-72.729 999 999 999 989 772 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -72.729 999 999 999 989 772 8(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
-72.729 999 999 999 989 772 8(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:

|-72.729 999 999 999 989 772 8| = 72.729 999 999 999 989 772 8


2. First, convert to binary (in base 2) the integer part: 72.
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;
  • 72 ÷ 2 = 36 + 0;
  • 36 ÷ 2 = 18 + 0;
  • 18 ÷ 2 = 9 + 0;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

72(10) =

100 1000(2)


4. Convert to binary (base 2) the fractional part: 0.729 999 999 999 989 772 8.

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) 0.729 999 999 999 989 772 8 × 2 = 1 + 0.459 999 999 999 979 545 6;
  • 2) 0.459 999 999 999 979 545 6 × 2 = 0 + 0.919 999 999 999 959 091 2;
  • 3) 0.919 999 999 999 959 091 2 × 2 = 1 + 0.839 999 999 999 918 182 4;
  • 4) 0.839 999 999 999 918 182 4 × 2 = 1 + 0.679 999 999 999 836 364 8;
  • 5) 0.679 999 999 999 836 364 8 × 2 = 1 + 0.359 999 999 999 672 729 6;
  • 6) 0.359 999 999 999 672 729 6 × 2 = 0 + 0.719 999 999 999 345 459 2;
  • 7) 0.719 999 999 999 345 459 2 × 2 = 1 + 0.439 999 999 998 690 918 4;
  • 8) 0.439 999 999 998 690 918 4 × 2 = 0 + 0.879 999 999 997 381 836 8;
  • 9) 0.879 999 999 997 381 836 8 × 2 = 1 + 0.759 999 999 994 763 673 6;
  • 10) 0.759 999 999 994 763 673 6 × 2 = 1 + 0.519 999 999 989 527 347 2;
  • 11) 0.519 999 999 989 527 347 2 × 2 = 1 + 0.039 999 999 979 054 694 4;
  • 12) 0.039 999 999 979 054 694 4 × 2 = 0 + 0.079 999 999 958 109 388 8;
  • 13) 0.079 999 999 958 109 388 8 × 2 = 0 + 0.159 999 999 916 218 777 6;
  • 14) 0.159 999 999 916 218 777 6 × 2 = 0 + 0.319 999 999 832 437 555 2;
  • 15) 0.319 999 999 832 437 555 2 × 2 = 0 + 0.639 999 999 664 875 110 4;
  • 16) 0.639 999 999 664 875 110 4 × 2 = 1 + 0.279 999 999 329 750 220 8;
  • 17) 0.279 999 999 329 750 220 8 × 2 = 0 + 0.559 999 998 659 500 441 6;
  • 18) 0.559 999 998 659 500 441 6 × 2 = 1 + 0.119 999 997 319 000 883 2;
  • 19) 0.119 999 997 319 000 883 2 × 2 = 0 + 0.239 999 994 638 001 766 4;
  • 20) 0.239 999 994 638 001 766 4 × 2 = 0 + 0.479 999 989 276 003 532 8;
  • 21) 0.479 999 989 276 003 532 8 × 2 = 0 + 0.959 999 978 552 007 065 6;
  • 22) 0.959 999 978 552 007 065 6 × 2 = 1 + 0.919 999 957 104 014 131 2;
  • 23) 0.919 999 957 104 014 131 2 × 2 = 1 + 0.839 999 914 208 028 262 4;
  • 24) 0.839 999 914 208 028 262 4 × 2 = 1 + 0.679 999 828 416 056 524 8;
  • 25) 0.679 999 828 416 056 524 8 × 2 = 1 + 0.359 999 656 832 113 049 6;
  • 26) 0.359 999 656 832 113 049 6 × 2 = 0 + 0.719 999 313 664 226 099 2;
  • 27) 0.719 999 313 664 226 099 2 × 2 = 1 + 0.439 998 627 328 452 198 4;
  • 28) 0.439 998 627 328 452 198 4 × 2 = 0 + 0.879 997 254 656 904 396 8;
  • 29) 0.879 997 254 656 904 396 8 × 2 = 1 + 0.759 994 509 313 808 793 6;
  • 30) 0.759 994 509 313 808 793 6 × 2 = 1 + 0.519 989 018 627 617 587 2;
  • 31) 0.519 989 018 627 617 587 2 × 2 = 1 + 0.039 978 037 255 235 174 4;
  • 32) 0.039 978 037 255 235 174 4 × 2 = 0 + 0.079 956 074 510 470 348 8;
  • 33) 0.079 956 074 510 470 348 8 × 2 = 0 + 0.159 912 149 020 940 697 6;
  • 34) 0.159 912 149 020 940 697 6 × 2 = 0 + 0.319 824 298 041 881 395 2;
  • 35) 0.319 824 298 041 881 395 2 × 2 = 0 + 0.639 648 596 083 762 790 4;
  • 36) 0.639 648 596 083 762 790 4 × 2 = 1 + 0.279 297 192 167 525 580 8;
  • 37) 0.279 297 192 167 525 580 8 × 2 = 0 + 0.558 594 384 335 051 161 6;
  • 38) 0.558 594 384 335 051 161 6 × 2 = 1 + 0.117 188 768 670 102 323 2;
  • 39) 0.117 188 768 670 102 323 2 × 2 = 0 + 0.234 377 537 340 204 646 4;
  • 40) 0.234 377 537 340 204 646 4 × 2 = 0 + 0.468 755 074 680 409 292 8;
  • 41) 0.468 755 074 680 409 292 8 × 2 = 0 + 0.937 510 149 360 818 585 6;
  • 42) 0.937 510 149 360 818 585 6 × 2 = 1 + 0.875 020 298 721 637 171 2;
  • 43) 0.875 020 298 721 637 171 2 × 2 = 1 + 0.750 040 597 443 274 342 4;
  • 44) 0.750 040 597 443 274 342 4 × 2 = 1 + 0.500 081 194 886 548 684 8;
  • 45) 0.500 081 194 886 548 684 8 × 2 = 1 + 0.000 162 389 773 097 369 6;
  • 46) 0.000 162 389 773 097 369 6 × 2 = 0 + 0.000 324 779 546 194 739 2;
  • 47) 0.000 324 779 546 194 739 2 × 2 = 0 + 0.000 649 559 092 389 478 4;
  • 48) 0.000 649 559 092 389 478 4 × 2 = 0 + 0.001 299 118 184 778 956 8;
  • 49) 0.001 299 118 184 778 956 8 × 2 = 0 + 0.002 598 236 369 557 913 6;
  • 50) 0.002 598 236 369 557 913 6 × 2 = 0 + 0.005 196 472 739 115 827 2;
  • 51) 0.005 196 472 739 115 827 2 × 2 = 0 + 0.010 392 945 478 231 654 4;
  • 52) 0.010 392 945 478 231 654 4 × 2 = 0 + 0.020 785 890 956 463 308 8;
  • 53) 0.020 785 890 956 463 308 8 × 2 = 0 + 0.041 571 781 912 926 617 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:


0.729 999 999 999 989 772 8(10) =

0.1011 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1000 0000 0(2)

6. Positive number before normalization:

72.729 999 999 999 989 772 8(10) =

100 1000.1011 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1000 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


72.729 999 999 999 989 772 8(10) =

100 1000.1011 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1000 0000 0(2) =

100 1000.1011 1010 1110 0001 0100 0111 1010 1110 0001 0100 0111 1000 0000 0(2) × 20 =

1.0010 0010 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 0000 000(2) × 26


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):
1.0010 0010 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 0000 000


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) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1029(10) =

100 0000 0101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0010 0010 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 000 0000 =

0010 0010 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 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) =
100 0000 0101

Mantissa (52 bits) =
0010 0010 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110


Decimal number -72.729 999 999 999 989 772 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0101 - 0010 0010 1110 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110

» Convert decimal -72.729 999 999 999 989 779 4 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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