-0.016 738 891 601 562 496 521 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 521(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
-0.016 738 891 601 562 496 521(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:

|-0.016 738 891 601 562 496 521| = 0.016 738 891 601 562 496 521


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: 0.016 738 891 601 562 496 521.

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) 0.016 738 891 601 562 496 521 × 2 = 0 + 0.033 477 783 203 124 993 042;
  • 2) 0.033 477 783 203 124 993 042 × 2 = 0 + 0.066 955 566 406 249 986 084;
  • 3) 0.066 955 566 406 249 986 084 × 2 = 0 + 0.133 911 132 812 499 972 168;
  • 4) 0.133 911 132 812 499 972 168 × 2 = 0 + 0.267 822 265 624 999 944 336;
  • 5) 0.267 822 265 624 999 944 336 × 2 = 0 + 0.535 644 531 249 999 888 672;
  • 6) 0.535 644 531 249 999 888 672 × 2 = 1 + 0.071 289 062 499 999 777 344;
  • 7) 0.071 289 062 499 999 777 344 × 2 = 0 + 0.142 578 124 999 999 554 688;
  • 8) 0.142 578 124 999 999 554 688 × 2 = 0 + 0.285 156 249 999 999 109 376;
  • 9) 0.285 156 249 999 999 109 376 × 2 = 0 + 0.570 312 499 999 998 218 752;
  • 10) 0.570 312 499 999 998 218 752 × 2 = 1 + 0.140 624 999 999 996 437 504;
  • 11) 0.140 624 999 999 996 437 504 × 2 = 0 + 0.281 249 999 999 992 875 008;
  • 12) 0.281 249 999 999 992 875 008 × 2 = 0 + 0.562 499 999 999 985 750 016;
  • 13) 0.562 499 999 999 985 750 016 × 2 = 1 + 0.124 999 999 999 971 500 032;
  • 14) 0.124 999 999 999 971 500 032 × 2 = 0 + 0.249 999 999 999 943 000 064;
  • 15) 0.249 999 999 999 943 000 064 × 2 = 0 + 0.499 999 999 999 886 000 128;
  • 16) 0.499 999 999 999 886 000 128 × 2 = 0 + 0.999 999 999 999 772 000 256;
  • 17) 0.999 999 999 999 772 000 256 × 2 = 1 + 0.999 999 999 999 544 000 512;
  • 18) 0.999 999 999 999 544 000 512 × 2 = 1 + 0.999 999 999 999 088 001 024;
  • 19) 0.999 999 999 999 088 001 024 × 2 = 1 + 0.999 999 999 998 176 002 048;
  • 20) 0.999 999 999 998 176 002 048 × 2 = 1 + 0.999 999 999 996 352 004 096;
  • 21) 0.999 999 999 996 352 004 096 × 2 = 1 + 0.999 999 999 992 704 008 192;
  • 22) 0.999 999 999 992 704 008 192 × 2 = 1 + 0.999 999 999 985 408 016 384;
  • 23) 0.999 999 999 985 408 016 384 × 2 = 1 + 0.999 999 999 970 816 032 768;
  • 24) 0.999 999 999 970 816 032 768 × 2 = 1 + 0.999 999 999 941 632 065 536;
  • 25) 0.999 999 999 941 632 065 536 × 2 = 1 + 0.999 999 999 883 264 131 072;
  • 26) 0.999 999 999 883 264 131 072 × 2 = 1 + 0.999 999 999 766 528 262 144;
  • 27) 0.999 999 999 766 528 262 144 × 2 = 1 + 0.999 999 999 533 056 524 288;
  • 28) 0.999 999 999 533 056 524 288 × 2 = 1 + 0.999 999 999 066 113 048 576;
  • 29) 0.999 999 999 066 113 048 576 × 2 = 1 + 0.999 999 998 132 226 097 152;
  • 30) 0.999 999 998 132 226 097 152 × 2 = 1 + 0.999 999 996 264 452 194 304;
  • 31) 0.999 999 996 264 452 194 304 × 2 = 1 + 0.999 999 992 528 904 388 608;
  • 32) 0.999 999 992 528 904 388 608 × 2 = 1 + 0.999 999 985 057 808 777 216;
  • 33) 0.999 999 985 057 808 777 216 × 2 = 1 + 0.999 999 970 115 617 554 432;
  • 34) 0.999 999 970 115 617 554 432 × 2 = 1 + 0.999 999 940 231 235 108 864;
  • 35) 0.999 999 940 231 235 108 864 × 2 = 1 + 0.999 999 880 462 470 217 728;
  • 36) 0.999 999 880 462 470 217 728 × 2 = 1 + 0.999 999 760 924 940 435 456;
  • 37) 0.999 999 760 924 940 435 456 × 2 = 1 + 0.999 999 521 849 880 870 912;
  • 38) 0.999 999 521 849 880 870 912 × 2 = 1 + 0.999 999 043 699 761 741 824;
  • 39) 0.999 999 043 699 761 741 824 × 2 = 1 + 0.999 998 087 399 523 483 648;
  • 40) 0.999 998 087 399 523 483 648 × 2 = 1 + 0.999 996 174 799 046 967 296;
  • 41) 0.999 996 174 799 046 967 296 × 2 = 1 + 0.999 992 349 598 093 934 592;
  • 42) 0.999 992 349 598 093 934 592 × 2 = 1 + 0.999 984 699 196 187 869 184;
  • 43) 0.999 984 699 196 187 869 184 × 2 = 1 + 0.999 969 398 392 375 738 368;
  • 44) 0.999 969 398 392 375 738 368 × 2 = 1 + 0.999 938 796 784 751 476 736;
  • 45) 0.999 938 796 784 751 476 736 × 2 = 1 + 0.999 877 593 569 502 953 472;
  • 46) 0.999 877 593 569 502 953 472 × 2 = 1 + 0.999 755 187 139 005 906 944;
  • 47) 0.999 755 187 139 005 906 944 × 2 = 1 + 0.999 510 374 278 011 813 888;
  • 48) 0.999 510 374 278 011 813 888 × 2 = 1 + 0.999 020 748 556 023 627 776;
  • 49) 0.999 020 748 556 023 627 776 × 2 = 1 + 0.998 041 497 112 047 255 552;
  • 50) 0.998 041 497 112 047 255 552 × 2 = 1 + 0.996 082 994 224 094 511 104;
  • 51) 0.996 082 994 224 094 511 104 × 2 = 1 + 0.992 165 988 448 189 022 208;
  • 52) 0.992 165 988 448 189 022 208 × 2 = 1 + 0.984 331 976 896 378 044 416;
  • 53) 0.984 331 976 896 378 044 416 × 2 = 1 + 0.968 663 953 792 756 088 832;
  • 54) 0.968 663 953 792 756 088 832 × 2 = 1 + 0.937 327 907 585 512 177 664;
  • 55) 0.937 327 907 585 512 177 664 × 2 = 1 + 0.874 655 815 171 024 355 328;
  • 56) 0.874 655 815 171 024 355 328 × 2 = 1 + 0.749 311 630 342 048 710 656;
  • 57) 0.749 311 630 342 048 710 656 × 2 = 1 + 0.498 623 260 684 097 421 312;
  • 58) 0.498 623 260 684 097 421 312 × 2 = 0 + 0.997 246 521 368 194 842 624;

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:


0.016 738 891 601 562 496 521(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 521(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:


0.016 738 891 601 562 496 521(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 -0.016 738 891 601 562 496 521 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 -0.016 738 891 601 562 496 589 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 Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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