-0.016 738 891 601 562 496 529 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 529 69(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
-0.016 738 891 601 562 496 529 69(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:

|-0.016 738 891 601 562 496 529 69| = 0.016 738 891 601 562 496 529 69


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: 0.016 738 891 601 562 496 529 69.

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) 0.016 738 891 601 562 496 529 69 × 2 = 0 + 0.033 477 783 203 124 993 059 38;
  • 2) 0.033 477 783 203 124 993 059 38 × 2 = 0 + 0.066 955 566 406 249 986 118 76;
  • 3) 0.066 955 566 406 249 986 118 76 × 2 = 0 + 0.133 911 132 812 499 972 237 52;
  • 4) 0.133 911 132 812 499 972 237 52 × 2 = 0 + 0.267 822 265 624 999 944 475 04;
  • 5) 0.267 822 265 624 999 944 475 04 × 2 = 0 + 0.535 644 531 249 999 888 950 08;
  • 6) 0.535 644 531 249 999 888 950 08 × 2 = 1 + 0.071 289 062 499 999 777 900 16;
  • 7) 0.071 289 062 499 999 777 900 16 × 2 = 0 + 0.142 578 124 999 999 555 800 32;
  • 8) 0.142 578 124 999 999 555 800 32 × 2 = 0 + 0.285 156 249 999 999 111 600 64;
  • 9) 0.285 156 249 999 999 111 600 64 × 2 = 0 + 0.570 312 499 999 998 223 201 28;
  • 10) 0.570 312 499 999 998 223 201 28 × 2 = 1 + 0.140 624 999 999 996 446 402 56;
  • 11) 0.140 624 999 999 996 446 402 56 × 2 = 0 + 0.281 249 999 999 992 892 805 12;
  • 12) 0.281 249 999 999 992 892 805 12 × 2 = 0 + 0.562 499 999 999 985 785 610 24;
  • 13) 0.562 499 999 999 985 785 610 24 × 2 = 1 + 0.124 999 999 999 971 571 220 48;
  • 14) 0.124 999 999 999 971 571 220 48 × 2 = 0 + 0.249 999 999 999 943 142 440 96;
  • 15) 0.249 999 999 999 943 142 440 96 × 2 = 0 + 0.499 999 999 999 886 284 881 92;
  • 16) 0.499 999 999 999 886 284 881 92 × 2 = 0 + 0.999 999 999 999 772 569 763 84;
  • 17) 0.999 999 999 999 772 569 763 84 × 2 = 1 + 0.999 999 999 999 545 139 527 68;
  • 18) 0.999 999 999 999 545 139 527 68 × 2 = 1 + 0.999 999 999 999 090 279 055 36;
  • 19) 0.999 999 999 999 090 279 055 36 × 2 = 1 + 0.999 999 999 998 180 558 110 72;
  • 20) 0.999 999 999 998 180 558 110 72 × 2 = 1 + 0.999 999 999 996 361 116 221 44;
  • 21) 0.999 999 999 996 361 116 221 44 × 2 = 1 + 0.999 999 999 992 722 232 442 88;
  • 22) 0.999 999 999 992 722 232 442 88 × 2 = 1 + 0.999 999 999 985 444 464 885 76;
  • 23) 0.999 999 999 985 444 464 885 76 × 2 = 1 + 0.999 999 999 970 888 929 771 52;
  • 24) 0.999 999 999 970 888 929 771 52 × 2 = 1 + 0.999 999 999 941 777 859 543 04;
  • 25) 0.999 999 999 941 777 859 543 04 × 2 = 1 + 0.999 999 999 883 555 719 086 08;
  • 26) 0.999 999 999 883 555 719 086 08 × 2 = 1 + 0.999 999 999 767 111 438 172 16;
  • 27) 0.999 999 999 767 111 438 172 16 × 2 = 1 + 0.999 999 999 534 222 876 344 32;
  • 28) 0.999 999 999 534 222 876 344 32 × 2 = 1 + 0.999 999 999 068 445 752 688 64;
  • 29) 0.999 999 999 068 445 752 688 64 × 2 = 1 + 0.999 999 998 136 891 505 377 28;
  • 30) 0.999 999 998 136 891 505 377 28 × 2 = 1 + 0.999 999 996 273 783 010 754 56;
  • 31) 0.999 999 996 273 783 010 754 56 × 2 = 1 + 0.999 999 992 547 566 021 509 12;
  • 32) 0.999 999 992 547 566 021 509 12 × 2 = 1 + 0.999 999 985 095 132 043 018 24;
  • 33) 0.999 999 985 095 132 043 018 24 × 2 = 1 + 0.999 999 970 190 264 086 036 48;
  • 34) 0.999 999 970 190 264 086 036 48 × 2 = 1 + 0.999 999 940 380 528 172 072 96;
  • 35) 0.999 999 940 380 528 172 072 96 × 2 = 1 + 0.999 999 880 761 056 344 145 92;
  • 36) 0.999 999 880 761 056 344 145 92 × 2 = 1 + 0.999 999 761 522 112 688 291 84;
  • 37) 0.999 999 761 522 112 688 291 84 × 2 = 1 + 0.999 999 523 044 225 376 583 68;
  • 38) 0.999 999 523 044 225 376 583 68 × 2 = 1 + 0.999 999 046 088 450 753 167 36;
  • 39) 0.999 999 046 088 450 753 167 36 × 2 = 1 + 0.999 998 092 176 901 506 334 72;
  • 40) 0.999 998 092 176 901 506 334 72 × 2 = 1 + 0.999 996 184 353 803 012 669 44;
  • 41) 0.999 996 184 353 803 012 669 44 × 2 = 1 + 0.999 992 368 707 606 025 338 88;
  • 42) 0.999 992 368 707 606 025 338 88 × 2 = 1 + 0.999 984 737 415 212 050 677 76;
  • 43) 0.999 984 737 415 212 050 677 76 × 2 = 1 + 0.999 969 474 830 424 101 355 52;
  • 44) 0.999 969 474 830 424 101 355 52 × 2 = 1 + 0.999 938 949 660 848 202 711 04;
  • 45) 0.999 938 949 660 848 202 711 04 × 2 = 1 + 0.999 877 899 321 696 405 422 08;
  • 46) 0.999 877 899 321 696 405 422 08 × 2 = 1 + 0.999 755 798 643 392 810 844 16;
  • 47) 0.999 755 798 643 392 810 844 16 × 2 = 1 + 0.999 511 597 286 785 621 688 32;
  • 48) 0.999 511 597 286 785 621 688 32 × 2 = 1 + 0.999 023 194 573 571 243 376 64;
  • 49) 0.999 023 194 573 571 243 376 64 × 2 = 1 + 0.998 046 389 147 142 486 753 28;
  • 50) 0.998 046 389 147 142 486 753 28 × 2 = 1 + 0.996 092 778 294 284 973 506 56;
  • 51) 0.996 092 778 294 284 973 506 56 × 2 = 1 + 0.992 185 556 588 569 947 013 12;
  • 52) 0.992 185 556 588 569 947 013 12 × 2 = 1 + 0.984 371 113 177 139 894 026 24;
  • 53) 0.984 371 113 177 139 894 026 24 × 2 = 1 + 0.968 742 226 354 279 788 052 48;
  • 54) 0.968 742 226 354 279 788 052 48 × 2 = 1 + 0.937 484 452 708 559 576 104 96;
  • 55) 0.937 484 452 708 559 576 104 96 × 2 = 1 + 0.874 968 905 417 119 152 209 92;
  • 56) 0.874 968 905 417 119 152 209 92 × 2 = 1 + 0.749 937 810 834 238 304 419 84;
  • 57) 0.749 937 810 834 238 304 419 84 × 2 = 1 + 0.499 875 621 668 476 608 839 68;
  • 58) 0.499 875 621 668 476 608 839 68 × 2 = 0 + 0.999 751 243 336 953 217 679 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:


0.016 738 891 601 562 496 529 69(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 529 69(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:


0.016 738 891 601 562 496 529 69(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 -0.016 738 891 601 562 496 529 69 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

Share

» Convert decimal -0.016 738 891 601 562 496 530 07 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: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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