-0.016 738 891 601 562 496 530 553 065 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 065(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
-0.016 738 891 601 562 496 530 553 065(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:

|-0.016 738 891 601 562 496 530 553 065| = 0.016 738 891 601 562 496 530 553 065


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: 0.016 738 891 601 562 496 530 553 065.

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) 0.016 738 891 601 562 496 530 553 065 × 2 = 0 + 0.033 477 783 203 124 993 061 106 13;
  • 2) 0.033 477 783 203 124 993 061 106 13 × 2 = 0 + 0.066 955 566 406 249 986 122 212 26;
  • 3) 0.066 955 566 406 249 986 122 212 26 × 2 = 0 + 0.133 911 132 812 499 972 244 424 52;
  • 4) 0.133 911 132 812 499 972 244 424 52 × 2 = 0 + 0.267 822 265 624 999 944 488 849 04;
  • 5) 0.267 822 265 624 999 944 488 849 04 × 2 = 0 + 0.535 644 531 249 999 888 977 698 08;
  • 6) 0.535 644 531 249 999 888 977 698 08 × 2 = 1 + 0.071 289 062 499 999 777 955 396 16;
  • 7) 0.071 289 062 499 999 777 955 396 16 × 2 = 0 + 0.142 578 124 999 999 555 910 792 32;
  • 8) 0.142 578 124 999 999 555 910 792 32 × 2 = 0 + 0.285 156 249 999 999 111 821 584 64;
  • 9) 0.285 156 249 999 999 111 821 584 64 × 2 = 0 + 0.570 312 499 999 998 223 643 169 28;
  • 10) 0.570 312 499 999 998 223 643 169 28 × 2 = 1 + 0.140 624 999 999 996 447 286 338 56;
  • 11) 0.140 624 999 999 996 447 286 338 56 × 2 = 0 + 0.281 249 999 999 992 894 572 677 12;
  • 12) 0.281 249 999 999 992 894 572 677 12 × 2 = 0 + 0.562 499 999 999 985 789 145 354 24;
  • 13) 0.562 499 999 999 985 789 145 354 24 × 2 = 1 + 0.124 999 999 999 971 578 290 708 48;
  • 14) 0.124 999 999 999 971 578 290 708 48 × 2 = 0 + 0.249 999 999 999 943 156 581 416 96;
  • 15) 0.249 999 999 999 943 156 581 416 96 × 2 = 0 + 0.499 999 999 999 886 313 162 833 92;
  • 16) 0.499 999 999 999 886 313 162 833 92 × 2 = 0 + 0.999 999 999 999 772 626 325 667 84;
  • 17) 0.999 999 999 999 772 626 325 667 84 × 2 = 1 + 0.999 999 999 999 545 252 651 335 68;
  • 18) 0.999 999 999 999 545 252 651 335 68 × 2 = 1 + 0.999 999 999 999 090 505 302 671 36;
  • 19) 0.999 999 999 999 090 505 302 671 36 × 2 = 1 + 0.999 999 999 998 181 010 605 342 72;
  • 20) 0.999 999 999 998 181 010 605 342 72 × 2 = 1 + 0.999 999 999 996 362 021 210 685 44;
  • 21) 0.999 999 999 996 362 021 210 685 44 × 2 = 1 + 0.999 999 999 992 724 042 421 370 88;
  • 22) 0.999 999 999 992 724 042 421 370 88 × 2 = 1 + 0.999 999 999 985 448 084 842 741 76;
  • 23) 0.999 999 999 985 448 084 842 741 76 × 2 = 1 + 0.999 999 999 970 896 169 685 483 52;
  • 24) 0.999 999 999 970 896 169 685 483 52 × 2 = 1 + 0.999 999 999 941 792 339 370 967 04;
  • 25) 0.999 999 999 941 792 339 370 967 04 × 2 = 1 + 0.999 999 999 883 584 678 741 934 08;
  • 26) 0.999 999 999 883 584 678 741 934 08 × 2 = 1 + 0.999 999 999 767 169 357 483 868 16;
  • 27) 0.999 999 999 767 169 357 483 868 16 × 2 = 1 + 0.999 999 999 534 338 714 967 736 32;
  • 28) 0.999 999 999 534 338 714 967 736 32 × 2 = 1 + 0.999 999 999 068 677 429 935 472 64;
  • 29) 0.999 999 999 068 677 429 935 472 64 × 2 = 1 + 0.999 999 998 137 354 859 870 945 28;
  • 30) 0.999 999 998 137 354 859 870 945 28 × 2 = 1 + 0.999 999 996 274 709 719 741 890 56;
  • 31) 0.999 999 996 274 709 719 741 890 56 × 2 = 1 + 0.999 999 992 549 419 439 483 781 12;
  • 32) 0.999 999 992 549 419 439 483 781 12 × 2 = 1 + 0.999 999 985 098 838 878 967 562 24;
  • 33) 0.999 999 985 098 838 878 967 562 24 × 2 = 1 + 0.999 999 970 197 677 757 935 124 48;
  • 34) 0.999 999 970 197 677 757 935 124 48 × 2 = 1 + 0.999 999 940 395 355 515 870 248 96;
  • 35) 0.999 999 940 395 355 515 870 248 96 × 2 = 1 + 0.999 999 880 790 711 031 740 497 92;
  • 36) 0.999 999 880 790 711 031 740 497 92 × 2 = 1 + 0.999 999 761 581 422 063 480 995 84;
  • 37) 0.999 999 761 581 422 063 480 995 84 × 2 = 1 + 0.999 999 523 162 844 126 961 991 68;
  • 38) 0.999 999 523 162 844 126 961 991 68 × 2 = 1 + 0.999 999 046 325 688 253 923 983 36;
  • 39) 0.999 999 046 325 688 253 923 983 36 × 2 = 1 + 0.999 998 092 651 376 507 847 966 72;
  • 40) 0.999 998 092 651 376 507 847 966 72 × 2 = 1 + 0.999 996 185 302 753 015 695 933 44;
  • 41) 0.999 996 185 302 753 015 695 933 44 × 2 = 1 + 0.999 992 370 605 506 031 391 866 88;
  • 42) 0.999 992 370 605 506 031 391 866 88 × 2 = 1 + 0.999 984 741 211 012 062 783 733 76;
  • 43) 0.999 984 741 211 012 062 783 733 76 × 2 = 1 + 0.999 969 482 422 024 125 567 467 52;
  • 44) 0.999 969 482 422 024 125 567 467 52 × 2 = 1 + 0.999 938 964 844 048 251 134 935 04;
  • 45) 0.999 938 964 844 048 251 134 935 04 × 2 = 1 + 0.999 877 929 688 096 502 269 870 08;
  • 46) 0.999 877 929 688 096 502 269 870 08 × 2 = 1 + 0.999 755 859 376 193 004 539 740 16;
  • 47) 0.999 755 859 376 193 004 539 740 16 × 2 = 1 + 0.999 511 718 752 386 009 079 480 32;
  • 48) 0.999 511 718 752 386 009 079 480 32 × 2 = 1 + 0.999 023 437 504 772 018 158 960 64;
  • 49) 0.999 023 437 504 772 018 158 960 64 × 2 = 1 + 0.998 046 875 009 544 036 317 921 28;
  • 50) 0.998 046 875 009 544 036 317 921 28 × 2 = 1 + 0.996 093 750 019 088 072 635 842 56;
  • 51) 0.996 093 750 019 088 072 635 842 56 × 2 = 1 + 0.992 187 500 038 176 145 271 685 12;
  • 52) 0.992 187 500 038 176 145 271 685 12 × 2 = 1 + 0.984 375 000 076 352 290 543 370 24;
  • 53) 0.984 375 000 076 352 290 543 370 24 × 2 = 1 + 0.968 750 000 152 704 581 086 740 48;
  • 54) 0.968 750 000 152 704 581 086 740 48 × 2 = 1 + 0.937 500 000 305 409 162 173 480 96;
  • 55) 0.937 500 000 305 409 162 173 480 96 × 2 = 1 + 0.875 000 000 610 818 324 346 961 92;
  • 56) 0.875 000 000 610 818 324 346 961 92 × 2 = 1 + 0.750 000 001 221 636 648 693 923 84;
  • 57) 0.750 000 001 221 636 648 693 923 84 × 2 = 1 + 0.500 000 002 443 273 297 387 847 68;
  • 58) 0.500 000 002 443 273 297 387 847 68 × 2 = 1 + 0.000 000 004 886 546 594 775 695 36;

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:


0.016 738 891 601 562 496 530 553 065(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 065(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:


0.016 738 891 601 562 496 530 553 065(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 -0.016 738 891 601 562 496 530 553 065 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