0.000 046 473 642 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 642(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
0.000 046 473 642(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: 0.000 046 473 642.

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) 0.000 046 473 642 × 2 = 0 + 0.000 092 947 284;
  • 2) 0.000 092 947 284 × 2 = 0 + 0.000 185 894 568;
  • 3) 0.000 185 894 568 × 2 = 0 + 0.000 371 789 136;
  • 4) 0.000 371 789 136 × 2 = 0 + 0.000 743 578 272;
  • 5) 0.000 743 578 272 × 2 = 0 + 0.001 487 156 544;
  • 6) 0.001 487 156 544 × 2 = 0 + 0.002 974 313 088;
  • 7) 0.002 974 313 088 × 2 = 0 + 0.005 948 626 176;
  • 8) 0.005 948 626 176 × 2 = 0 + 0.011 897 252 352;
  • 9) 0.011 897 252 352 × 2 = 0 + 0.023 794 504 704;
  • 10) 0.023 794 504 704 × 2 = 0 + 0.047 589 009 408;
  • 11) 0.047 589 009 408 × 2 = 0 + 0.095 178 018 816;
  • 12) 0.095 178 018 816 × 2 = 0 + 0.190 356 037 632;
  • 13) 0.190 356 037 632 × 2 = 0 + 0.380 712 075 264;
  • 14) 0.380 712 075 264 × 2 = 0 + 0.761 424 150 528;
  • 15) 0.761 424 150 528 × 2 = 1 + 0.522 848 301 056;
  • 16) 0.522 848 301 056 × 2 = 1 + 0.045 696 602 112;
  • 17) 0.045 696 602 112 × 2 = 0 + 0.091 393 204 224;
  • 18) 0.091 393 204 224 × 2 = 0 + 0.182 786 408 448;
  • 19) 0.182 786 408 448 × 2 = 0 + 0.365 572 816 896;
  • 20) 0.365 572 816 896 × 2 = 0 + 0.731 145 633 792;
  • 21) 0.731 145 633 792 × 2 = 1 + 0.462 291 267 584;
  • 22) 0.462 291 267 584 × 2 = 0 + 0.924 582 535 168;
  • 23) 0.924 582 535 168 × 2 = 1 + 0.849 165 070 336;
  • 24) 0.849 165 070 336 × 2 = 1 + 0.698 330 140 672;
  • 25) 0.698 330 140 672 × 2 = 1 + 0.396 660 281 344;
  • 26) 0.396 660 281 344 × 2 = 0 + 0.793 320 562 688;
  • 27) 0.793 320 562 688 × 2 = 1 + 0.586 641 125 376;
  • 28) 0.586 641 125 376 × 2 = 1 + 0.173 282 250 752;
  • 29) 0.173 282 250 752 × 2 = 0 + 0.346 564 501 504;
  • 30) 0.346 564 501 504 × 2 = 0 + 0.693 129 003 008;
  • 31) 0.693 129 003 008 × 2 = 1 + 0.386 258 006 016;
  • 32) 0.386 258 006 016 × 2 = 0 + 0.772 516 012 032;
  • 33) 0.772 516 012 032 × 2 = 1 + 0.545 032 024 064;
  • 34) 0.545 032 024 064 × 2 = 1 + 0.090 064 048 128;
  • 35) 0.090 064 048 128 × 2 = 0 + 0.180 128 096 256;
  • 36) 0.180 128 096 256 × 2 = 0 + 0.360 256 192 512;
  • 37) 0.360 256 192 512 × 2 = 0 + 0.720 512 385 024;
  • 38) 0.720 512 385 024 × 2 = 1 + 0.441 024 770 048;
  • 39) 0.441 024 770 048 × 2 = 0 + 0.882 049 540 096;
  • 40) 0.882 049 540 096 × 2 = 1 + 0.764 099 080 192;
  • 41) 0.764 099 080 192 × 2 = 1 + 0.528 198 160 384;
  • 42) 0.528 198 160 384 × 2 = 1 + 0.056 396 320 768;
  • 43) 0.056 396 320 768 × 2 = 0 + 0.112 792 641 536;
  • 44) 0.112 792 641 536 × 2 = 0 + 0.225 585 283 072;
  • 45) 0.225 585 283 072 × 2 = 0 + 0.451 170 566 144;
  • 46) 0.451 170 566 144 × 2 = 0 + 0.902 341 132 288;
  • 47) 0.902 341 132 288 × 2 = 1 + 0.804 682 264 576;
  • 48) 0.804 682 264 576 × 2 = 1 + 0.609 364 529 152;
  • 49) 0.609 364 529 152 × 2 = 1 + 0.218 729 058 304;
  • 50) 0.218 729 058 304 × 2 = 0 + 0.437 458 116 608;
  • 51) 0.437 458 116 608 × 2 = 0 + 0.874 916 233 216;
  • 52) 0.874 916 233 216 × 2 = 1 + 0.749 832 466 432;
  • 53) 0.749 832 466 432 × 2 = 1 + 0.499 664 932 864;
  • 54) 0.499 664 932 864 × 2 = 0 + 0.999 329 865 728;
  • 55) 0.999 329 865 728 × 2 = 1 + 0.998 659 731 456;
  • 56) 0.998 659 731 456 × 2 = 1 + 0.997 319 462 912;
  • 57) 0.997 319 462 912 × 2 = 1 + 0.994 638 925 824;
  • 58) 0.994 638 925 824 × 2 = 1 + 0.989 277 851 648;
  • 59) 0.989 277 851 648 × 2 = 1 + 0.978 555 703 296;
  • 60) 0.978 555 703 296 × 2 = 1 + 0.957 111 406 592;
  • 61) 0.957 111 406 592 × 2 = 1 + 0.914 222 813 184;
  • 62) 0.914 222 813 184 × 2 = 1 + 0.828 445 626 368;
  • 63) 0.828 445 626 368 × 2 = 1 + 0.656 891 252 736;
  • 64) 0.656 891 252 736 × 2 = 1 + 0.313 782 505 472;
  • 65) 0.313 782 505 472 × 2 = 0 + 0.627 565 010 944;
  • 66) 0.627 565 010 944 × 2 = 1 + 0.255 130 021 888;
  • 67) 0.255 130 021 888 × 2 = 0 + 0.510 260 043 776;

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:


0.000 046 473 642(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 0101 1100 0011 1001 1011 1111 1111 010(2)

5. Positive number before normalization:

0.000 046 473 642(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 0101 1100 0011 1001 1011 1111 1111 010(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:


0.000 046 473 642(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 0101 1100 0011 1001 1011 1111 1111 010(2) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1100 0101 1100 0011 1001 1011 1111 1111 010(2) × 20 =

1.1000 0101 1101 1001 0110 0010 1110 0001 1100 1101 1111 1111 1010(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):
1.1000 0101 1101 1001 0110 0010 1110 0001 1100 1101 1111 1111 1010


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. 1000 0101 1101 1001 0110 0010 1110 0001 1100 1101 1111 1111 1010 =

1000 0101 1101 1001 0110 0010 1110 0001 1100 1101 1111 1111 1010


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) =
1000 0101 1101 1001 0110 0010 1110 0001 1100 1101 1111 1111 1010


Decimal number 0.000 046 473 642 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 0110 0010 1110 0001 1100 1101 1111 1111 1010

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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