0.000 053 174 915 595 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 174 915 595 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
0.000 053 174 915 595 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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 053 174 915 595 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) 0.000 053 174 915 595 9 × 2 = 0 + 0.000 106 349 831 191 8;
  • 2) 0.000 106 349 831 191 8 × 2 = 0 + 0.000 212 699 662 383 6;
  • 3) 0.000 212 699 662 383 6 × 2 = 0 + 0.000 425 399 324 767 2;
  • 4) 0.000 425 399 324 767 2 × 2 = 0 + 0.000 850 798 649 534 4;
  • 5) 0.000 850 798 649 534 4 × 2 = 0 + 0.001 701 597 299 068 8;
  • 6) 0.001 701 597 299 068 8 × 2 = 0 + 0.003 403 194 598 137 6;
  • 7) 0.003 403 194 598 137 6 × 2 = 0 + 0.006 806 389 196 275 2;
  • 8) 0.006 806 389 196 275 2 × 2 = 0 + 0.013 612 778 392 550 4;
  • 9) 0.013 612 778 392 550 4 × 2 = 0 + 0.027 225 556 785 100 8;
  • 10) 0.027 225 556 785 100 8 × 2 = 0 + 0.054 451 113 570 201 6;
  • 11) 0.054 451 113 570 201 6 × 2 = 0 + 0.108 902 227 140 403 2;
  • 12) 0.108 902 227 140 403 2 × 2 = 0 + 0.217 804 454 280 806 4;
  • 13) 0.217 804 454 280 806 4 × 2 = 0 + 0.435 608 908 561 612 8;
  • 14) 0.435 608 908 561 612 8 × 2 = 0 + 0.871 217 817 123 225 6;
  • 15) 0.871 217 817 123 225 6 × 2 = 1 + 0.742 435 634 246 451 2;
  • 16) 0.742 435 634 246 451 2 × 2 = 1 + 0.484 871 268 492 902 4;
  • 17) 0.484 871 268 492 902 4 × 2 = 0 + 0.969 742 536 985 804 8;
  • 18) 0.969 742 536 985 804 8 × 2 = 1 + 0.939 485 073 971 609 6;
  • 19) 0.939 485 073 971 609 6 × 2 = 1 + 0.878 970 147 943 219 2;
  • 20) 0.878 970 147 943 219 2 × 2 = 1 + 0.757 940 295 886 438 4;
  • 21) 0.757 940 295 886 438 4 × 2 = 1 + 0.515 880 591 772 876 8;
  • 22) 0.515 880 591 772 876 8 × 2 = 1 + 0.031 761 183 545 753 6;
  • 23) 0.031 761 183 545 753 6 × 2 = 0 + 0.063 522 367 091 507 2;
  • 24) 0.063 522 367 091 507 2 × 2 = 0 + 0.127 044 734 183 014 4;
  • 25) 0.127 044 734 183 014 4 × 2 = 0 + 0.254 089 468 366 028 8;
  • 26) 0.254 089 468 366 028 8 × 2 = 0 + 0.508 178 936 732 057 6;
  • 27) 0.508 178 936 732 057 6 × 2 = 1 + 0.016 357 873 464 115 2;
  • 28) 0.016 357 873 464 115 2 × 2 = 0 + 0.032 715 746 928 230 4;
  • 29) 0.032 715 746 928 230 4 × 2 = 0 + 0.065 431 493 856 460 8;
  • 30) 0.065 431 493 856 460 8 × 2 = 0 + 0.130 862 987 712 921 6;
  • 31) 0.130 862 987 712 921 6 × 2 = 0 + 0.261 725 975 425 843 2;
  • 32) 0.261 725 975 425 843 2 × 2 = 0 + 0.523 451 950 851 686 4;
  • 33) 0.523 451 950 851 686 4 × 2 = 1 + 0.046 903 901 703 372 8;
  • 34) 0.046 903 901 703 372 8 × 2 = 0 + 0.093 807 803 406 745 6;
  • 35) 0.093 807 803 406 745 6 × 2 = 0 + 0.187 615 606 813 491 2;
  • 36) 0.187 615 606 813 491 2 × 2 = 0 + 0.375 231 213 626 982 4;
  • 37) 0.375 231 213 626 982 4 × 2 = 0 + 0.750 462 427 253 964 8;
  • 38) 0.750 462 427 253 964 8 × 2 = 1 + 0.500 924 854 507 929 6;
  • 39) 0.500 924 854 507 929 6 × 2 = 1 + 0.001 849 709 015 859 2;
  • 40) 0.001 849 709 015 859 2 × 2 = 0 + 0.003 699 418 031 718 4;
  • 41) 0.003 699 418 031 718 4 × 2 = 0 + 0.007 398 836 063 436 8;
  • 42) 0.007 398 836 063 436 8 × 2 = 0 + 0.014 797 672 126 873 6;
  • 43) 0.014 797 672 126 873 6 × 2 = 0 + 0.029 595 344 253 747 2;
  • 44) 0.029 595 344 253 747 2 × 2 = 0 + 0.059 190 688 507 494 4;
  • 45) 0.059 190 688 507 494 4 × 2 = 0 + 0.118 381 377 014 988 8;
  • 46) 0.118 381 377 014 988 8 × 2 = 0 + 0.236 762 754 029 977 6;
  • 47) 0.236 762 754 029 977 6 × 2 = 0 + 0.473 525 508 059 955 2;
  • 48) 0.473 525 508 059 955 2 × 2 = 0 + 0.947 051 016 119 910 4;
  • 49) 0.947 051 016 119 910 4 × 2 = 1 + 0.894 102 032 239 820 8;
  • 50) 0.894 102 032 239 820 8 × 2 = 1 + 0.788 204 064 479 641 6;
  • 51) 0.788 204 064 479 641 6 × 2 = 1 + 0.576 408 128 959 283 2;
  • 52) 0.576 408 128 959 283 2 × 2 = 1 + 0.152 816 257 918 566 4;
  • 53) 0.152 816 257 918 566 4 × 2 = 0 + 0.305 632 515 837 132 8;
  • 54) 0.305 632 515 837 132 8 × 2 = 0 + 0.611 265 031 674 265 6;
  • 55) 0.611 265 031 674 265 6 × 2 = 1 + 0.222 530 063 348 531 2;
  • 56) 0.222 530 063 348 531 2 × 2 = 0 + 0.445 060 126 697 062 4;
  • 57) 0.445 060 126 697 062 4 × 2 = 0 + 0.890 120 253 394 124 8;
  • 58) 0.890 120 253 394 124 8 × 2 = 1 + 0.780 240 506 788 249 6;
  • 59) 0.780 240 506 788 249 6 × 2 = 1 + 0.560 481 013 576 499 2;
  • 60) 0.560 481 013 576 499 2 × 2 = 1 + 0.120 962 027 152 998 4;
  • 61) 0.120 962 027 152 998 4 × 2 = 0 + 0.241 924 054 305 996 8;
  • 62) 0.241 924 054 305 996 8 × 2 = 0 + 0.483 848 108 611 993 6;
  • 63) 0.483 848 108 611 993 6 × 2 = 0 + 0.967 696 217 223 987 2;
  • 64) 0.967 696 217 223 987 2 × 2 = 1 + 0.935 392 434 447 974 4;
  • 65) 0.935 392 434 447 974 4 × 2 = 1 + 0.870 784 868 895 948 8;
  • 66) 0.870 784 868 895 948 8 × 2 = 1 + 0.741 569 737 791 897 6;
  • 67) 0.741 569 737 791 897 6 × 2 = 1 + 0.483 139 475 583 795 2;

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).


4. 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:


0.000 053 174 915 595 9(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0110 0000 0000 1111 0010 0111 0001 111(2)

5. Positive number before normalization:

0.000 053 174 915 595 9(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0110 0000 0000 1111 0010 0111 0001 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 053 174 915 595 9(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0110 0000 0000 1111 0010 0111 0001 111(2) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0110 0000 0000 1111 0010 0111 0001 111(2) × 20 =

1.1011 1110 0001 0000 0100 0011 0000 0000 0111 1001 0011 1000 1111(2) × 2-15


7. 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 0 (a positive number)

Exponent (unadjusted): -15

Mantissa (not normalized):
1.1011 1110 0001 0000 0100 0011 0000 0000 0111 1001 0011 1000 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(10)


9. 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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1008(10) =

011 1111 0000(2)


11. 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. 1011 1110 0001 0000 0100 0011 0000 0000 0111 1001 0011 1000 1111 =

1011 1110 0001 0000 0100 0011 0000 0000 0111 1001 0011 1000 1111


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 0000

Mantissa (52 bits) =
1011 1110 0001 0000 0100 0011 0000 0000 0111 1001 0011 1000 1111


Decimal number 0.000 053 174 915 595 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1110 0001 0000 0100 0011 0000 0000 0111 1001 0011 1000 1111

Share

» Convert decimal 0.000 053 174 915 598 8 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 01, 2026 [January]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: January - 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