-0.016 738 891 601 562 496 530 527 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 527 1(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
-0.016 738 891 601 562 496 530 527 1(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:

|-0.016 738 891 601 562 496 530 527 1| = 0.016 738 891 601 562 496 530 527 1


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: 0.016 738 891 601 562 496 530 527 1.

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) 0.016 738 891 601 562 496 530 527 1 × 2 = 0 + 0.033 477 783 203 124 993 061 054 2;
  • 2) 0.033 477 783 203 124 993 061 054 2 × 2 = 0 + 0.066 955 566 406 249 986 122 108 4;
  • 3) 0.066 955 566 406 249 986 122 108 4 × 2 = 0 + 0.133 911 132 812 499 972 244 216 8;
  • 4) 0.133 911 132 812 499 972 244 216 8 × 2 = 0 + 0.267 822 265 624 999 944 488 433 6;
  • 5) 0.267 822 265 624 999 944 488 433 6 × 2 = 0 + 0.535 644 531 249 999 888 976 867 2;
  • 6) 0.535 644 531 249 999 888 976 867 2 × 2 = 1 + 0.071 289 062 499 999 777 953 734 4;
  • 7) 0.071 289 062 499 999 777 953 734 4 × 2 = 0 + 0.142 578 124 999 999 555 907 468 8;
  • 8) 0.142 578 124 999 999 555 907 468 8 × 2 = 0 + 0.285 156 249 999 999 111 814 937 6;
  • 9) 0.285 156 249 999 999 111 814 937 6 × 2 = 0 + 0.570 312 499 999 998 223 629 875 2;
  • 10) 0.570 312 499 999 998 223 629 875 2 × 2 = 1 + 0.140 624 999 999 996 447 259 750 4;
  • 11) 0.140 624 999 999 996 447 259 750 4 × 2 = 0 + 0.281 249 999 999 992 894 519 500 8;
  • 12) 0.281 249 999 999 992 894 519 500 8 × 2 = 0 + 0.562 499 999 999 985 789 039 001 6;
  • 13) 0.562 499 999 999 985 789 039 001 6 × 2 = 1 + 0.124 999 999 999 971 578 078 003 2;
  • 14) 0.124 999 999 999 971 578 078 003 2 × 2 = 0 + 0.249 999 999 999 943 156 156 006 4;
  • 15) 0.249 999 999 999 943 156 156 006 4 × 2 = 0 + 0.499 999 999 999 886 312 312 012 8;
  • 16) 0.499 999 999 999 886 312 312 012 8 × 2 = 0 + 0.999 999 999 999 772 624 624 025 6;
  • 17) 0.999 999 999 999 772 624 624 025 6 × 2 = 1 + 0.999 999 999 999 545 249 248 051 2;
  • 18) 0.999 999 999 999 545 249 248 051 2 × 2 = 1 + 0.999 999 999 999 090 498 496 102 4;
  • 19) 0.999 999 999 999 090 498 496 102 4 × 2 = 1 + 0.999 999 999 998 180 996 992 204 8;
  • 20) 0.999 999 999 998 180 996 992 204 8 × 2 = 1 + 0.999 999 999 996 361 993 984 409 6;
  • 21) 0.999 999 999 996 361 993 984 409 6 × 2 = 1 + 0.999 999 999 992 723 987 968 819 2;
  • 22) 0.999 999 999 992 723 987 968 819 2 × 2 = 1 + 0.999 999 999 985 447 975 937 638 4;
  • 23) 0.999 999 999 985 447 975 937 638 4 × 2 = 1 + 0.999 999 999 970 895 951 875 276 8;
  • 24) 0.999 999 999 970 895 951 875 276 8 × 2 = 1 + 0.999 999 999 941 791 903 750 553 6;
  • 25) 0.999 999 999 941 791 903 750 553 6 × 2 = 1 + 0.999 999 999 883 583 807 501 107 2;
  • 26) 0.999 999 999 883 583 807 501 107 2 × 2 = 1 + 0.999 999 999 767 167 615 002 214 4;
  • 27) 0.999 999 999 767 167 615 002 214 4 × 2 = 1 + 0.999 999 999 534 335 230 004 428 8;
  • 28) 0.999 999 999 534 335 230 004 428 8 × 2 = 1 + 0.999 999 999 068 670 460 008 857 6;
  • 29) 0.999 999 999 068 670 460 008 857 6 × 2 = 1 + 0.999 999 998 137 340 920 017 715 2;
  • 30) 0.999 999 998 137 340 920 017 715 2 × 2 = 1 + 0.999 999 996 274 681 840 035 430 4;
  • 31) 0.999 999 996 274 681 840 035 430 4 × 2 = 1 + 0.999 999 992 549 363 680 070 860 8;
  • 32) 0.999 999 992 549 363 680 070 860 8 × 2 = 1 + 0.999 999 985 098 727 360 141 721 6;
  • 33) 0.999 999 985 098 727 360 141 721 6 × 2 = 1 + 0.999 999 970 197 454 720 283 443 2;
  • 34) 0.999 999 970 197 454 720 283 443 2 × 2 = 1 + 0.999 999 940 394 909 440 566 886 4;
  • 35) 0.999 999 940 394 909 440 566 886 4 × 2 = 1 + 0.999 999 880 789 818 881 133 772 8;
  • 36) 0.999 999 880 789 818 881 133 772 8 × 2 = 1 + 0.999 999 761 579 637 762 267 545 6;
  • 37) 0.999 999 761 579 637 762 267 545 6 × 2 = 1 + 0.999 999 523 159 275 524 535 091 2;
  • 38) 0.999 999 523 159 275 524 535 091 2 × 2 = 1 + 0.999 999 046 318 551 049 070 182 4;
  • 39) 0.999 999 046 318 551 049 070 182 4 × 2 = 1 + 0.999 998 092 637 102 098 140 364 8;
  • 40) 0.999 998 092 637 102 098 140 364 8 × 2 = 1 + 0.999 996 185 274 204 196 280 729 6;
  • 41) 0.999 996 185 274 204 196 280 729 6 × 2 = 1 + 0.999 992 370 548 408 392 561 459 2;
  • 42) 0.999 992 370 548 408 392 561 459 2 × 2 = 1 + 0.999 984 741 096 816 785 122 918 4;
  • 43) 0.999 984 741 096 816 785 122 918 4 × 2 = 1 + 0.999 969 482 193 633 570 245 836 8;
  • 44) 0.999 969 482 193 633 570 245 836 8 × 2 = 1 + 0.999 938 964 387 267 140 491 673 6;
  • 45) 0.999 938 964 387 267 140 491 673 6 × 2 = 1 + 0.999 877 928 774 534 280 983 347 2;
  • 46) 0.999 877 928 774 534 280 983 347 2 × 2 = 1 + 0.999 755 857 549 068 561 966 694 4;
  • 47) 0.999 755 857 549 068 561 966 694 4 × 2 = 1 + 0.999 511 715 098 137 123 933 388 8;
  • 48) 0.999 511 715 098 137 123 933 388 8 × 2 = 1 + 0.999 023 430 196 274 247 866 777 6;
  • 49) 0.999 023 430 196 274 247 866 777 6 × 2 = 1 + 0.998 046 860 392 548 495 733 555 2;
  • 50) 0.998 046 860 392 548 495 733 555 2 × 2 = 1 + 0.996 093 720 785 096 991 467 110 4;
  • 51) 0.996 093 720 785 096 991 467 110 4 × 2 = 1 + 0.992 187 441 570 193 982 934 220 8;
  • 52) 0.992 187 441 570 193 982 934 220 8 × 2 = 1 + 0.984 374 883 140 387 965 868 441 6;
  • 53) 0.984 374 883 140 387 965 868 441 6 × 2 = 1 + 0.968 749 766 280 775 931 736 883 2;
  • 54) 0.968 749 766 280 775 931 736 883 2 × 2 = 1 + 0.937 499 532 561 551 863 473 766 4;
  • 55) 0.937 499 532 561 551 863 473 766 4 × 2 = 1 + 0.874 999 065 123 103 726 947 532 8;
  • 56) 0.874 999 065 123 103 726 947 532 8 × 2 = 1 + 0.749 998 130 246 207 453 895 065 6;
  • 57) 0.749 998 130 246 207 453 895 065 6 × 2 = 1 + 0.499 996 260 492 414 907 790 131 2;
  • 58) 0.499 996 260 492 414 907 790 131 2 × 2 = 0 + 0.999 992 520 984 829 815 580 262 4;

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:


0.016 738 891 601 562 496 530 527 1(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 527 1(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:


0.016 738 891 601 562 496 530 527 1(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 -0.016 738 891 601 562 496 530 527 1 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