-0.016 738 891 601 562 496 529 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 529 9(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 9(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 9| = 0.016 738 891 601 562 496 529 9


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 9.

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 9 × 2 = 0 + 0.033 477 783 203 124 993 059 8;
  • 2) 0.033 477 783 203 124 993 059 8 × 2 = 0 + 0.066 955 566 406 249 986 119 6;
  • 3) 0.066 955 566 406 249 986 119 6 × 2 = 0 + 0.133 911 132 812 499 972 239 2;
  • 4) 0.133 911 132 812 499 972 239 2 × 2 = 0 + 0.267 822 265 624 999 944 478 4;
  • 5) 0.267 822 265 624 999 944 478 4 × 2 = 0 + 0.535 644 531 249 999 888 956 8;
  • 6) 0.535 644 531 249 999 888 956 8 × 2 = 1 + 0.071 289 062 499 999 777 913 6;
  • 7) 0.071 289 062 499 999 777 913 6 × 2 = 0 + 0.142 578 124 999 999 555 827 2;
  • 8) 0.142 578 124 999 999 555 827 2 × 2 = 0 + 0.285 156 249 999 999 111 654 4;
  • 9) 0.285 156 249 999 999 111 654 4 × 2 = 0 + 0.570 312 499 999 998 223 308 8;
  • 10) 0.570 312 499 999 998 223 308 8 × 2 = 1 + 0.140 624 999 999 996 446 617 6;
  • 11) 0.140 624 999 999 996 446 617 6 × 2 = 0 + 0.281 249 999 999 992 893 235 2;
  • 12) 0.281 249 999 999 992 893 235 2 × 2 = 0 + 0.562 499 999 999 985 786 470 4;
  • 13) 0.562 499 999 999 985 786 470 4 × 2 = 1 + 0.124 999 999 999 971 572 940 8;
  • 14) 0.124 999 999 999 971 572 940 8 × 2 = 0 + 0.249 999 999 999 943 145 881 6;
  • 15) 0.249 999 999 999 943 145 881 6 × 2 = 0 + 0.499 999 999 999 886 291 763 2;
  • 16) 0.499 999 999 999 886 291 763 2 × 2 = 0 + 0.999 999 999 999 772 583 526 4;
  • 17) 0.999 999 999 999 772 583 526 4 × 2 = 1 + 0.999 999 999 999 545 167 052 8;
  • 18) 0.999 999 999 999 545 167 052 8 × 2 = 1 + 0.999 999 999 999 090 334 105 6;
  • 19) 0.999 999 999 999 090 334 105 6 × 2 = 1 + 0.999 999 999 998 180 668 211 2;
  • 20) 0.999 999 999 998 180 668 211 2 × 2 = 1 + 0.999 999 999 996 361 336 422 4;
  • 21) 0.999 999 999 996 361 336 422 4 × 2 = 1 + 0.999 999 999 992 722 672 844 8;
  • 22) 0.999 999 999 992 722 672 844 8 × 2 = 1 + 0.999 999 999 985 445 345 689 6;
  • 23) 0.999 999 999 985 445 345 689 6 × 2 = 1 + 0.999 999 999 970 890 691 379 2;
  • 24) 0.999 999 999 970 890 691 379 2 × 2 = 1 + 0.999 999 999 941 781 382 758 4;
  • 25) 0.999 999 999 941 781 382 758 4 × 2 = 1 + 0.999 999 999 883 562 765 516 8;
  • 26) 0.999 999 999 883 562 765 516 8 × 2 = 1 + 0.999 999 999 767 125 531 033 6;
  • 27) 0.999 999 999 767 125 531 033 6 × 2 = 1 + 0.999 999 999 534 251 062 067 2;
  • 28) 0.999 999 999 534 251 062 067 2 × 2 = 1 + 0.999 999 999 068 502 124 134 4;
  • 29) 0.999 999 999 068 502 124 134 4 × 2 = 1 + 0.999 999 998 137 004 248 268 8;
  • 30) 0.999 999 998 137 004 248 268 8 × 2 = 1 + 0.999 999 996 274 008 496 537 6;
  • 31) 0.999 999 996 274 008 496 537 6 × 2 = 1 + 0.999 999 992 548 016 993 075 2;
  • 32) 0.999 999 992 548 016 993 075 2 × 2 = 1 + 0.999 999 985 096 033 986 150 4;
  • 33) 0.999 999 985 096 033 986 150 4 × 2 = 1 + 0.999 999 970 192 067 972 300 8;
  • 34) 0.999 999 970 192 067 972 300 8 × 2 = 1 + 0.999 999 940 384 135 944 601 6;
  • 35) 0.999 999 940 384 135 944 601 6 × 2 = 1 + 0.999 999 880 768 271 889 203 2;
  • 36) 0.999 999 880 768 271 889 203 2 × 2 = 1 + 0.999 999 761 536 543 778 406 4;
  • 37) 0.999 999 761 536 543 778 406 4 × 2 = 1 + 0.999 999 523 073 087 556 812 8;
  • 38) 0.999 999 523 073 087 556 812 8 × 2 = 1 + 0.999 999 046 146 175 113 625 6;
  • 39) 0.999 999 046 146 175 113 625 6 × 2 = 1 + 0.999 998 092 292 350 227 251 2;
  • 40) 0.999 998 092 292 350 227 251 2 × 2 = 1 + 0.999 996 184 584 700 454 502 4;
  • 41) 0.999 996 184 584 700 454 502 4 × 2 = 1 + 0.999 992 369 169 400 909 004 8;
  • 42) 0.999 992 369 169 400 909 004 8 × 2 = 1 + 0.999 984 738 338 801 818 009 6;
  • 43) 0.999 984 738 338 801 818 009 6 × 2 = 1 + 0.999 969 476 677 603 636 019 2;
  • 44) 0.999 969 476 677 603 636 019 2 × 2 = 1 + 0.999 938 953 355 207 272 038 4;
  • 45) 0.999 938 953 355 207 272 038 4 × 2 = 1 + 0.999 877 906 710 414 544 076 8;
  • 46) 0.999 877 906 710 414 544 076 8 × 2 = 1 + 0.999 755 813 420 829 088 153 6;
  • 47) 0.999 755 813 420 829 088 153 6 × 2 = 1 + 0.999 511 626 841 658 176 307 2;
  • 48) 0.999 511 626 841 658 176 307 2 × 2 = 1 + 0.999 023 253 683 316 352 614 4;
  • 49) 0.999 023 253 683 316 352 614 4 × 2 = 1 + 0.998 046 507 366 632 705 228 8;
  • 50) 0.998 046 507 366 632 705 228 8 × 2 = 1 + 0.996 093 014 733 265 410 457 6;
  • 51) 0.996 093 014 733 265 410 457 6 × 2 = 1 + 0.992 186 029 466 530 820 915 2;
  • 52) 0.992 186 029 466 530 820 915 2 × 2 = 1 + 0.984 372 058 933 061 641 830 4;
  • 53) 0.984 372 058 933 061 641 830 4 × 2 = 1 + 0.968 744 117 866 123 283 660 8;
  • 54) 0.968 744 117 866 123 283 660 8 × 2 = 1 + 0.937 488 235 732 246 567 321 6;
  • 55) 0.937 488 235 732 246 567 321 6 × 2 = 1 + 0.874 976 471 464 493 134 643 2;
  • 56) 0.874 976 471 464 493 134 643 2 × 2 = 1 + 0.749 952 942 928 986 269 286 4;
  • 57) 0.749 952 942 928 986 269 286 4 × 2 = 1 + 0.499 905 885 857 972 538 572 8;
  • 58) 0.499 905 885 857 972 538 572 8 × 2 = 0 + 0.999 811 771 715 945 077 145 6;

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 9(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 9(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 9(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 9 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 -0.016 738 891 601 562 496 522 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: Magnetic Leadership: The Invisible Power of the Pull. NotebookLM YouTube Video of 11.31 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