-0.016 738 891 601 562 496 531 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 531 3(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
-0.016 738 891 601 562 496 531 3(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:

|-0.016 738 891 601 562 496 531 3| = 0.016 738 891 601 562 496 531 3


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: 0.016 738 891 601 562 496 531 3.

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) 0.016 738 891 601 562 496 531 3 × 2 = 0 + 0.033 477 783 203 124 993 062 6;
  • 2) 0.033 477 783 203 124 993 062 6 × 2 = 0 + 0.066 955 566 406 249 986 125 2;
  • 3) 0.066 955 566 406 249 986 125 2 × 2 = 0 + 0.133 911 132 812 499 972 250 4;
  • 4) 0.133 911 132 812 499 972 250 4 × 2 = 0 + 0.267 822 265 624 999 944 500 8;
  • 5) 0.267 822 265 624 999 944 500 8 × 2 = 0 + 0.535 644 531 249 999 889 001 6;
  • 6) 0.535 644 531 249 999 889 001 6 × 2 = 1 + 0.071 289 062 499 999 778 003 2;
  • 7) 0.071 289 062 499 999 778 003 2 × 2 = 0 + 0.142 578 124 999 999 556 006 4;
  • 8) 0.142 578 124 999 999 556 006 4 × 2 = 0 + 0.285 156 249 999 999 112 012 8;
  • 9) 0.285 156 249 999 999 112 012 8 × 2 = 0 + 0.570 312 499 999 998 224 025 6;
  • 10) 0.570 312 499 999 998 224 025 6 × 2 = 1 + 0.140 624 999 999 996 448 051 2;
  • 11) 0.140 624 999 999 996 448 051 2 × 2 = 0 + 0.281 249 999 999 992 896 102 4;
  • 12) 0.281 249 999 999 992 896 102 4 × 2 = 0 + 0.562 499 999 999 985 792 204 8;
  • 13) 0.562 499 999 999 985 792 204 8 × 2 = 1 + 0.124 999 999 999 971 584 409 6;
  • 14) 0.124 999 999 999 971 584 409 6 × 2 = 0 + 0.249 999 999 999 943 168 819 2;
  • 15) 0.249 999 999 999 943 168 819 2 × 2 = 0 + 0.499 999 999 999 886 337 638 4;
  • 16) 0.499 999 999 999 886 337 638 4 × 2 = 0 + 0.999 999 999 999 772 675 276 8;
  • 17) 0.999 999 999 999 772 675 276 8 × 2 = 1 + 0.999 999 999 999 545 350 553 6;
  • 18) 0.999 999 999 999 545 350 553 6 × 2 = 1 + 0.999 999 999 999 090 701 107 2;
  • 19) 0.999 999 999 999 090 701 107 2 × 2 = 1 + 0.999 999 999 998 181 402 214 4;
  • 20) 0.999 999 999 998 181 402 214 4 × 2 = 1 + 0.999 999 999 996 362 804 428 8;
  • 21) 0.999 999 999 996 362 804 428 8 × 2 = 1 + 0.999 999 999 992 725 608 857 6;
  • 22) 0.999 999 999 992 725 608 857 6 × 2 = 1 + 0.999 999 999 985 451 217 715 2;
  • 23) 0.999 999 999 985 451 217 715 2 × 2 = 1 + 0.999 999 999 970 902 435 430 4;
  • 24) 0.999 999 999 970 902 435 430 4 × 2 = 1 + 0.999 999 999 941 804 870 860 8;
  • 25) 0.999 999 999 941 804 870 860 8 × 2 = 1 + 0.999 999 999 883 609 741 721 6;
  • 26) 0.999 999 999 883 609 741 721 6 × 2 = 1 + 0.999 999 999 767 219 483 443 2;
  • 27) 0.999 999 999 767 219 483 443 2 × 2 = 1 + 0.999 999 999 534 438 966 886 4;
  • 28) 0.999 999 999 534 438 966 886 4 × 2 = 1 + 0.999 999 999 068 877 933 772 8;
  • 29) 0.999 999 999 068 877 933 772 8 × 2 = 1 + 0.999 999 998 137 755 867 545 6;
  • 30) 0.999 999 998 137 755 867 545 6 × 2 = 1 + 0.999 999 996 275 511 735 091 2;
  • 31) 0.999 999 996 275 511 735 091 2 × 2 = 1 + 0.999 999 992 551 023 470 182 4;
  • 32) 0.999 999 992 551 023 470 182 4 × 2 = 1 + 0.999 999 985 102 046 940 364 8;
  • 33) 0.999 999 985 102 046 940 364 8 × 2 = 1 + 0.999 999 970 204 093 880 729 6;
  • 34) 0.999 999 970 204 093 880 729 6 × 2 = 1 + 0.999 999 940 408 187 761 459 2;
  • 35) 0.999 999 940 408 187 761 459 2 × 2 = 1 + 0.999 999 880 816 375 522 918 4;
  • 36) 0.999 999 880 816 375 522 918 4 × 2 = 1 + 0.999 999 761 632 751 045 836 8;
  • 37) 0.999 999 761 632 751 045 836 8 × 2 = 1 + 0.999 999 523 265 502 091 673 6;
  • 38) 0.999 999 523 265 502 091 673 6 × 2 = 1 + 0.999 999 046 531 004 183 347 2;
  • 39) 0.999 999 046 531 004 183 347 2 × 2 = 1 + 0.999 998 093 062 008 366 694 4;
  • 40) 0.999 998 093 062 008 366 694 4 × 2 = 1 + 0.999 996 186 124 016 733 388 8;
  • 41) 0.999 996 186 124 016 733 388 8 × 2 = 1 + 0.999 992 372 248 033 466 777 6;
  • 42) 0.999 992 372 248 033 466 777 6 × 2 = 1 + 0.999 984 744 496 066 933 555 2;
  • 43) 0.999 984 744 496 066 933 555 2 × 2 = 1 + 0.999 969 488 992 133 867 110 4;
  • 44) 0.999 969 488 992 133 867 110 4 × 2 = 1 + 0.999 938 977 984 267 734 220 8;
  • 45) 0.999 938 977 984 267 734 220 8 × 2 = 1 + 0.999 877 955 968 535 468 441 6;
  • 46) 0.999 877 955 968 535 468 441 6 × 2 = 1 + 0.999 755 911 937 070 936 883 2;
  • 47) 0.999 755 911 937 070 936 883 2 × 2 = 1 + 0.999 511 823 874 141 873 766 4;
  • 48) 0.999 511 823 874 141 873 766 4 × 2 = 1 + 0.999 023 647 748 283 747 532 8;
  • 49) 0.999 023 647 748 283 747 532 8 × 2 = 1 + 0.998 047 295 496 567 495 065 6;
  • 50) 0.998 047 295 496 567 495 065 6 × 2 = 1 + 0.996 094 590 993 134 990 131 2;
  • 51) 0.996 094 590 993 134 990 131 2 × 2 = 1 + 0.992 189 181 986 269 980 262 4;
  • 52) 0.992 189 181 986 269 980 262 4 × 2 = 1 + 0.984 378 363 972 539 960 524 8;
  • 53) 0.984 378 363 972 539 960 524 8 × 2 = 1 + 0.968 756 727 945 079 921 049 6;
  • 54) 0.968 756 727 945 079 921 049 6 × 2 = 1 + 0.937 513 455 890 159 842 099 2;
  • 55) 0.937 513 455 890 159 842 099 2 × 2 = 1 + 0.875 026 911 780 319 684 198 4;
  • 56) 0.875 026 911 780 319 684 198 4 × 2 = 1 + 0.750 053 823 560 639 368 396 8;
  • 57) 0.750 053 823 560 639 368 396 8 × 2 = 1 + 0.500 107 647 121 278 736 793 6;
  • 58) 0.500 107 647 121 278 736 793 6 × 2 = 1 + 0.000 215 294 242 557 473 587 2;

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:


0.016 738 891 601 562 496 531 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 531 3(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:


0.016 738 891 601 562 496 531 3(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 -0.016 738 891 601 562 496 531 3 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

Share

» Convert decimal -0.016 738 891 601 562 496 534 7 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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