0.000 053 174 916 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 053 174 916 7(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
0.000 053 174 916 7(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: 0.000 053 174 916 7.

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) 0.000 053 174 916 7 × 2 = 0 + 0.000 106 349 833 4;
  • 2) 0.000 106 349 833 4 × 2 = 0 + 0.000 212 699 666 8;
  • 3) 0.000 212 699 666 8 × 2 = 0 + 0.000 425 399 333 6;
  • 4) 0.000 425 399 333 6 × 2 = 0 + 0.000 850 798 667 2;
  • 5) 0.000 850 798 667 2 × 2 = 0 + 0.001 701 597 334 4;
  • 6) 0.001 701 597 334 4 × 2 = 0 + 0.003 403 194 668 8;
  • 7) 0.003 403 194 668 8 × 2 = 0 + 0.006 806 389 337 6;
  • 8) 0.006 806 389 337 6 × 2 = 0 + 0.013 612 778 675 2;
  • 9) 0.013 612 778 675 2 × 2 = 0 + 0.027 225 557 350 4;
  • 10) 0.027 225 557 350 4 × 2 = 0 + 0.054 451 114 700 8;
  • 11) 0.054 451 114 700 8 × 2 = 0 + 0.108 902 229 401 6;
  • 12) 0.108 902 229 401 6 × 2 = 0 + 0.217 804 458 803 2;
  • 13) 0.217 804 458 803 2 × 2 = 0 + 0.435 608 917 606 4;
  • 14) 0.435 608 917 606 4 × 2 = 0 + 0.871 217 835 212 8;
  • 15) 0.871 217 835 212 8 × 2 = 1 + 0.742 435 670 425 6;
  • 16) 0.742 435 670 425 6 × 2 = 1 + 0.484 871 340 851 2;
  • 17) 0.484 871 340 851 2 × 2 = 0 + 0.969 742 681 702 4;
  • 18) 0.969 742 681 702 4 × 2 = 1 + 0.939 485 363 404 8;
  • 19) 0.939 485 363 404 8 × 2 = 1 + 0.878 970 726 809 6;
  • 20) 0.878 970 726 809 6 × 2 = 1 + 0.757 941 453 619 2;
  • 21) 0.757 941 453 619 2 × 2 = 1 + 0.515 882 907 238 4;
  • 22) 0.515 882 907 238 4 × 2 = 1 + 0.031 765 814 476 8;
  • 23) 0.031 765 814 476 8 × 2 = 0 + 0.063 531 628 953 6;
  • 24) 0.063 531 628 953 6 × 2 = 0 + 0.127 063 257 907 2;
  • 25) 0.127 063 257 907 2 × 2 = 0 + 0.254 126 515 814 4;
  • 26) 0.254 126 515 814 4 × 2 = 0 + 0.508 253 031 628 8;
  • 27) 0.508 253 031 628 8 × 2 = 1 + 0.016 506 063 257 6;
  • 28) 0.016 506 063 257 6 × 2 = 0 + 0.033 012 126 515 2;
  • 29) 0.033 012 126 515 2 × 2 = 0 + 0.066 024 253 030 4;
  • 30) 0.066 024 253 030 4 × 2 = 0 + 0.132 048 506 060 8;
  • 31) 0.132 048 506 060 8 × 2 = 0 + 0.264 097 012 121 6;
  • 32) 0.264 097 012 121 6 × 2 = 0 + 0.528 194 024 243 2;
  • 33) 0.528 194 024 243 2 × 2 = 1 + 0.056 388 048 486 4;
  • 34) 0.056 388 048 486 4 × 2 = 0 + 0.112 776 096 972 8;
  • 35) 0.112 776 096 972 8 × 2 = 0 + 0.225 552 193 945 6;
  • 36) 0.225 552 193 945 6 × 2 = 0 + 0.451 104 387 891 2;
  • 37) 0.451 104 387 891 2 × 2 = 0 + 0.902 208 775 782 4;
  • 38) 0.902 208 775 782 4 × 2 = 1 + 0.804 417 551 564 8;
  • 39) 0.804 417 551 564 8 × 2 = 1 + 0.608 835 103 129 6;
  • 40) 0.608 835 103 129 6 × 2 = 1 + 0.217 670 206 259 2;
  • 41) 0.217 670 206 259 2 × 2 = 0 + 0.435 340 412 518 4;
  • 42) 0.435 340 412 518 4 × 2 = 0 + 0.870 680 825 036 8;
  • 43) 0.870 680 825 036 8 × 2 = 1 + 0.741 361 650 073 6;
  • 44) 0.741 361 650 073 6 × 2 = 1 + 0.482 723 300 147 2;
  • 45) 0.482 723 300 147 2 × 2 = 0 + 0.965 446 600 294 4;
  • 46) 0.965 446 600 294 4 × 2 = 1 + 0.930 893 200 588 8;
  • 47) 0.930 893 200 588 8 × 2 = 1 + 0.861 786 401 177 6;
  • 48) 0.861 786 401 177 6 × 2 = 1 + 0.723 572 802 355 2;
  • 49) 0.723 572 802 355 2 × 2 = 1 + 0.447 145 604 710 4;
  • 50) 0.447 145 604 710 4 × 2 = 0 + 0.894 291 209 420 8;
  • 51) 0.894 291 209 420 8 × 2 = 1 + 0.788 582 418 841 6;
  • 52) 0.788 582 418 841 6 × 2 = 1 + 0.577 164 837 683 2;
  • 53) 0.577 164 837 683 2 × 2 = 1 + 0.154 329 675 366 4;
  • 54) 0.154 329 675 366 4 × 2 = 0 + 0.308 659 350 732 8;
  • 55) 0.308 659 350 732 8 × 2 = 0 + 0.617 318 701 465 6;
  • 56) 0.617 318 701 465 6 × 2 = 1 + 0.234 637 402 931 2;
  • 57) 0.234 637 402 931 2 × 2 = 0 + 0.469 274 805 862 4;
  • 58) 0.469 274 805 862 4 × 2 = 0 + 0.938 549 611 724 8;
  • 59) 0.938 549 611 724 8 × 2 = 1 + 0.877 099 223 449 6;
  • 60) 0.877 099 223 449 6 × 2 = 1 + 0.754 198 446 899 2;
  • 61) 0.754 198 446 899 2 × 2 = 1 + 0.508 396 893 798 4;
  • 62) 0.508 396 893 798 4 × 2 = 1 + 0.016 793 787 596 8;
  • 63) 0.016 793 787 596 8 × 2 = 0 + 0.033 587 575 193 6;
  • 64) 0.033 587 575 193 6 × 2 = 0 + 0.067 175 150 387 2;
  • 65) 0.067 175 150 387 2 × 2 = 0 + 0.134 350 300 774 4;
  • 66) 0.134 350 300 774 4 × 2 = 0 + 0.268 700 601 548 8;
  • 67) 0.268 700 601 548 8 × 2 = 0 + 0.537 401 203 097 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).


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:


0.000 053 174 916 7(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0111 0011 0111 1011 1001 0011 1100 000(2)

5. Positive number before normalization:

0.000 053 174 916 7(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0111 0011 0111 1011 1001 0011 1100 000(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:


0.000 053 174 916 7(10) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0111 0011 0111 1011 1001 0011 1100 000(2) =

0.0000 0000 0000 0011 0111 1100 0010 0000 1000 0111 0011 0111 1011 1001 0011 1100 000(2) × 20 =

1.1011 1110 0001 0000 0100 0011 1001 1011 1101 1100 1001 1110 0000(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 1001 1011 1101 1100 1001 1110 0000


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 1001 1011 1101 1100 1001 1110 0000 =

1011 1110 0001 0000 0100 0011 1001 1011 1101 1100 1001 1110 0000


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 1001 1011 1101 1100 1001 1110 0000


Decimal number 0.000 053 174 916 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1011 1110 0001 0000 0100 0011 1001 1011 1101 1100 1001 1110 0000

» Convert decimal 0.000 053 174 914 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