-0.016 738 891 601 562 496 530 553 040 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 040 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
-0.016 738 891 601 562 496 530 553 040 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:

|-0.016 738 891 601 562 496 530 553 040 4| = 0.016 738 891 601 562 496 530 553 040 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: 0.016 738 891 601 562 496 530 553 040 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) 0.016 738 891 601 562 496 530 553 040 4 × 2 = 0 + 0.033 477 783 203 124 993 061 106 080 8;
  • 2) 0.033 477 783 203 124 993 061 106 080 8 × 2 = 0 + 0.066 955 566 406 249 986 122 212 161 6;
  • 3) 0.066 955 566 406 249 986 122 212 161 6 × 2 = 0 + 0.133 911 132 812 499 972 244 424 323 2;
  • 4) 0.133 911 132 812 499 972 244 424 323 2 × 2 = 0 + 0.267 822 265 624 999 944 488 848 646 4;
  • 5) 0.267 822 265 624 999 944 488 848 646 4 × 2 = 0 + 0.535 644 531 249 999 888 977 697 292 8;
  • 6) 0.535 644 531 249 999 888 977 697 292 8 × 2 = 1 + 0.071 289 062 499 999 777 955 394 585 6;
  • 7) 0.071 289 062 499 999 777 955 394 585 6 × 2 = 0 + 0.142 578 124 999 999 555 910 789 171 2;
  • 8) 0.142 578 124 999 999 555 910 789 171 2 × 2 = 0 + 0.285 156 249 999 999 111 821 578 342 4;
  • 9) 0.285 156 249 999 999 111 821 578 342 4 × 2 = 0 + 0.570 312 499 999 998 223 643 156 684 8;
  • 10) 0.570 312 499 999 998 223 643 156 684 8 × 2 = 1 + 0.140 624 999 999 996 447 286 313 369 6;
  • 11) 0.140 624 999 999 996 447 286 313 369 6 × 2 = 0 + 0.281 249 999 999 992 894 572 626 739 2;
  • 12) 0.281 249 999 999 992 894 572 626 739 2 × 2 = 0 + 0.562 499 999 999 985 789 145 253 478 4;
  • 13) 0.562 499 999 999 985 789 145 253 478 4 × 2 = 1 + 0.124 999 999 999 971 578 290 506 956 8;
  • 14) 0.124 999 999 999 971 578 290 506 956 8 × 2 = 0 + 0.249 999 999 999 943 156 581 013 913 6;
  • 15) 0.249 999 999 999 943 156 581 013 913 6 × 2 = 0 + 0.499 999 999 999 886 313 162 027 827 2;
  • 16) 0.499 999 999 999 886 313 162 027 827 2 × 2 = 0 + 0.999 999 999 999 772 626 324 055 654 4;
  • 17) 0.999 999 999 999 772 626 324 055 654 4 × 2 = 1 + 0.999 999 999 999 545 252 648 111 308 8;
  • 18) 0.999 999 999 999 545 252 648 111 308 8 × 2 = 1 + 0.999 999 999 999 090 505 296 222 617 6;
  • 19) 0.999 999 999 999 090 505 296 222 617 6 × 2 = 1 + 0.999 999 999 998 181 010 592 445 235 2;
  • 20) 0.999 999 999 998 181 010 592 445 235 2 × 2 = 1 + 0.999 999 999 996 362 021 184 890 470 4;
  • 21) 0.999 999 999 996 362 021 184 890 470 4 × 2 = 1 + 0.999 999 999 992 724 042 369 780 940 8;
  • 22) 0.999 999 999 992 724 042 369 780 940 8 × 2 = 1 + 0.999 999 999 985 448 084 739 561 881 6;
  • 23) 0.999 999 999 985 448 084 739 561 881 6 × 2 = 1 + 0.999 999 999 970 896 169 479 123 763 2;
  • 24) 0.999 999 999 970 896 169 479 123 763 2 × 2 = 1 + 0.999 999 999 941 792 338 958 247 526 4;
  • 25) 0.999 999 999 941 792 338 958 247 526 4 × 2 = 1 + 0.999 999 999 883 584 677 916 495 052 8;
  • 26) 0.999 999 999 883 584 677 916 495 052 8 × 2 = 1 + 0.999 999 999 767 169 355 832 990 105 6;
  • 27) 0.999 999 999 767 169 355 832 990 105 6 × 2 = 1 + 0.999 999 999 534 338 711 665 980 211 2;
  • 28) 0.999 999 999 534 338 711 665 980 211 2 × 2 = 1 + 0.999 999 999 068 677 423 331 960 422 4;
  • 29) 0.999 999 999 068 677 423 331 960 422 4 × 2 = 1 + 0.999 999 998 137 354 846 663 920 844 8;
  • 30) 0.999 999 998 137 354 846 663 920 844 8 × 2 = 1 + 0.999 999 996 274 709 693 327 841 689 6;
  • 31) 0.999 999 996 274 709 693 327 841 689 6 × 2 = 1 + 0.999 999 992 549 419 386 655 683 379 2;
  • 32) 0.999 999 992 549 419 386 655 683 379 2 × 2 = 1 + 0.999 999 985 098 838 773 311 366 758 4;
  • 33) 0.999 999 985 098 838 773 311 366 758 4 × 2 = 1 + 0.999 999 970 197 677 546 622 733 516 8;
  • 34) 0.999 999 970 197 677 546 622 733 516 8 × 2 = 1 + 0.999 999 940 395 355 093 245 467 033 6;
  • 35) 0.999 999 940 395 355 093 245 467 033 6 × 2 = 1 + 0.999 999 880 790 710 186 490 934 067 2;
  • 36) 0.999 999 880 790 710 186 490 934 067 2 × 2 = 1 + 0.999 999 761 581 420 372 981 868 134 4;
  • 37) 0.999 999 761 581 420 372 981 868 134 4 × 2 = 1 + 0.999 999 523 162 840 745 963 736 268 8;
  • 38) 0.999 999 523 162 840 745 963 736 268 8 × 2 = 1 + 0.999 999 046 325 681 491 927 472 537 6;
  • 39) 0.999 999 046 325 681 491 927 472 537 6 × 2 = 1 + 0.999 998 092 651 362 983 854 945 075 2;
  • 40) 0.999 998 092 651 362 983 854 945 075 2 × 2 = 1 + 0.999 996 185 302 725 967 709 890 150 4;
  • 41) 0.999 996 185 302 725 967 709 890 150 4 × 2 = 1 + 0.999 992 370 605 451 935 419 780 300 8;
  • 42) 0.999 992 370 605 451 935 419 780 300 8 × 2 = 1 + 0.999 984 741 210 903 870 839 560 601 6;
  • 43) 0.999 984 741 210 903 870 839 560 601 6 × 2 = 1 + 0.999 969 482 421 807 741 679 121 203 2;
  • 44) 0.999 969 482 421 807 741 679 121 203 2 × 2 = 1 + 0.999 938 964 843 615 483 358 242 406 4;
  • 45) 0.999 938 964 843 615 483 358 242 406 4 × 2 = 1 + 0.999 877 929 687 230 966 716 484 812 8;
  • 46) 0.999 877 929 687 230 966 716 484 812 8 × 2 = 1 + 0.999 755 859 374 461 933 432 969 625 6;
  • 47) 0.999 755 859 374 461 933 432 969 625 6 × 2 = 1 + 0.999 511 718 748 923 866 865 939 251 2;
  • 48) 0.999 511 718 748 923 866 865 939 251 2 × 2 = 1 + 0.999 023 437 497 847 733 731 878 502 4;
  • 49) 0.999 023 437 497 847 733 731 878 502 4 × 2 = 1 + 0.998 046 874 995 695 467 463 757 004 8;
  • 50) 0.998 046 874 995 695 467 463 757 004 8 × 2 = 1 + 0.996 093 749 991 390 934 927 514 009 6;
  • 51) 0.996 093 749 991 390 934 927 514 009 6 × 2 = 1 + 0.992 187 499 982 781 869 855 028 019 2;
  • 52) 0.992 187 499 982 781 869 855 028 019 2 × 2 = 1 + 0.984 374 999 965 563 739 710 056 038 4;
  • 53) 0.984 374 999 965 563 739 710 056 038 4 × 2 = 1 + 0.968 749 999 931 127 479 420 112 076 8;
  • 54) 0.968 749 999 931 127 479 420 112 076 8 × 2 = 1 + 0.937 499 999 862 254 958 840 224 153 6;
  • 55) 0.937 499 999 862 254 958 840 224 153 6 × 2 = 1 + 0.874 999 999 724 509 917 680 448 307 2;
  • 56) 0.874 999 999 724 509 917 680 448 307 2 × 2 = 1 + 0.749 999 999 449 019 835 360 896 614 4;
  • 57) 0.749 999 999 449 019 835 360 896 614 4 × 2 = 1 + 0.499 999 998 898 039 670 721 793 228 8;
  • 58) 0.499 999 998 898 039 670 721 793 228 8 × 2 = 0 + 0.999 999 997 796 079 341 443 586 457 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:


0.016 738 891 601 562 496 530 553 040 4(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 040 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:


0.016 738 891 601 562 496 530 553 040 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 -0.016 738 891 601 562 496 530 553 040 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