Convert 0.000 003 669 410 95 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.000 003 669 410 95(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 003 669 410 95.

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) 0.000 003 669 410 95 × 2 = 0 + 0.000 007 338 821 9;
  • 2) 0.000 007 338 821 9 × 2 = 0 + 0.000 014 677 643 8;
  • 3) 0.000 014 677 643 8 × 2 = 0 + 0.000 029 355 287 6;
  • 4) 0.000 029 355 287 6 × 2 = 0 + 0.000 058 710 575 2;
  • 5) 0.000 058 710 575 2 × 2 = 0 + 0.000 117 421 150 4;
  • 6) 0.000 117 421 150 4 × 2 = 0 + 0.000 234 842 300 8;
  • 7) 0.000 234 842 300 8 × 2 = 0 + 0.000 469 684 601 6;
  • 8) 0.000 469 684 601 6 × 2 = 0 + 0.000 939 369 203 2;
  • 9) 0.000 939 369 203 2 × 2 = 0 + 0.001 878 738 406 4;
  • 10) 0.001 878 738 406 4 × 2 = 0 + 0.003 757 476 812 8;
  • 11) 0.003 757 476 812 8 × 2 = 0 + 0.007 514 953 625 6;
  • 12) 0.007 514 953 625 6 × 2 = 0 + 0.015 029 907 251 2;
  • 13) 0.015 029 907 251 2 × 2 = 0 + 0.030 059 814 502 4;
  • 14) 0.030 059 814 502 4 × 2 = 0 + 0.060 119 629 004 8;
  • 15) 0.060 119 629 004 8 × 2 = 0 + 0.120 239 258 009 6;
  • 16) 0.120 239 258 009 6 × 2 = 0 + 0.240 478 516 019 2;
  • 17) 0.240 478 516 019 2 × 2 = 0 + 0.480 957 032 038 4;
  • 18) 0.480 957 032 038 4 × 2 = 0 + 0.961 914 064 076 8;
  • 19) 0.961 914 064 076 8 × 2 = 1 + 0.923 828 128 153 6;
  • 20) 0.923 828 128 153 6 × 2 = 1 + 0.847 656 256 307 2;
  • 21) 0.847 656 256 307 2 × 2 = 1 + 0.695 312 512 614 4;
  • 22) 0.695 312 512 614 4 × 2 = 1 + 0.390 625 025 228 8;
  • 23) 0.390 625 025 228 8 × 2 = 0 + 0.781 250 050 457 6;
  • 24) 0.781 250 050 457 6 × 2 = 1 + 0.562 500 100 915 2;
  • 25) 0.562 500 100 915 2 × 2 = 1 + 0.125 000 201 830 4;
  • 26) 0.125 000 201 830 4 × 2 = 0 + 0.250 000 403 660 8;
  • 27) 0.250 000 403 660 8 × 2 = 0 + 0.500 000 807 321 6;
  • 28) 0.500 000 807 321 6 × 2 = 1 + 0.000 001 614 643 2;
  • 29) 0.000 001 614 643 2 × 2 = 0 + 0.000 003 229 286 4;
  • 30) 0.000 003 229 286 4 × 2 = 0 + 0.000 006 458 572 8;
  • 31) 0.000 006 458 572 8 × 2 = 0 + 0.000 012 917 145 6;
  • 32) 0.000 012 917 145 6 × 2 = 0 + 0.000 025 834 291 2;
  • 33) 0.000 025 834 291 2 × 2 = 0 + 0.000 051 668 582 4;
  • 34) 0.000 051 668 582 4 × 2 = 0 + 0.000 103 337 164 8;
  • 35) 0.000 103 337 164 8 × 2 = 0 + 0.000 206 674 329 6;
  • 36) 0.000 206 674 329 6 × 2 = 0 + 0.000 413 348 659 2;
  • 37) 0.000 413 348 659 2 × 2 = 0 + 0.000 826 697 318 4;
  • 38) 0.000 826 697 318 4 × 2 = 0 + 0.001 653 394 636 8;
  • 39) 0.001 653 394 636 8 × 2 = 0 + 0.003 306 789 273 6;
  • 40) 0.003 306 789 273 6 × 2 = 0 + 0.006 613 578 547 2;
  • 41) 0.006 613 578 547 2 × 2 = 0 + 0.013 227 157 094 4;
  • 42) 0.013 227 157 094 4 × 2 = 0 + 0.026 454 314 188 8;
  • 43) 0.026 454 314 188 8 × 2 = 0 + 0.052 908 628 377 6;
  • 44) 0.052 908 628 377 6 × 2 = 0 + 0.105 817 256 755 2;
  • 45) 0.105 817 256 755 2 × 2 = 0 + 0.211 634 513 510 4;
  • 46) 0.211 634 513 510 4 × 2 = 0 + 0.423 269 027 020 8;
  • 47) 0.423 269 027 020 8 × 2 = 0 + 0.846 538 054 041 6;
  • 48) 0.846 538 054 041 6 × 2 = 1 + 0.693 076 108 083 2;
  • 49) 0.693 076 108 083 2 × 2 = 1 + 0.386 152 216 166 4;
  • 50) 0.386 152 216 166 4 × 2 = 0 + 0.772 304 432 332 8;
  • 51) 0.772 304 432 332 8 × 2 = 1 + 0.544 608 864 665 6;
  • 52) 0.544 608 864 665 6 × 2 = 1 + 0.089 217 729 331 2;
  • 53) 0.089 217 729 331 2 × 2 = 0 + 0.178 435 458 662 4;
  • 54) 0.178 435 458 662 4 × 2 = 0 + 0.356 870 917 324 8;
  • 55) 0.356 870 917 324 8 × 2 = 0 + 0.713 741 834 649 6;
  • 56) 0.713 741 834 649 6 × 2 = 1 + 0.427 483 669 299 2;
  • 57) 0.427 483 669 299 2 × 2 = 0 + 0.854 967 338 598 4;
  • 58) 0.854 967 338 598 4 × 2 = 1 + 0.709 934 677 196 8;
  • 59) 0.709 934 677 196 8 × 2 = 1 + 0.419 869 354 393 6;
  • 60) 0.419 869 354 393 6 × 2 = 0 + 0.839 738 708 787 2;
  • 61) 0.839 738 708 787 2 × 2 = 1 + 0.679 477 417 574 4;
  • 62) 0.679 477 417 574 4 × 2 = 1 + 0.358 954 835 148 8;
  • 63) 0.358 954 835 148 8 × 2 = 0 + 0.717 909 670 297 6;
  • 64) 0.717 909 670 297 6 × 2 = 1 + 0.435 819 340 595 2;
  • 65) 0.435 819 340 595 2 × 2 = 0 + 0.871 638 681 190 4;
  • 66) 0.871 638 681 190 4 × 2 = 1 + 0.743 277 362 380 8;
  • 67) 0.743 277 362 380 8 × 2 = 1 + 0.486 554 724 761 6;
  • 68) 0.486 554 724 761 6 × 2 = 0 + 0.973 109 449 523 2;
  • 69) 0.973 109 449 523 2 × 2 = 1 + 0.946 218 899 046 4;
  • 70) 0.946 218 899 046 4 × 2 = 1 + 0.892 437 798 092 8;
  • 71) 0.892 437 798 092 8 × 2 = 1 + 0.784 875 596 185 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...)


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:

0.000 003 669 410 95(10) =


0.0000 0000 0000 0000 0011 1101 1001 0000 0000 0000 0000 0001 1011 0001 0110 1101 0110 111(2)


5. Positive number before normalization:

0.000 003 669 410 95(10) =


0.0000 0000 0000 0000 0011 1101 1001 0000 0000 0000 0000 0001 1011 0001 0110 1101 0110 111(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the right so that only one non zero digit remains to the left of it:

0.000 003 669 410 95(10) =


0.0000 0000 0000 0000 0011 1101 1001 0000 0000 0000 0000 0001 1011 0001 0110 1101 0110 111(2) =


0.0000 0000 0000 0000 0011 1101 1001 0000 0000 0000 0000 0001 1011 0001 0110 1101 0110 111(2) × 20 =


1.1110 1100 1000 0000 0000 0000 0000 1101 1000 1011 0110 1011 0111(2) × 2-19


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): -19


Mantissa (not normalized):
1.1110 1100 1000 0000 0000 0000 0000 1101 1000 1011 0110 1011 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-19 + 2(11-1) - 1 =


(-19 + 1 023)(10) =


1 004(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;
  • 1 004 ÷ 2 = 502 + 0;
  • 502 ÷ 2 = 251 + 0;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =


1004(10) =


011 1110 1100(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. 1110 1100 1000 0000 0000 0000 0000 1101 1000 1011 0110 1011 0111 =


1110 1100 1000 0000 0000 0000 0000 1101 1000 1011 0110 1011 0111


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 1110 1100


Mantissa (52 bits) =
1110 1100 1000 0000 0000 0000 0000 1101 1000 1011 0110 1011 0111


Number 0.000 003 669 410 95 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1110 1100 - 1110 1100 1000 0000 0000 0000 0000 1101 1000 1011 0110 1011 0111

(64 bits IEEE 754)

More operations of this kind:

0.000 003 669 410 94 = ? ... 0.000 003 669 410 96 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

0.000 003 669 410 95 to 64 bit double precision IEEE 754 binary floating point = ? Dec 02 23:22 UTC (GMT)
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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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