-0.016 738 891 601 562 496 554 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 554(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
-0.016 738 891 601 562 496 554(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:

|-0.016 738 891 601 562 496 554| = 0.016 738 891 601 562 496 554


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: 0.016 738 891 601 562 496 554.

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) 0.016 738 891 601 562 496 554 × 2 = 0 + 0.033 477 783 203 124 993 108;
  • 2) 0.033 477 783 203 124 993 108 × 2 = 0 + 0.066 955 566 406 249 986 216;
  • 3) 0.066 955 566 406 249 986 216 × 2 = 0 + 0.133 911 132 812 499 972 432;
  • 4) 0.133 911 132 812 499 972 432 × 2 = 0 + 0.267 822 265 624 999 944 864;
  • 5) 0.267 822 265 624 999 944 864 × 2 = 0 + 0.535 644 531 249 999 889 728;
  • 6) 0.535 644 531 249 999 889 728 × 2 = 1 + 0.071 289 062 499 999 779 456;
  • 7) 0.071 289 062 499 999 779 456 × 2 = 0 + 0.142 578 124 999 999 558 912;
  • 8) 0.142 578 124 999 999 558 912 × 2 = 0 + 0.285 156 249 999 999 117 824;
  • 9) 0.285 156 249 999 999 117 824 × 2 = 0 + 0.570 312 499 999 998 235 648;
  • 10) 0.570 312 499 999 998 235 648 × 2 = 1 + 0.140 624 999 999 996 471 296;
  • 11) 0.140 624 999 999 996 471 296 × 2 = 0 + 0.281 249 999 999 992 942 592;
  • 12) 0.281 249 999 999 992 942 592 × 2 = 0 + 0.562 499 999 999 985 885 184;
  • 13) 0.562 499 999 999 985 885 184 × 2 = 1 + 0.124 999 999 999 971 770 368;
  • 14) 0.124 999 999 999 971 770 368 × 2 = 0 + 0.249 999 999 999 943 540 736;
  • 15) 0.249 999 999 999 943 540 736 × 2 = 0 + 0.499 999 999 999 887 081 472;
  • 16) 0.499 999 999 999 887 081 472 × 2 = 0 + 0.999 999 999 999 774 162 944;
  • 17) 0.999 999 999 999 774 162 944 × 2 = 1 + 0.999 999 999 999 548 325 888;
  • 18) 0.999 999 999 999 548 325 888 × 2 = 1 + 0.999 999 999 999 096 651 776;
  • 19) 0.999 999 999 999 096 651 776 × 2 = 1 + 0.999 999 999 998 193 303 552;
  • 20) 0.999 999 999 998 193 303 552 × 2 = 1 + 0.999 999 999 996 386 607 104;
  • 21) 0.999 999 999 996 386 607 104 × 2 = 1 + 0.999 999 999 992 773 214 208;
  • 22) 0.999 999 999 992 773 214 208 × 2 = 1 + 0.999 999 999 985 546 428 416;
  • 23) 0.999 999 999 985 546 428 416 × 2 = 1 + 0.999 999 999 971 092 856 832;
  • 24) 0.999 999 999 971 092 856 832 × 2 = 1 + 0.999 999 999 942 185 713 664;
  • 25) 0.999 999 999 942 185 713 664 × 2 = 1 + 0.999 999 999 884 371 427 328;
  • 26) 0.999 999 999 884 371 427 328 × 2 = 1 + 0.999 999 999 768 742 854 656;
  • 27) 0.999 999 999 768 742 854 656 × 2 = 1 + 0.999 999 999 537 485 709 312;
  • 28) 0.999 999 999 537 485 709 312 × 2 = 1 + 0.999 999 999 074 971 418 624;
  • 29) 0.999 999 999 074 971 418 624 × 2 = 1 + 0.999 999 998 149 942 837 248;
  • 30) 0.999 999 998 149 942 837 248 × 2 = 1 + 0.999 999 996 299 885 674 496;
  • 31) 0.999 999 996 299 885 674 496 × 2 = 1 + 0.999 999 992 599 771 348 992;
  • 32) 0.999 999 992 599 771 348 992 × 2 = 1 + 0.999 999 985 199 542 697 984;
  • 33) 0.999 999 985 199 542 697 984 × 2 = 1 + 0.999 999 970 399 085 395 968;
  • 34) 0.999 999 970 399 085 395 968 × 2 = 1 + 0.999 999 940 798 170 791 936;
  • 35) 0.999 999 940 798 170 791 936 × 2 = 1 + 0.999 999 881 596 341 583 872;
  • 36) 0.999 999 881 596 341 583 872 × 2 = 1 + 0.999 999 763 192 683 167 744;
  • 37) 0.999 999 763 192 683 167 744 × 2 = 1 + 0.999 999 526 385 366 335 488;
  • 38) 0.999 999 526 385 366 335 488 × 2 = 1 + 0.999 999 052 770 732 670 976;
  • 39) 0.999 999 052 770 732 670 976 × 2 = 1 + 0.999 998 105 541 465 341 952;
  • 40) 0.999 998 105 541 465 341 952 × 2 = 1 + 0.999 996 211 082 930 683 904;
  • 41) 0.999 996 211 082 930 683 904 × 2 = 1 + 0.999 992 422 165 861 367 808;
  • 42) 0.999 992 422 165 861 367 808 × 2 = 1 + 0.999 984 844 331 722 735 616;
  • 43) 0.999 984 844 331 722 735 616 × 2 = 1 + 0.999 969 688 663 445 471 232;
  • 44) 0.999 969 688 663 445 471 232 × 2 = 1 + 0.999 939 377 326 890 942 464;
  • 45) 0.999 939 377 326 890 942 464 × 2 = 1 + 0.999 878 754 653 781 884 928;
  • 46) 0.999 878 754 653 781 884 928 × 2 = 1 + 0.999 757 509 307 563 769 856;
  • 47) 0.999 757 509 307 563 769 856 × 2 = 1 + 0.999 515 018 615 127 539 712;
  • 48) 0.999 515 018 615 127 539 712 × 2 = 1 + 0.999 030 037 230 255 079 424;
  • 49) 0.999 030 037 230 255 079 424 × 2 = 1 + 0.998 060 074 460 510 158 848;
  • 50) 0.998 060 074 460 510 158 848 × 2 = 1 + 0.996 120 148 921 020 317 696;
  • 51) 0.996 120 148 921 020 317 696 × 2 = 1 + 0.992 240 297 842 040 635 392;
  • 52) 0.992 240 297 842 040 635 392 × 2 = 1 + 0.984 480 595 684 081 270 784;
  • 53) 0.984 480 595 684 081 270 784 × 2 = 1 + 0.968 961 191 368 162 541 568;
  • 54) 0.968 961 191 368 162 541 568 × 2 = 1 + 0.937 922 382 736 325 083 136;
  • 55) 0.937 922 382 736 325 083 136 × 2 = 1 + 0.875 844 765 472 650 166 272;
  • 56) 0.875 844 765 472 650 166 272 × 2 = 1 + 0.751 689 530 945 300 332 544;
  • 57) 0.751 689 530 945 300 332 544 × 2 = 1 + 0.503 379 061 890 600 665 088;
  • 58) 0.503 379 061 890 600 665 088 × 2 = 1 + 0.006 758 123 781 201 330 176;

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:


0.016 738 891 601 562 496 554(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 554(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:


0.016 738 891 601 562 496 554(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 -0.016 738 891 601 562 496 554 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