-0.016 738 891 601 562 496 530 524 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 524 2(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
-0.016 738 891 601 562 496 530 524 2(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:

|-0.016 738 891 601 562 496 530 524 2| = 0.016 738 891 601 562 496 530 524 2


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: 0.016 738 891 601 562 496 530 524 2.

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) 0.016 738 891 601 562 496 530 524 2 × 2 = 0 + 0.033 477 783 203 124 993 061 048 4;
  • 2) 0.033 477 783 203 124 993 061 048 4 × 2 = 0 + 0.066 955 566 406 249 986 122 096 8;
  • 3) 0.066 955 566 406 249 986 122 096 8 × 2 = 0 + 0.133 911 132 812 499 972 244 193 6;
  • 4) 0.133 911 132 812 499 972 244 193 6 × 2 = 0 + 0.267 822 265 624 999 944 488 387 2;
  • 5) 0.267 822 265 624 999 944 488 387 2 × 2 = 0 + 0.535 644 531 249 999 888 976 774 4;
  • 6) 0.535 644 531 249 999 888 976 774 4 × 2 = 1 + 0.071 289 062 499 999 777 953 548 8;
  • 7) 0.071 289 062 499 999 777 953 548 8 × 2 = 0 + 0.142 578 124 999 999 555 907 097 6;
  • 8) 0.142 578 124 999 999 555 907 097 6 × 2 = 0 + 0.285 156 249 999 999 111 814 195 2;
  • 9) 0.285 156 249 999 999 111 814 195 2 × 2 = 0 + 0.570 312 499 999 998 223 628 390 4;
  • 10) 0.570 312 499 999 998 223 628 390 4 × 2 = 1 + 0.140 624 999 999 996 447 256 780 8;
  • 11) 0.140 624 999 999 996 447 256 780 8 × 2 = 0 + 0.281 249 999 999 992 894 513 561 6;
  • 12) 0.281 249 999 999 992 894 513 561 6 × 2 = 0 + 0.562 499 999 999 985 789 027 123 2;
  • 13) 0.562 499 999 999 985 789 027 123 2 × 2 = 1 + 0.124 999 999 999 971 578 054 246 4;
  • 14) 0.124 999 999 999 971 578 054 246 4 × 2 = 0 + 0.249 999 999 999 943 156 108 492 8;
  • 15) 0.249 999 999 999 943 156 108 492 8 × 2 = 0 + 0.499 999 999 999 886 312 216 985 6;
  • 16) 0.499 999 999 999 886 312 216 985 6 × 2 = 0 + 0.999 999 999 999 772 624 433 971 2;
  • 17) 0.999 999 999 999 772 624 433 971 2 × 2 = 1 + 0.999 999 999 999 545 248 867 942 4;
  • 18) 0.999 999 999 999 545 248 867 942 4 × 2 = 1 + 0.999 999 999 999 090 497 735 884 8;
  • 19) 0.999 999 999 999 090 497 735 884 8 × 2 = 1 + 0.999 999 999 998 180 995 471 769 6;
  • 20) 0.999 999 999 998 180 995 471 769 6 × 2 = 1 + 0.999 999 999 996 361 990 943 539 2;
  • 21) 0.999 999 999 996 361 990 943 539 2 × 2 = 1 + 0.999 999 999 992 723 981 887 078 4;
  • 22) 0.999 999 999 992 723 981 887 078 4 × 2 = 1 + 0.999 999 999 985 447 963 774 156 8;
  • 23) 0.999 999 999 985 447 963 774 156 8 × 2 = 1 + 0.999 999 999 970 895 927 548 313 6;
  • 24) 0.999 999 999 970 895 927 548 313 6 × 2 = 1 + 0.999 999 999 941 791 855 096 627 2;
  • 25) 0.999 999 999 941 791 855 096 627 2 × 2 = 1 + 0.999 999 999 883 583 710 193 254 4;
  • 26) 0.999 999 999 883 583 710 193 254 4 × 2 = 1 + 0.999 999 999 767 167 420 386 508 8;
  • 27) 0.999 999 999 767 167 420 386 508 8 × 2 = 1 + 0.999 999 999 534 334 840 773 017 6;
  • 28) 0.999 999 999 534 334 840 773 017 6 × 2 = 1 + 0.999 999 999 068 669 681 546 035 2;
  • 29) 0.999 999 999 068 669 681 546 035 2 × 2 = 1 + 0.999 999 998 137 339 363 092 070 4;
  • 30) 0.999 999 998 137 339 363 092 070 4 × 2 = 1 + 0.999 999 996 274 678 726 184 140 8;
  • 31) 0.999 999 996 274 678 726 184 140 8 × 2 = 1 + 0.999 999 992 549 357 452 368 281 6;
  • 32) 0.999 999 992 549 357 452 368 281 6 × 2 = 1 + 0.999 999 985 098 714 904 736 563 2;
  • 33) 0.999 999 985 098 714 904 736 563 2 × 2 = 1 + 0.999 999 970 197 429 809 473 126 4;
  • 34) 0.999 999 970 197 429 809 473 126 4 × 2 = 1 + 0.999 999 940 394 859 618 946 252 8;
  • 35) 0.999 999 940 394 859 618 946 252 8 × 2 = 1 + 0.999 999 880 789 719 237 892 505 6;
  • 36) 0.999 999 880 789 719 237 892 505 6 × 2 = 1 + 0.999 999 761 579 438 475 785 011 2;
  • 37) 0.999 999 761 579 438 475 785 011 2 × 2 = 1 + 0.999 999 523 158 876 951 570 022 4;
  • 38) 0.999 999 523 158 876 951 570 022 4 × 2 = 1 + 0.999 999 046 317 753 903 140 044 8;
  • 39) 0.999 999 046 317 753 903 140 044 8 × 2 = 1 + 0.999 998 092 635 507 806 280 089 6;
  • 40) 0.999 998 092 635 507 806 280 089 6 × 2 = 1 + 0.999 996 185 271 015 612 560 179 2;
  • 41) 0.999 996 185 271 015 612 560 179 2 × 2 = 1 + 0.999 992 370 542 031 225 120 358 4;
  • 42) 0.999 992 370 542 031 225 120 358 4 × 2 = 1 + 0.999 984 741 084 062 450 240 716 8;
  • 43) 0.999 984 741 084 062 450 240 716 8 × 2 = 1 + 0.999 969 482 168 124 900 481 433 6;
  • 44) 0.999 969 482 168 124 900 481 433 6 × 2 = 1 + 0.999 938 964 336 249 800 962 867 2;
  • 45) 0.999 938 964 336 249 800 962 867 2 × 2 = 1 + 0.999 877 928 672 499 601 925 734 4;
  • 46) 0.999 877 928 672 499 601 925 734 4 × 2 = 1 + 0.999 755 857 344 999 203 851 468 8;
  • 47) 0.999 755 857 344 999 203 851 468 8 × 2 = 1 + 0.999 511 714 689 998 407 702 937 6;
  • 48) 0.999 511 714 689 998 407 702 937 6 × 2 = 1 + 0.999 023 429 379 996 815 405 875 2;
  • 49) 0.999 023 429 379 996 815 405 875 2 × 2 = 1 + 0.998 046 858 759 993 630 811 750 4;
  • 50) 0.998 046 858 759 993 630 811 750 4 × 2 = 1 + 0.996 093 717 519 987 261 623 500 8;
  • 51) 0.996 093 717 519 987 261 623 500 8 × 2 = 1 + 0.992 187 435 039 974 523 247 001 6;
  • 52) 0.992 187 435 039 974 523 247 001 6 × 2 = 1 + 0.984 374 870 079 949 046 494 003 2;
  • 53) 0.984 374 870 079 949 046 494 003 2 × 2 = 1 + 0.968 749 740 159 898 092 988 006 4;
  • 54) 0.968 749 740 159 898 092 988 006 4 × 2 = 1 + 0.937 499 480 319 796 185 976 012 8;
  • 55) 0.937 499 480 319 796 185 976 012 8 × 2 = 1 + 0.874 998 960 639 592 371 952 025 6;
  • 56) 0.874 998 960 639 592 371 952 025 6 × 2 = 1 + 0.749 997 921 279 184 743 904 051 2;
  • 57) 0.749 997 921 279 184 743 904 051 2 × 2 = 1 + 0.499 995 842 558 369 487 808 102 4;
  • 58) 0.499 995 842 558 369 487 808 102 4 × 2 = 0 + 0.999 991 685 116 738 975 616 204 8;

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:


0.016 738 891 601 562 496 530 524 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 524 2(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:


0.016 738 891 601 562 496 530 524 2(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 -0.016 738 891 601 562 496 530 524 2 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