0.000 046 473 737 953 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 046 473 737 953 4(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
0.000 046 473 737 953 4(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: 0.000 046 473 737 953 4.

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) 0.000 046 473 737 953 4 × 2 = 0 + 0.000 092 947 475 906 8;
  • 2) 0.000 092 947 475 906 8 × 2 = 0 + 0.000 185 894 951 813 6;
  • 3) 0.000 185 894 951 813 6 × 2 = 0 + 0.000 371 789 903 627 2;
  • 4) 0.000 371 789 903 627 2 × 2 = 0 + 0.000 743 579 807 254 4;
  • 5) 0.000 743 579 807 254 4 × 2 = 0 + 0.001 487 159 614 508 8;
  • 6) 0.001 487 159 614 508 8 × 2 = 0 + 0.002 974 319 229 017 6;
  • 7) 0.002 974 319 229 017 6 × 2 = 0 + 0.005 948 638 458 035 2;
  • 8) 0.005 948 638 458 035 2 × 2 = 0 + 0.011 897 276 916 070 4;
  • 9) 0.011 897 276 916 070 4 × 2 = 0 + 0.023 794 553 832 140 8;
  • 10) 0.023 794 553 832 140 8 × 2 = 0 + 0.047 589 107 664 281 6;
  • 11) 0.047 589 107 664 281 6 × 2 = 0 + 0.095 178 215 328 563 2;
  • 12) 0.095 178 215 328 563 2 × 2 = 0 + 0.190 356 430 657 126 4;
  • 13) 0.190 356 430 657 126 4 × 2 = 0 + 0.380 712 861 314 252 8;
  • 14) 0.380 712 861 314 252 8 × 2 = 0 + 0.761 425 722 628 505 6;
  • 15) 0.761 425 722 628 505 6 × 2 = 1 + 0.522 851 445 257 011 2;
  • 16) 0.522 851 445 257 011 2 × 2 = 1 + 0.045 702 890 514 022 4;
  • 17) 0.045 702 890 514 022 4 × 2 = 0 + 0.091 405 781 028 044 8;
  • 18) 0.091 405 781 028 044 8 × 2 = 0 + 0.182 811 562 056 089 6;
  • 19) 0.182 811 562 056 089 6 × 2 = 0 + 0.365 623 124 112 179 2;
  • 20) 0.365 623 124 112 179 2 × 2 = 0 + 0.731 246 248 224 358 4;
  • 21) 0.731 246 248 224 358 4 × 2 = 1 + 0.462 492 496 448 716 8;
  • 22) 0.462 492 496 448 716 8 × 2 = 0 + 0.924 984 992 897 433 6;
  • 23) 0.924 984 992 897 433 6 × 2 = 1 + 0.849 969 985 794 867 2;
  • 24) 0.849 969 985 794 867 2 × 2 = 1 + 0.699 939 971 589 734 4;
  • 25) 0.699 939 971 589 734 4 × 2 = 1 + 0.399 879 943 179 468 8;
  • 26) 0.399 879 943 179 468 8 × 2 = 0 + 0.799 759 886 358 937 6;
  • 27) 0.799 759 886 358 937 6 × 2 = 1 + 0.599 519 772 717 875 2;
  • 28) 0.599 519 772 717 875 2 × 2 = 1 + 0.199 039 545 435 750 4;
  • 29) 0.199 039 545 435 750 4 × 2 = 0 + 0.398 079 090 871 500 8;
  • 30) 0.398 079 090 871 500 8 × 2 = 0 + 0.796 158 181 743 001 6;
  • 31) 0.796 158 181 743 001 6 × 2 = 1 + 0.592 316 363 486 003 2;
  • 32) 0.592 316 363 486 003 2 × 2 = 1 + 0.184 632 726 972 006 4;
  • 33) 0.184 632 726 972 006 4 × 2 = 0 + 0.369 265 453 944 012 8;
  • 34) 0.369 265 453 944 012 8 × 2 = 0 + 0.738 530 907 888 025 6;
  • 35) 0.738 530 907 888 025 6 × 2 = 1 + 0.477 061 815 776 051 2;
  • 36) 0.477 061 815 776 051 2 × 2 = 0 + 0.954 123 631 552 102 4;
  • 37) 0.954 123 631 552 102 4 × 2 = 1 + 0.908 247 263 104 204 8;
  • 38) 0.908 247 263 104 204 8 × 2 = 1 + 0.816 494 526 208 409 6;
  • 39) 0.816 494 526 208 409 6 × 2 = 1 + 0.632 989 052 416 819 2;
  • 40) 0.632 989 052 416 819 2 × 2 = 1 + 0.265 978 104 833 638 4;
  • 41) 0.265 978 104 833 638 4 × 2 = 0 + 0.531 956 209 667 276 8;
  • 42) 0.531 956 209 667 276 8 × 2 = 1 + 0.063 912 419 334 553 6;
  • 43) 0.063 912 419 334 553 6 × 2 = 0 + 0.127 824 838 669 107 2;
  • 44) 0.127 824 838 669 107 2 × 2 = 0 + 0.255 649 677 338 214 4;
  • 45) 0.255 649 677 338 214 4 × 2 = 0 + 0.511 299 354 676 428 8;
  • 46) 0.511 299 354 676 428 8 × 2 = 1 + 0.022 598 709 352 857 6;
  • 47) 0.022 598 709 352 857 6 × 2 = 0 + 0.045 197 418 705 715 2;
  • 48) 0.045 197 418 705 715 2 × 2 = 0 + 0.090 394 837 411 430 4;
  • 49) 0.090 394 837 411 430 4 × 2 = 0 + 0.180 789 674 822 860 8;
  • 50) 0.180 789 674 822 860 8 × 2 = 0 + 0.361 579 349 645 721 6;
  • 51) 0.361 579 349 645 721 6 × 2 = 0 + 0.723 158 699 291 443 2;
  • 52) 0.723 158 699 291 443 2 × 2 = 1 + 0.446 317 398 582 886 4;
  • 53) 0.446 317 398 582 886 4 × 2 = 0 + 0.892 634 797 165 772 8;
  • 54) 0.892 634 797 165 772 8 × 2 = 1 + 0.785 269 594 331 545 6;
  • 55) 0.785 269 594 331 545 6 × 2 = 1 + 0.570 539 188 663 091 2;
  • 56) 0.570 539 188 663 091 2 × 2 = 1 + 0.141 078 377 326 182 4;
  • 57) 0.141 078 377 326 182 4 × 2 = 0 + 0.282 156 754 652 364 8;
  • 58) 0.282 156 754 652 364 8 × 2 = 0 + 0.564 313 509 304 729 6;
  • 59) 0.564 313 509 304 729 6 × 2 = 1 + 0.128 627 018 609 459 2;
  • 60) 0.128 627 018 609 459 2 × 2 = 0 + 0.257 254 037 218 918 4;
  • 61) 0.257 254 037 218 918 4 × 2 = 0 + 0.514 508 074 437 836 8;
  • 62) 0.514 508 074 437 836 8 × 2 = 1 + 0.029 016 148 875 673 6;
  • 63) 0.029 016 148 875 673 6 × 2 = 0 + 0.058 032 297 751 347 2;
  • 64) 0.058 032 297 751 347 2 × 2 = 0 + 0.116 064 595 502 694 4;
  • 65) 0.116 064 595 502 694 4 × 2 = 0 + 0.232 129 191 005 388 8;
  • 66) 0.232 129 191 005 388 8 × 2 = 0 + 0.464 258 382 010 777 6;
  • 67) 0.464 258 382 010 777 6 × 2 = 0 + 0.928 516 764 021 555 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:


0.000 046 473 737 953 4(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0001 0111 0010 0100 000(2)

5. Positive number before normalization:

0.000 046 473 737 953 4(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0001 0111 0010 0100 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:


0.000 046 473 737 953 4(10) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0001 0111 0010 0100 000(2) =

0.0000 0000 0000 0011 0000 1011 1011 0011 0010 1111 0100 0100 0001 0111 0010 0100 000(2) × 20 =

1.1000 0101 1101 1001 1001 0111 1010 0010 0000 1011 1001 0010 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):
1.1000 0101 1101 1001 1001 0111 1010 0010 0000 1011 1001 0010 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. 1000 0101 1101 1001 1001 0111 1010 0010 0000 1011 1001 0010 0000 =

1000 0101 1101 1001 1001 0111 1010 0010 0000 1011 1001 0010 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) =
1000 0101 1101 1001 1001 0111 1010 0010 0000 1011 1001 0010 0000


Decimal number 0.000 046 473 737 953 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 1000 0101 1101 1001 1001 0111 1010 0010 0000 1011 1001 0010 0000

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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