 313 6 × 2 = 1 + 0.999 999 999 941 685 752 627 2;
  • 25) 0.999 999 999 941 685 752 627 2 × 2 = 1 + 0.999 999 999 883 371 505 254 4;
  • 26) 0.999 999 999 883 371 505 254 4 × 2 = 1 + 0.999 999 999 766 743 010 508 8;
  • 27) 0.999 999 999 766 743 010 508 8 × 2 = 1 + 0.999 999 999 533 486 021 017 6;
  • 28) 0.999 999 999 533 486 021 017 6 × 2 = 1 + 0.999 999 999 066 972 042 035 2;
  • 29) 0.999 999 999 066 972 042 035 2 × 2 = 1 + 0.999 999 998 133 944 084 070 4;
  • 30) 0.999 999 998 133 944 084 070 4 × 2 = 1 + 0.999 999 996 267 888 168 140 8;
  • 31) 0.999 999 996 267 888 168 140 8 × 2 = 1 + 0.999 999 992 535 776 336 281 6;
  • 32) 0.999 999 992 535 776 336 281 6 × 2 = 1 + 0.999 999 985 071 552 672 563 2;
  • 33) 0.999 999 985 071 552 672 563 2 × 2 = 1 + 0.999 999 970 143 105 345 126 4;
  • 34) 0.999 999 970 143 105 345 126 4 × 2 = 1 + 0.999 999 940 286 210 690 252 8;
  • 35) 0.999 999 940 286 210 690 252 8 × 2 = 1 + 0.999 999 880 572 421 380 505 6;
  • 36) 0.999 999 880 572 421 380 505 6 × 2 = 1 + 0.999 999 761 144 842 761 011 2;
  • 37) 0.999 999 761 144 842 761 011 2 × 2 = 1 + 0.999 999 522 289 685 522 022 4;
  • 38) 0.999 999 522 289 685 522 022 4 × 2 = 1 + 0.999 999 044 579 371 044 044 8;
  • 39) 0.999 999 044 579 371 044 044 8 × 2 = 1 + 0.999 998 089 158 742 088 089 6;
  • 40) 0.999 998 089 158 742 088 089 6 × 2 = 1 + 0.999 996 178 317 484 176 179 2;
  • 41) 0.999 996 178 317 484 176 179 2 × 2 = 1 + 0.999 992 356 634 968 352 358 4;
  • 42) 0.999 992 356 634 968 352 358 4 × 2 = 1 + 0.999 984 713 269 936 704 716 8;
  • 43) 0.999 984 713 269 936 704 716 8 × 2 = 1 + 0.999 969 426 539 873 409 433 6;
  • 44) 0.999 969 426 539 873 409 433 6 × 2 = 1 + 0.999 938 853 079 746 818 867 2;
  • 45) 0.999 938 853 079 746 818 867 2 × 2 = 1 + 0.999 877 706 159 493 637 734 4;
  • 46) 0.999 877 706 159 493 637 734 4 × 2 = 1 + 0.999 755 412 318 987 275 468 8;
  • 47) 0.999 755 412 318 987 275 468 8 × 2 = 1 + 0.999 510 824 637 974 550 937 6;
  • 48) 0.999 510 824 637 974 550 937 6 × 2 = 1 + 0.999 021 649 275 949 101 875 2;
  • 49) 0.999 021 649 275 949 101 875 2 × 2 = 1 + 0.998 043 298 551 898 203 750 4;
  • 50) 0.998 043 298 551 898 203 750 4 × 2 = 1 + 0.996 086 597 103 796 407 500 8;
  • 51) 0.996 086 597 103 796 407 500 8 × 2 = 1 + 0.992 173 194 207 592 815 001 6;
  • 52) 0.992 173 194 207 592 815 001 6 × 2 = 1 + 0.984 346 388 415 185 630 003 2;
  • 53) 0.984 346 388 415 185 630 003 2 × 2 = 1 + 0.968 692 776 830 371 260 006 4;
  • 54) 0.968 692 776 830 371 260 006 4 × 2 = 1 + 0.937 385 553 660 742 520 012 8;
  • 55) 0.937 385 553 660 742 520 012 8 × 2 = 1 + 0.874 771 107 321 485 040 025 6;
  • 56) 0.874 771 107 321 485 040 025 6 × 2 = 1 + 0.749 542 214 642 970 080 051 2;
  • 57) 0.749 542 214 642 970 080 051 2 × 2 = 1 + 0.499 084 429 285 940 160 102 4;
  • 58) 0.499 084 429 285 940 160 102 4 × 2 = 0 + 0.998 168 858 571 880 320 204 8;

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:


0.016 738 891 601 562 496 524 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 524 2(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:


0.016 738 891 601 562 496 524 2(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 -0.016 738 891 601 562 496 524 2 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