-0.016 738 891 601 562 496 530 552 23 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 23(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
-0.016 738 891 601 562 496 530 552 23(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:

|-0.016 738 891 601 562 496 530 552 23| = 0.016 738 891 601 562 496 530 552 23


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: 0.016 738 891 601 562 496 530 552 23.

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) 0.016 738 891 601 562 496 530 552 23 × 2 = 0 + 0.033 477 783 203 124 993 061 104 46;
  • 2) 0.033 477 783 203 124 993 061 104 46 × 2 = 0 + 0.066 955 566 406 249 986 122 208 92;
  • 3) 0.066 955 566 406 249 986 122 208 92 × 2 = 0 + 0.133 911 132 812 499 972 244 417 84;
  • 4) 0.133 911 132 812 499 972 244 417 84 × 2 = 0 + 0.267 822 265 624 999 944 488 835 68;
  • 5) 0.267 822 265 624 999 944 488 835 68 × 2 = 0 + 0.535 644 531 249 999 888 977 671 36;
  • 6) 0.535 644 531 249 999 888 977 671 36 × 2 = 1 + 0.071 289 062 499 999 777 955 342 72;
  • 7) 0.071 289 062 499 999 777 955 342 72 × 2 = 0 + 0.142 578 124 999 999 555 910 685 44;
  • 8) 0.142 578 124 999 999 555 910 685 44 × 2 = 0 + 0.285 156 249 999 999 111 821 370 88;
  • 9) 0.285 156 249 999 999 111 821 370 88 × 2 = 0 + 0.570 312 499 999 998 223 642 741 76;
  • 10) 0.570 312 499 999 998 223 642 741 76 × 2 = 1 + 0.140 624 999 999 996 447 285 483 52;
  • 11) 0.140 624 999 999 996 447 285 483 52 × 2 = 0 + 0.281 249 999 999 992 894 570 967 04;
  • 12) 0.281 249 999 999 992 894 570 967 04 × 2 = 0 + 0.562 499 999 999 985 789 141 934 08;
  • 13) 0.562 499 999 999 985 789 141 934 08 × 2 = 1 + 0.124 999 999 999 971 578 283 868 16;
  • 14) 0.124 999 999 999 971 578 283 868 16 × 2 = 0 + 0.249 999 999 999 943 156 567 736 32;
  • 15) 0.249 999 999 999 943 156 567 736 32 × 2 = 0 + 0.499 999 999 999 886 313 135 472 64;
  • 16) 0.499 999 999 999 886 313 135 472 64 × 2 = 0 + 0.999 999 999 999 772 626 270 945 28;
  • 17) 0.999 999 999 999 772 626 270 945 28 × 2 = 1 + 0.999 999 999 999 545 252 541 890 56;
  • 18) 0.999 999 999 999 545 252 541 890 56 × 2 = 1 + 0.999 999 999 999 090 505 083 781 12;
  • 19) 0.999 999 999 999 090 505 083 781 12 × 2 = 1 + 0.999 999 999 998 181 010 167 562 24;
  • 20) 0.999 999 999 998 181 010 167 562 24 × 2 = 1 + 0.999 999 999 996 362 020 335 124 48;
  • 21) 0.999 999 999 996 362 020 335 124 48 × 2 = 1 + 0.999 999 999 992 724 040 670 248 96;
  • 22) 0.999 999 999 992 724 040 670 248 96 × 2 = 1 + 0.999 999 999 985 448 081 340 497 92;
  • 23) 0.999 999 999 985 448 081 340 497 92 × 2 = 1 + 0.999 999 999 970 896 162 680 995 84;
  • 24) 0.999 999 999 970 896 162 680 995 84 × 2 = 1 + 0.999 999 999 941 792 325 361 991 68;
  • 25) 0.999 999 999 941 792 325 361 991 68 × 2 = 1 + 0.999 999 999 883 584 650 723 983 36;
  • 26) 0.999 999 999 883 584 650 723 983 36 × 2 = 1 + 0.999 999 999 767 169 301 447 966 72;
  • 27) 0.999 999 999 767 169 301 447 966 72 × 2 = 1 + 0.999 999 999 534 338 602 895 933 44;
  • 28) 0.999 999 999 534 338 602 895 933 44 × 2 = 1 + 0.999 999 999 068 677 205 791 866 88;
  • 29) 0.999 999 999 068 677 205 791 866 88 × 2 = 1 + 0.999 999 998 137 354 411 583 733 76;
  • 30) 0.999 999 998 137 354 411 583 733 76 × 2 = 1 + 0.999 999 996 274 708 823 167 467 52;
  • 31) 0.999 999 996 274 708 823 167 467 52 × 2 = 1 + 0.999 999 992 549 417 646 334 935 04;
  • 32) 0.999 999 992 549 417 646 334 935 04 × 2 = 1 + 0.999 999 985 098 835 292 669 870 08;
  • 33) 0.999 999 985 098 835 292 669 870 08 × 2 = 1 + 0.999 999 970 197 670 585 339 740 16;
  • 34) 0.999 999 970 197 670 585 339 740 16 × 2 = 1 + 0.999 999 940 395 341 170 679 480 32;
  • 35) 0.999 999 940 395 341 170 679 480 32 × 2 = 1 + 0.999 999 880 790 682 341 358 960 64;
  • 36) 0.999 999 880 790 682 341 358 960 64 × 2 = 1 + 0.999 999 761 581 364 682 717 921 28;
  • 37) 0.999 999 761 581 364 682 717 921 28 × 2 = 1 + 0.999 999 523 162 729 365 435 842 56;
  • 38) 0.999 999 523 162 729 365 435 842 56 × 2 = 1 + 0.999 999 046 325 458 730 871 685 12;
  • 39) 0.999 999 046 325 458 730 871 685 12 × 2 = 1 + 0.999 998 092 650 917 461 743 370 24;
  • 40) 0.999 998 092 650 917 461 743 370 24 × 2 = 1 + 0.999 996 185 301 834 923 486 740 48;
  • 41) 0.999 996 185 301 834 923 486 740 48 × 2 = 1 + 0.999 992 370 603 669 846 973 480 96;
  • 42) 0.999 992 370 603 669 846 973 480 96 × 2 = 1 + 0.999 984 741 207 339 693 946 961 92;
  • 43) 0.999 984 741 207 339 693 946 961 92 × 2 = 1 + 0.999 969 482 414 679 387 893 923 84;
  • 44) 0.999 969 482 414 679 387 893 923 84 × 2 = 1 + 0.999 938 964 829 358 775 787 847 68;
  • 45) 0.999 938 964 829 358 775 787 847 68 × 2 = 1 + 0.999 877 929 658 717 551 575 695 36;
  • 46) 0.999 877 929 658 717 551 575 695 36 × 2 = 1 + 0.999 755 859 317 435 103 151 390 72;
  • 47) 0.999 755 859 317 435 103 151 390 72 × 2 = 1 + 0.999 511 718 634 870 206 302 781 44;
  • 48) 0.999 511 718 634 870 206 302 781 44 × 2 = 1 + 0.999 023 437 269 740 412 605 562 88;
  • 49) 0.999 023 437 269 740 412 605 562 88 × 2 = 1 + 0.998 046 874 539 480 825 211 125 76;
  • 50) 0.998 046 874 539 480 825 211 125 76 × 2 = 1 + 0.996 093 749 078 961 650 422 251 52;
  • 51) 0.996 093 749 078 961 650 422 251 52 × 2 = 1 + 0.992 187 498 157 923 300 844 503 04;
  • 52) 0.992 187 498 157 923 300 844 503 04 × 2 = 1 + 0.984 374 996 315 846 601 689 006 08;
  • 53) 0.984 374 996 315 846 601 689 006 08 × 2 = 1 + 0.968 749 992 631 693 203 378 012 16;
  • 54) 0.968 749 992 631 693 203 378 012 16 × 2 = 1 + 0.937 499 985 263 386 406 756 024 32;
  • 55) 0.937 499 985 263 386 406 756 024 32 × 2 = 1 + 0.874 999 970 526 772 813 512 048 64;
  • 56) 0.874 999 970 526 772 813 512 048 64 × 2 = 1 + 0.749 999 941 053 545 627 024 097 28;
  • 57) 0.749 999 941 053 545 627 024 097 28 × 2 = 1 + 0.499 999 882 107 091 254 048 194 56;
  • 58) 0.499 999 882 107 091 254 048 194 56 × 2 = 0 + 0.999 999 764 214 182 508 096 389 12;

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:


0.016 738 891 601 562 496 530 552 23(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 23(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:


0.016 738 891 601 562 496 530 552 23(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 -0.016 738 891 601 562 496 530 552 23 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