-0.016 738 891 601 562 496 530 676 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 676(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
-0.016 738 891 601 562 496 530 676(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:

|-0.016 738 891 601 562 496 530 676| = 0.016 738 891 601 562 496 530 676


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: 0.016 738 891 601 562 496 530 676.

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) 0.016 738 891 601 562 496 530 676 × 2 = 0 + 0.033 477 783 203 124 993 061 352;
  • 2) 0.033 477 783 203 124 993 061 352 × 2 = 0 + 0.066 955 566 406 249 986 122 704;
  • 3) 0.066 955 566 406 249 986 122 704 × 2 = 0 + 0.133 911 132 812 499 972 245 408;
  • 4) 0.133 911 132 812 499 972 245 408 × 2 = 0 + 0.267 822 265 624 999 944 490 816;
  • 5) 0.267 822 265 624 999 944 490 816 × 2 = 0 + 0.535 644 531 249 999 888 981 632;
  • 6) 0.535 644 531 249 999 888 981 632 × 2 = 1 + 0.071 289 062 499 999 777 963 264;
  • 7) 0.071 289 062 499 999 777 963 264 × 2 = 0 + 0.142 578 124 999 999 555 926 528;
  • 8) 0.142 578 124 999 999 555 926 528 × 2 = 0 + 0.285 156 249 999 999 111 853 056;
  • 9) 0.285 156 249 999 999 111 853 056 × 2 = 0 + 0.570 312 499 999 998 223 706 112;
  • 10) 0.570 312 499 999 998 223 706 112 × 2 = 1 + 0.140 624 999 999 996 447 412 224;
  • 11) 0.140 624 999 999 996 447 412 224 × 2 = 0 + 0.281 249 999 999 992 894 824 448;
  • 12) 0.281 249 999 999 992 894 824 448 × 2 = 0 + 0.562 499 999 999 985 789 648 896;
  • 13) 0.562 499 999 999 985 789 648 896 × 2 = 1 + 0.124 999 999 999 971 579 297 792;
  • 14) 0.124 999 999 999 971 579 297 792 × 2 = 0 + 0.249 999 999 999 943 158 595 584;
  • 15) 0.249 999 999 999 943 158 595 584 × 2 = 0 + 0.499 999 999 999 886 317 191 168;
  • 16) 0.499 999 999 999 886 317 191 168 × 2 = 0 + 0.999 999 999 999 772 634 382 336;
  • 17) 0.999 999 999 999 772 634 382 336 × 2 = 1 + 0.999 999 999 999 545 268 764 672;
  • 18) 0.999 999 999 999 545 268 764 672 × 2 = 1 + 0.999 999 999 999 090 537 529 344;
  • 19) 0.999 999 999 999 090 537 529 344 × 2 = 1 + 0.999 999 999 998 181 075 058 688;
  • 20) 0.999 999 999 998 181 075 058 688 × 2 = 1 + 0.999 999 999 996 362 150 117 376;
  • 21) 0.999 999 999 996 362 150 117 376 × 2 = 1 + 0.999 999 999 992 724 300 234 752;
  • 22) 0.999 999 999 992 724 300 234 752 × 2 = 1 + 0.999 999 999 985 448 600 469 504;
  • 23) 0.999 999 999 985 448 600 469 504 × 2 = 1 + 0.999 999 999 970 897 200 939 008;
  • 24) 0.999 999 999 970 897 200 939 008 × 2 = 1 + 0.999 999 999 941 794 401 878 016;
  • 25) 0.999 999 999 941 794 401 878 016 × 2 = 1 + 0.999 999 999 883 588 803 756 032;
  • 26) 0.999 999 999 883 588 803 756 032 × 2 = 1 + 0.999 999 999 767 177 607 512 064;
  • 27) 0.999 999 999 767 177 607 512 064 × 2 = 1 + 0.999 999 999 534 355 215 024 128;
  • 28) 0.999 999 999 534 355 215 024 128 × 2 = 1 + 0.999 999 999 068 710 430 048 256;
  • 29) 0.999 999 999 068 710 430 048 256 × 2 = 1 + 0.999 999 998 137 420 860 096 512;
  • 30) 0.999 999 998 137 420 860 096 512 × 2 = 1 + 0.999 999 996 274 841 720 193 024;
  • 31) 0.999 999 996 274 841 720 193 024 × 2 = 1 + 0.999 999 992 549 683 440 386 048;
  • 32) 0.999 999 992 549 683 440 386 048 × 2 = 1 + 0.999 999 985 099 366 880 772 096;
  • 33) 0.999 999 985 099 366 880 772 096 × 2 = 1 + 0.999 999 970 198 733 761 544 192;
  • 34) 0.999 999 970 198 733 761 544 192 × 2 = 1 + 0.999 999 940 397 467 523 088 384;
  • 35) 0.999 999 940 397 467 523 088 384 × 2 = 1 + 0.999 999 880 794 935 046 176 768;
  • 36) 0.999 999 880 794 935 046 176 768 × 2 = 1 + 0.999 999 761 589 870 092 353 536;
  • 37) 0.999 999 761 589 870 092 353 536 × 2 = 1 + 0.999 999 523 179 740 184 707 072;
  • 38) 0.999 999 523 179 740 184 707 072 × 2 = 1 + 0.999 999 046 359 480 369 414 144;
  • 39) 0.999 999 046 359 480 369 414 144 × 2 = 1 + 0.999 998 092 718 960 738 828 288;
  • 40) 0.999 998 092 718 960 738 828 288 × 2 = 1 + 0.999 996 185 437 921 477 656 576;
  • 41) 0.999 996 185 437 921 477 656 576 × 2 = 1 + 0.999 992 370 875 842 955 313 152;
  • 42) 0.999 992 370 875 842 955 313 152 × 2 = 1 + 0.999 984 741 751 685 910 626 304;
  • 43) 0.999 984 741 751 685 910 626 304 × 2 = 1 + 0.999 969 483 503 371 821 252 608;
  • 44) 0.999 969 483 503 371 821 252 608 × 2 = 1 + 0.999 938 967 006 743 642 505 216;
  • 45) 0.999 938 967 006 743 642 505 216 × 2 = 1 + 0.999 877 934 013 487 285 010 432;
  • 46) 0.999 877 934 013 487 285 010 432 × 2 = 1 + 0.999 755 868 026 974 570 020 864;
  • 47) 0.999 755 868 026 974 570 020 864 × 2 = 1 + 0.999 511 736 053 949 140 041 728;
  • 48) 0.999 511 736 053 949 140 041 728 × 2 = 1 + 0.999 023 472 107 898 280 083 456;
  • 49) 0.999 023 472 107 898 280 083 456 × 2 = 1 + 0.998 046 944 215 796 560 166 912;
  • 50) 0.998 046 944 215 796 560 166 912 × 2 = 1 + 0.996 093 888 431 593 120 333 824;
  • 51) 0.996 093 888 431 593 120 333 824 × 2 = 1 + 0.992 187 776 863 186 240 667 648;
  • 52) 0.992 187 776 863 186 240 667 648 × 2 = 1 + 0.984 375 553 726 372 481 335 296;
  • 53) 0.984 375 553 726 372 481 335 296 × 2 = 1 + 0.968 751 107 452 744 962 670 592;
  • 54) 0.968 751 107 452 744 962 670 592 × 2 = 1 + 0.937 502 214 905 489 925 341 184;
  • 55) 0.937 502 214 905 489 925 341 184 × 2 = 1 + 0.875 004 429 810 979 850 682 368;
  • 56) 0.875 004 429 810 979 850 682 368 × 2 = 1 + 0.750 008 859 621 959 701 364 736;
  • 57) 0.750 008 859 621 959 701 364 736 × 2 = 1 + 0.500 017 719 243 919 402 729 472;
  • 58) 0.500 017 719 243 919 402 729 472 × 2 = 1 + 0.000 035 438 487 838 805 458 944;

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:


0.016 738 891 601 562 496 530 676(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 676(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:


0.016 738 891 601 562 496 530 676(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 -0.016 738 891 601 562 496 530 676 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