0.000 046 473 596 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 596(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
0.000 046 473 596(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: 0.000 046 473 596.

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) 0.000 046 473 596 × 2 = 0 + 0.000 092 947 192;
  • 2) 0.000 092 947 192 × 2 = 0 + 0.000 185 894 384;
  • 3) 0.000 185 894 384 × 2 = 0 + 0.000 371 788 768;
  • 4) 0.000 371 788 768 × 2 = 0 + 0.000 743 577 536;
  • 5) 0.000 743 577 536 × 2 = 0 + 0.001 487 155 072;
  • 6) 0.001 487 155 072 × 2 = 0 + 0.002 974 310 144;
  • 7) 0.002 974 310 144 × 2 = 0 + 0.005 948 620 288;
  • 8) 0.005 948 620 288 × 2 = 0 + 0.011 897 240 576;
  • 9) 0.011 897 240 576 × 2 = 0 + 0.023 794 481 152;
  • 10) 0.023 794 481 152 × 2 = 0 + 0.047 588 962 304;
  • 11) 0.047 588 962 304 × 2 = 0 + 0.095 177 924 608;
  • 12) 0.095 177 924 608 × 2 = 0 + 0.190 355 849 216;
  • 13) 0.190 355 849 216 × 2 = 0 + 0.380 711 698 432;
  • 14) 0.380 711 698 432 × 2 = 0 + 0.761 423 396 864;
  • 15) 0.761 423 396 864 × 2 = 1 + 0.522 846 793 728;
  • 16) 0.522 846 793 728 × 2 = 1 + 0.045 693 587 456;
  • 17) 0.045 693 587 456 × 2 = 0 + 0.091 387 174 912;
  • 18) 0.091 387 174 912 × 2 = 0 + 0.182 774 349 824;
  • 19) 0.182 774 349 824 × 2 = 0 + 0.365 548 699 648;
  • 20) 0.365 548 699 648 × 2 = 0 + 0.731 097 399 296;
  • 21) 0.731 097 399 296 × 2 = 1 + 0.462 194 798 592;
  • 22) 0.462 194 798 592 × 2 = 0 + 0.924 389 597 184;
  • 23) 0.924 389 597 184 × 2 = 1 + 0.848 779 194 368;
  • 24) 0.848 779 194 368 × 2 = 1 + 0.697 558 388 736;
  • 25) 0.697 558 388 736 × 2 = 1 + 0.395 116 777 472;
  • 26) 0.395 116 777 472 × 2 = 0 + 0.790 233 554 944;
  • 27) 0.790 233 554 944 × 2 = 1 + 0.580 467 109 888;
  • 28) 0.580 467 109 888 × 2 = 1 + 0.160 934 219 776;
  • 29) 0.160 934 219 776 × 2 = 0 + 0.321 868 439 552;
  • 30) 0.321 868 439 552 × 2 = 0 + 0.643 736 879 104;
  • 31) 0.643 736 879 104 × 2 = 1 + 0.287 473 758 208;
  • 32) 0.287 473 758 208 × 2 = 0 + 0.574 947 516 416;
  • 33) 0.574 947 516 416 × 2 = 1 + 0.149 895 032 832;
  • 34) 0.149 895 032 832 × 2 = 0 + 0.299 790 065 664;
  • 35) 0.299 790 065 664 × 2 = 0 + 0.599 580 131 328;
  • 36) 0.599 580 131 328 × 2 = 1 + 0.199 160 262 656;
  • 37) 0.199 160 262 656 × 2 = 0 + 0.398 320 525 312;
  • 38) 0.398 320 525 312 × 2 = 0 + 0.796 641 050 624;
  • 39) 0.796 641 050 624 × 2 = 1 + 0.593 282 101 248;
  • 40) 0.593 282 101 248 × 2 = 1 + 0.186 564 202 496;
  • 41) 0.186 564 202 496 × 2 = 0 + 0.373 128 404 992;
  • 42) 0.373 128 404 992 × 2 = 0 + 0.746 256 809 984;
  • 43) 0.746 256 809 984 × 2 = 1 + 0.492 513 619 968;
  • 44) 0.492 513 619 968 × 2 = 0 + 0.985 027 239 936;
  • 45) 0.985 027 239 936 × 2 = 1 + 0.970 054 479 872;
  • 46) 0.970 054 479 872 × 2 = 1 + 0.940 108 959 744;
  • 47) 0.940 108 959 744 × 2 = 1 + 0.880 217 919 488;
  • 48) 0.880 217 919 488 × 2 = 1 + 0.760 435 838 976;
  • 49) 0.760 435 838 976 × 2 = 1 + 0.520 871 677 952;
  • 50) 0.520 871 677 952 × 2 = 1 + 0.041 743 355 904;
  • 51) 0.041 743 355 904 × 2 = 0 + 0.083 486 711 808;
  • 52) 0.083 486 711 808 × 2 = 0 + 0.166 973 423 616;
  • 53) 0.166 973 423 616 × 2 = 0 + 0.333 946 847 232;
  • 54) 0.333 946 847 232 × 2 = 0 + 0.667 893 694 464;
  • 55) 0.667 893 694 464 × 2 = 1 + 0.335 787 388 928;
  • 56) 0.335 787 388 928 × 2 = 0 + 0.671 574 777 856;
  • 57) 0.671 574 777 856 × 2 = 1 + 0.343 149 555 712;
  • 58) 0.343 149 555 712 × 2 = 0 + 0.686 299 111 424;
  • 59) 0.686 299 111 424 × 2 = 1 + 0.372 598 222 848;
  • 60) 0.372 598 222 848 × 2 = 0 + 0.745 196 445 696;
  • 61) 0.745 196 445 696 × 2 = 1 + 0.490 392 891 392;
  • 62) 0.490 392 891 392 × 2 = 0 + 0.980 785 782 784;
  • 63) 0.980 785 782 784 × 2 = 1 + 0.961 571 565 568;
  • 64) 0.961 571 565 568 × 2 = 1 + 0.923 143 131 136;
  • 65) 0.923 143 131 136 × 2 = 1 + 0.846 286 262 272;
  • 66) 0.846 286 262 272 × 2 = 1 + 0.692 572 524 544;
  • 67) 0.692 572 524 544 × 2 = 1 + 0.385 145 049 088;

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:


0.000 046 473 596(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 0011 0010 1111 1100 0010 1010 1011 111(2)

5. Positive number before normalization:

0.000 046 473 596(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 0011 0010 1111 1100 0010 1010 1011 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:


0.000 046 473 596(10) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 0011 0010 1111 1100 0010 1010 1011 111(2) =

0.0000 0000 0000 0011 0000 1011 1011 0010 1001 0011 0010 1111 1100 0010 1010 1011 111(2) × 20 =

1.1000 0101 1101 1001 0100 1001 1001 0111 1110 0001 0101 0101 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):
1.1000 0101 1101 1001 0100 1001 1001 0111 1110 0001 0101 0101 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. 1000 0101 1101 1001 0100 1001 1001 0111 1110 0001 0101 0101 1111 =

1000 0101 1101 1001 0100 1001 1001 0111 1110 0001 0101 0101 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) =
1000 0101 1101 1001 0100 1001 1001 0111 1110 0001 0101 0101 1111


Decimal number 0.000 046 473 596 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 0100 1001 1001 0111 1110 0001 0101 0101 1111

» Convert decimal 0.000 046 473 607 to 64 bit double precision IEEE 754 binary floating point representation standard

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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