-0.016 738 891 601 562 496 456 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 456(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
-0.016 738 891 601 562 496 456(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:

|-0.016 738 891 601 562 496 456| = 0.016 738 891 601 562 496 456


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: 0.016 738 891 601 562 496 456.

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) 0.016 738 891 601 562 496 456 × 2 = 0 + 0.033 477 783 203 124 992 912;
  • 2) 0.033 477 783 203 124 992 912 × 2 = 0 + 0.066 955 566 406 249 985 824;
  • 3) 0.066 955 566 406 249 985 824 × 2 = 0 + 0.133 911 132 812 499 971 648;
  • 4) 0.133 911 132 812 499 971 648 × 2 = 0 + 0.267 822 265 624 999 943 296;
  • 5) 0.267 822 265 624 999 943 296 × 2 = 0 + 0.535 644 531 249 999 886 592;
  • 6) 0.535 644 531 249 999 886 592 × 2 = 1 + 0.071 289 062 499 999 773 184;
  • 7) 0.071 289 062 499 999 773 184 × 2 = 0 + 0.142 578 124 999 999 546 368;
  • 8) 0.142 578 124 999 999 546 368 × 2 = 0 + 0.285 156 249 999 999 092 736;
  • 9) 0.285 156 249 999 999 092 736 × 2 = 0 + 0.570 312 499 999 998 185 472;
  • 10) 0.570 312 499 999 998 185 472 × 2 = 1 + 0.140 624 999 999 996 370 944;
  • 11) 0.140 624 999 999 996 370 944 × 2 = 0 + 0.281 249 999 999 992 741 888;
  • 12) 0.281 249 999 999 992 741 888 × 2 = 0 + 0.562 499 999 999 985 483 776;
  • 13) 0.562 499 999 999 985 483 776 × 2 = 1 + 0.124 999 999 999 970 967 552;
  • 14) 0.124 999 999 999 970 967 552 × 2 = 0 + 0.249 999 999 999 941 935 104;
  • 15) 0.249 999 999 999 941 935 104 × 2 = 0 + 0.499 999 999 999 883 870 208;
  • 16) 0.499 999 999 999 883 870 208 × 2 = 0 + 0.999 999 999 999 767 740 416;
  • 17) 0.999 999 999 999 767 740 416 × 2 = 1 + 0.999 999 999 999 535 480 832;
  • 18) 0.999 999 999 999 535 480 832 × 2 = 1 + 0.999 999 999 999 070 961 664;
  • 19) 0.999 999 999 999 070 961 664 × 2 = 1 + 0.999 999 999 998 141 923 328;
  • 20) 0.999 999 999 998 141 923 328 × 2 = 1 + 0.999 999 999 996 283 846 656;
  • 21) 0.999 999 999 996 283 846 656 × 2 = 1 + 0.999 999 999 992 567 693 312;
  • 22) 0.999 999 999 992 567 693 312 × 2 = 1 + 0.999 999 999 985 135 386 624;
  • 23) 0.999 999 999 985 135 386 624 × 2 = 1 + 0.999 999 999 970 270 773 248;
  • 24) 0.999 999 999 970 270 773 248 × 2 = 1 + 0.999 999 999 940 541 546 496;
  • 25) 0.999 999 999 940 541 546 496 × 2 = 1 + 0.999 999 999 881 083 092 992;
  • 26) 0.999 999 999 881 083 092 992 × 2 = 1 + 0.999 999 999 762 166 185 984;
  • 27) 0.999 999 999 762 166 185 984 × 2 = 1 + 0.999 999 999 524 332 371 968;
  • 28) 0.999 999 999 524 332 371 968 × 2 = 1 + 0.999 999 999 048 664 743 936;
  • 29) 0.999 999 999 048 664 743 936 × 2 = 1 + 0.999 999 998 097 329 487 872;
  • 30) 0.999 999 998 097 329 487 872 × 2 = 1 + 0.999 999 996 194 658 975 744;
  • 31) 0.999 999 996 194 658 975 744 × 2 = 1 + 0.999 999 992 389 317 951 488;
  • 32) 0.999 999 992 389 317 951 488 × 2 = 1 + 0.999 999 984 778 635 902 976;
  • 33) 0.999 999 984 778 635 902 976 × 2 = 1 + 0.999 999 969 557 271 805 952;
  • 34) 0.999 999 969 557 271 805 952 × 2 = 1 + 0.999 999 939 114 543 611 904;
  • 35) 0.999 999 939 114 543 611 904 × 2 = 1 + 0.999 999 878 229 087 223 808;
  • 36) 0.999 999 878 229 087 223 808 × 2 = 1 + 0.999 999 756 458 174 447 616;
  • 37) 0.999 999 756 458 174 447 616 × 2 = 1 + 0.999 999 512 916 348 895 232;
  • 38) 0.999 999 512 916 348 895 232 × 2 = 1 + 0.999 999 025 832 697 790 464;
  • 39) 0.999 999 025 832 697 790 464 × 2 = 1 + 0.999 998 051 665 395 580 928;
  • 40) 0.999 998 051 665 395 580 928 × 2 = 1 + 0.999 996 103 330 791 161 856;
  • 41) 0.999 996 103 330 791 161 856 × 2 = 1 + 0.999 992 206 661 582 323 712;
  • 42) 0.999 992 206 661 582 323 712 × 2 = 1 + 0.999 984 413 323 164 647 424;
  • 43) 0.999 984 413 323 164 647 424 × 2 = 1 + 0.999 968 826 646 329 294 848;
  • 44) 0.999 968 826 646 329 294 848 × 2 = 1 + 0.999 937 653 292 658 589 696;
  • 45) 0.999 937 653 292 658 589 696 × 2 = 1 + 0.999 875 306 585 317 179 392;
  • 46) 0.999 875 306 585 317 179 392 × 2 = 1 + 0.999 750 613 170 634 358 784;
  • 47) 0.999 750 613 170 634 358 784 × 2 = 1 + 0.999 501 226 341 268 717 568;
  • 48) 0.999 501 226 341 268 717 568 × 2 = 1 + 0.999 002 452 682 537 435 136;
  • 49) 0.999 002 452 682 537 435 136 × 2 = 1 + 0.998 004 905 365 074 870 272;
  • 50) 0.998 004 905 365 074 870 272 × 2 = 1 + 0.996 009 810 730 149 740 544;
  • 51) 0.996 009 810 730 149 740 544 × 2 = 1 + 0.992 019 621 460 299 481 088;
  • 52) 0.992 019 621 460 299 481 088 × 2 = 1 + 0.984 039 242 920 598 962 176;
  • 53) 0.984 039 242 920 598 962 176 × 2 = 1 + 0.968 078 485 841 197 924 352;
  • 54) 0.968 078 485 841 197 924 352 × 2 = 1 + 0.936 156 971 682 395 848 704;
  • 55) 0.936 156 971 682 395 848 704 × 2 = 1 + 0.872 313 943 364 791 697 408;
  • 56) 0.872 313 943 364 791 697 408 × 2 = 1 + 0.744 627 886 729 583 394 816;
  • 57) 0.744 627 886 729 583 394 816 × 2 = 1 + 0.489 255 773 459 166 789 632;
  • 58) 0.489 255 773 459 166 789 632 × 2 = 0 + 0.978 511 546 918 333 579 264;

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:


0.016 738 891 601 562 496 456(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 456(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:


0.016 738 891 601 562 496 456(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 -0.016 738 891 601 562 496 456 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 -0.016 738 891 601 562 496 549 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: Pull vs. Push Leadership: The Invisible Power. 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