-0.016 738 891 601 562 496 517 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 517 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
-0.016 738 891 601 562 496 517 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:

|-0.016 738 891 601 562 496 517 4| = 0.016 738 891 601 562 496 517 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: 0.016 738 891 601 562 496 517 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) 0.016 738 891 601 562 496 517 4 × 2 = 0 + 0.033 477 783 203 124 993 034 8;
  • 2) 0.033 477 783 203 124 993 034 8 × 2 = 0 + 0.066 955 566 406 249 986 069 6;
  • 3) 0.066 955 566 406 249 986 069 6 × 2 = 0 + 0.133 911 132 812 499 972 139 2;
  • 4) 0.133 911 132 812 499 972 139 2 × 2 = 0 + 0.267 822 265 624 999 944 278 4;
  • 5) 0.267 822 265 624 999 944 278 4 × 2 = 0 + 0.535 644 531 249 999 888 556 8;
  • 6) 0.535 644 531 249 999 888 556 8 × 2 = 1 + 0.071 289 062 499 999 777 113 6;
  • 7) 0.071 289 062 499 999 777 113 6 × 2 = 0 + 0.142 578 124 999 999 554 227 2;
  • 8) 0.142 578 124 999 999 554 227 2 × 2 = 0 + 0.285 156 249 999 999 108 454 4;
  • 9) 0.285 156 249 999 999 108 454 4 × 2 = 0 + 0.570 312 499 999 998 216 908 8;
  • 10) 0.570 312 499 999 998 216 908 8 × 2 = 1 + 0.140 624 999 999 996 433 817 6;
  • 11) 0.140 624 999 999 996 433 817 6 × 2 = 0 + 0.281 249 999 999 992 867 635 2;
  • 12) 0.281 249 999 999 992 867 635 2 × 2 = 0 + 0.562 499 999 999 985 735 270 4;
  • 13) 0.562 499 999 999 985 735 270 4 × 2 = 1 + 0.124 999 999 999 971 470 540 8;
  • 14) 0.124 999 999 999 971 470 540 8 × 2 = 0 + 0.249 999 999 999 942 941 081 6;
  • 15) 0.249 999 999 999 942 941 081 6 × 2 = 0 + 0.499 999 999 999 885 882 163 2;
  • 16) 0.499 999 999 999 885 882 163 2 × 2 = 0 + 0.999 999 999 999 771 764 326 4;
  • 17) 0.999 999 999 999 771 764 326 4 × 2 = 1 + 0.999 999 999 999 543 528 652 8;
  • 18) 0.999 999 999 999 543 528 652 8 × 2 = 1 + 0.999 999 999 999 087 057 305 6;
  • 19) 0.999 999 999 999 087 057 305 6 × 2 = 1 + 0.999 999 999 998 174 114 611 2;
  • 20) 0.999 999 999 998 174 114 611 2 × 2 = 1 + 0.999 999 999 996 348 229 222 4;
  • 21) 0.999 999 999 996 348 229 222 4 × 2 = 1 + 0.999 999 999 992 696 458 444 8;
  • 22) 0.999 999 999 992 696 458 444 8 × 2 = 1 + 0.999 999 999 985 392 916 889 6;
  • 23) 0.999 999 999 985 392 916 889 6 × 2 = 1 + 0.999 999 999 970 785 833 779 2;
  • 24) 0.999 999 999 970 785 833 779 2 × 2 = 1 + 0.999 999 999 941 571 667 558 4;
  • 25) 0.999 999 999 941 571 667 558 4 × 2 = 1 + 0.999 999 999 883 143 335 116 8;
  • 26) 0.999 999 999 883 143 335 116 8 × 2 = 1 + 0.999 999 999 766 286 670 233 6;
  • 27) 0.999 999 999 766 286 670 233 6 × 2 = 1 + 0.999 999 999 532 573 340 467 2;
  • 28) 0.999 999 999 532 573 340 467 2 × 2 = 1 + 0.999 999 999 065 146 680 934 4;
  • 29) 0.999 999 999 065 146 680 934 4 × 2 = 1 + 0.999 999 998 130 293 361 868 8;
  • 30) 0.999 999 998 130 293 361 868 8 × 2 = 1 + 0.999 999 996 260 586 723 737 6;
  • 31) 0.999 999 996 260 586 723 737 6 × 2 = 1 + 0.999 999 992 521 173 447 475 2;
  • 32) 0.999 999 992 521 173 447 475 2 × 2 = 1 + 0.999 999 985 042 346 894 950 4;
  • 33) 0.999 999 985 042 346 894 950 4 × 2 = 1 + 0.999 999 970 084 693 789 900 8;
  • 34) 0.999 999 970 084 693 789 900 8 × 2 = 1 + 0.999 999 940 169 387 579 801 6;
  • 35) 0.999 999 940 169 387 579 801 6 × 2 = 1 + 0.999 999 880 338 775 159 603 2;
  • 36) 0.999 999 880 338 775 159 603 2 × 2 = 1 + 0.999 999 760 677 550 319 206 4;
  • 37) 0.999 999 760 677 550 319 206 4 × 2 = 1 + 0.999 999 521 355 100 638 412 8;
  • 38) 0.999 999 521 355 100 638 412 8 × 2 = 1 + 0.999 999 042 710 201 276 825 6;
  • 39) 0.999 999 042 710 201 276 825 6 × 2 = 1 + 0.999 998 085 420 402 553 651 2;
  • 40) 0.999 998 085 420 402 553 651 2 × 2 = 1 + 0.999 996 170 840 805 107 302 4;
  • 41) 0.999 996 170 840 805 107 302 4 × 2 = 1 + 0.999 992 341 681 610 214 604 8;
  • 42) 0.999 992 341 681 610 214 604 8 × 2 = 1 + 0.999 984 683 363 220 429 209 6;
  • 43) 0.999 984 683 363 220 429 209 6 × 2 = 1 + 0.999 969 366 726 440 858 419 2;
  • 44) 0.999 969 366 726 440 858 419 2 × 2 = 1 + 0.999 938 733 452 881 716 838 4;
  • 45) 0.999 938 733 452 881 716 838 4 × 2 = 1 + 0.999 877 466 905 763 433 676 8;
  • 46) 0.999 877 466 905 763 433 676 8 × 2 = 1 + 0.999 754 933 811 526 867 353 6;
  • 47) 0.999 754 933 811 526 867 353 6 × 2 = 1 + 0.999 509 867 623 053 734 707 2;
  • 48) 0.999 509 867 623 053 734 707 2 × 2 = 1 + 0.999 019 735 246 107 469 414 4;
  • 49) 0.999 019 735 246 107 469 414 4 × 2 = 1 + 0.998 039 470 492 214 938 828 8;
  • 50) 0.998 039 470 492 214 938 828 8 × 2 = 1 + 0.996 078 940 984 429 877 657 6;
  • 51) 0.996 078 940 984 429 877 657 6 × 2 = 1 + 0.992 157 881 968 859 755 315 2;
  • 52) 0.992 157 881 968 859 755 315 2 × 2 = 1 + 0.984 315 763 937 719 510 630 4;
  • 53) 0.984 315 763 937 719 510 630 4 × 2 = 1 + 0.968 631 527 875 439 021 260 8;
  • 54) 0.968 631 527 875 439 021 260 8 × 2 = 1 + 0.937 263 055 750 878 042 521 6;
  • 55) 0.937 263 055 750 878 042 521 6 × 2 = 1 + 0.874 526 111 501 756 085 043 2;
  • 56) 0.874 526 111 501 756 085 043 2 × 2 = 1 + 0.749 052 223 003 512 170 086 4;
  • 57) 0.749 052 223 003 512 170 086 4 × 2 = 1 + 0.498 104 446 007 024 340 172 8;
  • 58) 0.498 104 446 007 024 340 172 8 × 2 = 0 + 0.996 208 892 014 048 680 345 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:


0.016 738 891 601 562 496 517 4(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 517 4(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:


0.016 738 891 601 562 496 517 4(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 -0.016 738 891 601 562 496 517 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 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