-0.016 738 891 601 562 496 530 561 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 561 9(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
-0.016 738 891 601 562 496 530 561 9(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:

|-0.016 738 891 601 562 496 530 561 9| = 0.016 738 891 601 562 496 530 561 9


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: 0.016 738 891 601 562 496 530 561 9.

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) 0.016 738 891 601 562 496 530 561 9 × 2 = 0 + 0.033 477 783 203 124 993 061 123 8;
  • 2) 0.033 477 783 203 124 993 061 123 8 × 2 = 0 + 0.066 955 566 406 249 986 122 247 6;
  • 3) 0.066 955 566 406 249 986 122 247 6 × 2 = 0 + 0.133 911 132 812 499 972 244 495 2;
  • 4) 0.133 911 132 812 499 972 244 495 2 × 2 = 0 + 0.267 822 265 624 999 944 488 990 4;
  • 5) 0.267 822 265 624 999 944 488 990 4 × 2 = 0 + 0.535 644 531 249 999 888 977 980 8;
  • 6) 0.535 644 531 249 999 888 977 980 8 × 2 = 1 + 0.071 289 062 499 999 777 955 961 6;
  • 7) 0.071 289 062 499 999 777 955 961 6 × 2 = 0 + 0.142 578 124 999 999 555 911 923 2;
  • 8) 0.142 578 124 999 999 555 911 923 2 × 2 = 0 + 0.285 156 249 999 999 111 823 846 4;
  • 9) 0.285 156 249 999 999 111 823 846 4 × 2 = 0 + 0.570 312 499 999 998 223 647 692 8;
  • 10) 0.570 312 499 999 998 223 647 692 8 × 2 = 1 + 0.140 624 999 999 996 447 295 385 6;
  • 11) 0.140 624 999 999 996 447 295 385 6 × 2 = 0 + 0.281 249 999 999 992 894 590 771 2;
  • 12) 0.281 249 999 999 992 894 590 771 2 × 2 = 0 + 0.562 499 999 999 985 789 181 542 4;
  • 13) 0.562 499 999 999 985 789 181 542 4 × 2 = 1 + 0.124 999 999 999 971 578 363 084 8;
  • 14) 0.124 999 999 999 971 578 363 084 8 × 2 = 0 + 0.249 999 999 999 943 156 726 169 6;
  • 15) 0.249 999 999 999 943 156 726 169 6 × 2 = 0 + 0.499 999 999 999 886 313 452 339 2;
  • 16) 0.499 999 999 999 886 313 452 339 2 × 2 = 0 + 0.999 999 999 999 772 626 904 678 4;
  • 17) 0.999 999 999 999 772 626 904 678 4 × 2 = 1 + 0.999 999 999 999 545 253 809 356 8;
  • 18) 0.999 999 999 999 545 253 809 356 8 × 2 = 1 + 0.999 999 999 999 090 507 618 713 6;
  • 19) 0.999 999 999 999 090 507 618 713 6 × 2 = 1 + 0.999 999 999 998 181 015 237 427 2;
  • 20) 0.999 999 999 998 181 015 237 427 2 × 2 = 1 + 0.999 999 999 996 362 030 474 854 4;
  • 21) 0.999 999 999 996 362 030 474 854 4 × 2 = 1 + 0.999 999 999 992 724 060 949 708 8;
  • 22) 0.999 999 999 992 724 060 949 708 8 × 2 = 1 + 0.999 999 999 985 448 121 899 417 6;
  • 23) 0.999 999 999 985 448 121 899 417 6 × 2 = 1 + 0.999 999 999 970 896 243 798 835 2;
  • 24) 0.999 999 999 970 896 243 798 835 2 × 2 = 1 + 0.999 999 999 941 792 487 597 670 4;
  • 25) 0.999 999 999 941 792 487 597 670 4 × 2 = 1 + 0.999 999 999 883 584 975 195 340 8;
  • 26) 0.999 999 999 883 584 975 195 340 8 × 2 = 1 + 0.999 999 999 767 169 950 390 681 6;
  • 27) 0.999 999 999 767 169 950 390 681 6 × 2 = 1 + 0.999 999 999 534 339 900 781 363 2;
  • 28) 0.999 999 999 534 339 900 781 363 2 × 2 = 1 + 0.999 999 999 068 679 801 562 726 4;
  • 29) 0.999 999 999 068 679 801 562 726 4 × 2 = 1 + 0.999 999 998 137 359 603 125 452 8;
  • 30) 0.999 999 998 137 359 603 125 452 8 × 2 = 1 + 0.999 999 996 274 719 206 250 905 6;
  • 31) 0.999 999 996 274 719 206 250 905 6 × 2 = 1 + 0.999 999 992 549 438 412 501 811 2;
  • 32) 0.999 999 992 549 438 412 501 811 2 × 2 = 1 + 0.999 999 985 098 876 825 003 622 4;
  • 33) 0.999 999 985 098 876 825 003 622 4 × 2 = 1 + 0.999 999 970 197 753 650 007 244 8;
  • 34) 0.999 999 970 197 753 650 007 244 8 × 2 = 1 + 0.999 999 940 395 507 300 014 489 6;
  • 35) 0.999 999 940 395 507 300 014 489 6 × 2 = 1 + 0.999 999 880 791 014 600 028 979 2;
  • 36) 0.999 999 880 791 014 600 028 979 2 × 2 = 1 + 0.999 999 761 582 029 200 057 958 4;
  • 37) 0.999 999 761 582 029 200 057 958 4 × 2 = 1 + 0.999 999 523 164 058 400 115 916 8;
  • 38) 0.999 999 523 164 058 400 115 916 8 × 2 = 1 + 0.999 999 046 328 116 800 231 833 6;
  • 39) 0.999 999 046 328 116 800 231 833 6 × 2 = 1 + 0.999 998 092 656 233 600 463 667 2;
  • 40) 0.999 998 092 656 233 600 463 667 2 × 2 = 1 + 0.999 996 185 312 467 200 927 334 4;
  • 41) 0.999 996 185 312 467 200 927 334 4 × 2 = 1 + 0.999 992 370 624 934 401 854 668 8;
  • 42) 0.999 992 370 624 934 401 854 668 8 × 2 = 1 + 0.999 984 741 249 868 803 709 337 6;
  • 43) 0.999 984 741 249 868 803 709 337 6 × 2 = 1 + 0.999 969 482 499 737 607 418 675 2;
  • 44) 0.999 969 482 499 737 607 418 675 2 × 2 = 1 + 0.999 938 964 999 475 214 837 350 4;
  • 45) 0.999 938 964 999 475 214 837 350 4 × 2 = 1 + 0.999 877 929 998 950 429 674 700 8;
  • 46) 0.999 877 929 998 950 429 674 700 8 × 2 = 1 + 0.999 755 859 997 900 859 349 401 6;
  • 47) 0.999 755 859 997 900 859 349 401 6 × 2 = 1 + 0.999 511 719 995 801 718 698 803 2;
  • 48) 0.999 511 719 995 801 718 698 803 2 × 2 = 1 + 0.999 023 439 991 603 437 397 606 4;
  • 49) 0.999 023 439 991 603 437 397 606 4 × 2 = 1 + 0.998 046 879 983 206 874 795 212 8;
  • 50) 0.998 046 879 983 206 874 795 212 8 × 2 = 1 + 0.996 093 759 966 413 749 590 425 6;
  • 51) 0.996 093 759 966 413 749 590 425 6 × 2 = 1 + 0.992 187 519 932 827 499 180 851 2;
  • 52) 0.992 187 519 932 827 499 180 851 2 × 2 = 1 + 0.984 375 039 865 654 998 361 702 4;
  • 53) 0.984 375 039 865 654 998 361 702 4 × 2 = 1 + 0.968 750 079 731 309 996 723 404 8;
  • 54) 0.968 750 079 731 309 996 723 404 8 × 2 = 1 + 0.937 500 159 462 619 993 446 809 6;
  • 55) 0.937 500 159 462 619 993 446 809 6 × 2 = 1 + 0.875 000 318 925 239 986 893 619 2;
  • 56) 0.875 000 318 925 239 986 893 619 2 × 2 = 1 + 0.750 000 637 850 479 973 787 238 4;
  • 57) 0.750 000 637 850 479 973 787 238 4 × 2 = 1 + 0.500 001 275 700 959 947 574 476 8;
  • 58) 0.500 001 275 700 959 947 574 476 8 × 2 = 1 + 0.000 002 551 401 919 895 148 953 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:


0.016 738 891 601 562 496 530 561 9(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 561 9(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:


0.016 738 891 601 562 496 530 561 9(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 -0.016 738 891 601 562 496 530 561 9 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