0.000 000 000 436 572 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 436 572 6(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
0.000 000 000 436 572 6(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: 0.000 000 000 436 572 6.

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) 0.000 000 000 436 572 6 × 2 = 0 + 0.000 000 000 873 145 2;
  • 2) 0.000 000 000 873 145 2 × 2 = 0 + 0.000 000 001 746 290 4;
  • 3) 0.000 000 001 746 290 4 × 2 = 0 + 0.000 000 003 492 580 8;
  • 4) 0.000 000 003 492 580 8 × 2 = 0 + 0.000 000 006 985 161 6;
  • 5) 0.000 000 006 985 161 6 × 2 = 0 + 0.000 000 013 970 323 2;
  • 6) 0.000 000 013 970 323 2 × 2 = 0 + 0.000 000 027 940 646 4;
  • 7) 0.000 000 027 940 646 4 × 2 = 0 + 0.000 000 055 881 292 8;
  • 8) 0.000 000 055 881 292 8 × 2 = 0 + 0.000 000 111 762 585 6;
  • 9) 0.000 000 111 762 585 6 × 2 = 0 + 0.000 000 223 525 171 2;
  • 10) 0.000 000 223 525 171 2 × 2 = 0 + 0.000 000 447 050 342 4;
  • 11) 0.000 000 447 050 342 4 × 2 = 0 + 0.000 000 894 100 684 8;
  • 12) 0.000 000 894 100 684 8 × 2 = 0 + 0.000 001 788 201 369 6;
  • 13) 0.000 001 788 201 369 6 × 2 = 0 + 0.000 003 576 402 739 2;
  • 14) 0.000 003 576 402 739 2 × 2 = 0 + 0.000 007 152 805 478 4;
  • 15) 0.000 007 152 805 478 4 × 2 = 0 + 0.000 014 305 610 956 8;
  • 16) 0.000 014 305 610 956 8 × 2 = 0 + 0.000 028 611 221 913 6;
  • 17) 0.000 028 611 221 913 6 × 2 = 0 + 0.000 057 222 443 827 2;
  • 18) 0.000 057 222 443 827 2 × 2 = 0 + 0.000 114 444 887 654 4;
  • 19) 0.000 114 444 887 654 4 × 2 = 0 + 0.000 228 889 775 308 8;
  • 20) 0.000 228 889 775 308 8 × 2 = 0 + 0.000 457 779 550 617 6;
  • 21) 0.000 457 779 550 617 6 × 2 = 0 + 0.000 915 559 101 235 2;
  • 22) 0.000 915 559 101 235 2 × 2 = 0 + 0.001 831 118 202 470 4;
  • 23) 0.001 831 118 202 470 4 × 2 = 0 + 0.003 662 236 404 940 8;
  • 24) 0.003 662 236 404 940 8 × 2 = 0 + 0.007 324 472 809 881 6;
  • 25) 0.007 324 472 809 881 6 × 2 = 0 + 0.014 648 945 619 763 2;
  • 26) 0.014 648 945 619 763 2 × 2 = 0 + 0.029 297 891 239 526 4;
  • 27) 0.029 297 891 239 526 4 × 2 = 0 + 0.058 595 782 479 052 8;
  • 28) 0.058 595 782 479 052 8 × 2 = 0 + 0.117 191 564 958 105 6;
  • 29) 0.117 191 564 958 105 6 × 2 = 0 + 0.234 383 129 916 211 2;
  • 30) 0.234 383 129 916 211 2 × 2 = 0 + 0.468 766 259 832 422 4;
  • 31) 0.468 766 259 832 422 4 × 2 = 0 + 0.937 532 519 664 844 8;
  • 32) 0.937 532 519 664 844 8 × 2 = 1 + 0.875 065 039 329 689 6;
  • 33) 0.875 065 039 329 689 6 × 2 = 1 + 0.750 130 078 659 379 2;
  • 34) 0.750 130 078 659 379 2 × 2 = 1 + 0.500 260 157 318 758 4;
  • 35) 0.500 260 157 318 758 4 × 2 = 1 + 0.000 520 314 637 516 8;
  • 36) 0.000 520 314 637 516 8 × 2 = 0 + 0.001 040 629 275 033 6;
  • 37) 0.001 040 629 275 033 6 × 2 = 0 + 0.002 081 258 550 067 2;
  • 38) 0.002 081 258 550 067 2 × 2 = 0 + 0.004 162 517 100 134 4;
  • 39) 0.004 162 517 100 134 4 × 2 = 0 + 0.008 325 034 200 268 8;
  • 40) 0.008 325 034 200 268 8 × 2 = 0 + 0.016 650 068 400 537 6;
  • 41) 0.016 650 068 400 537 6 × 2 = 0 + 0.033 300 136 801 075 2;
  • 42) 0.033 300 136 801 075 2 × 2 = 0 + 0.066 600 273 602 150 4;
  • 43) 0.066 600 273 602 150 4 × 2 = 0 + 0.133 200 547 204 300 8;
  • 44) 0.133 200 547 204 300 8 × 2 = 0 + 0.266 401 094 408 601 6;
  • 45) 0.266 401 094 408 601 6 × 2 = 0 + 0.532 802 188 817 203 2;
  • 46) 0.532 802 188 817 203 2 × 2 = 1 + 0.065 604 377 634 406 4;
  • 47) 0.065 604 377 634 406 4 × 2 = 0 + 0.131 208 755 268 812 8;
  • 48) 0.131 208 755 268 812 8 × 2 = 0 + 0.262 417 510 537 625 6;
  • 49) 0.262 417 510 537 625 6 × 2 = 0 + 0.524 835 021 075 251 2;
  • 50) 0.524 835 021 075 251 2 × 2 = 1 + 0.049 670 042 150 502 4;
  • 51) 0.049 670 042 150 502 4 × 2 = 0 + 0.099 340 084 301 004 8;
  • 52) 0.099 340 084 301 004 8 × 2 = 0 + 0.198 680 168 602 009 6;
  • 53) 0.198 680 168 602 009 6 × 2 = 0 + 0.397 360 337 204 019 2;
  • 54) 0.397 360 337 204 019 2 × 2 = 0 + 0.794 720 674 408 038 4;
  • 55) 0.794 720 674 408 038 4 × 2 = 1 + 0.589 441 348 816 076 8;
  • 56) 0.589 441 348 816 076 8 × 2 = 1 + 0.178 882 697 632 153 6;
  • 57) 0.178 882 697 632 153 6 × 2 = 0 + 0.357 765 395 264 307 2;
  • 58) 0.357 765 395 264 307 2 × 2 = 0 + 0.715 530 790 528 614 4;
  • 59) 0.715 530 790 528 614 4 × 2 = 1 + 0.431 061 581 057 228 8;
  • 60) 0.431 061 581 057 228 8 × 2 = 0 + 0.862 123 162 114 457 6;
  • 61) 0.862 123 162 114 457 6 × 2 = 1 + 0.724 246 324 228 915 2;
  • 62) 0.724 246 324 228 915 2 × 2 = 1 + 0.448 492 648 457 830 4;
  • 63) 0.448 492 648 457 830 4 × 2 = 0 + 0.896 985 296 915 660 8;
  • 64) 0.896 985 296 915 660 8 × 2 = 1 + 0.793 970 593 831 321 6;
  • 65) 0.793 970 593 831 321 6 × 2 = 1 + 0.587 941 187 662 643 2;
  • 66) 0.587 941 187 662 643 2 × 2 = 1 + 0.175 882 375 325 286 4;
  • 67) 0.175 882 375 325 286 4 × 2 = 0 + 0.351 764 750 650 572 8;
  • 68) 0.351 764 750 650 572 8 × 2 = 0 + 0.703 529 501 301 145 6;
  • 69) 0.703 529 501 301 145 6 × 2 = 1 + 0.407 059 002 602 291 2;
  • 70) 0.407 059 002 602 291 2 × 2 = 0 + 0.814 118 005 204 582 4;
  • 71) 0.814 118 005 204 582 4 × 2 = 1 + 0.628 236 010 409 164 8;
  • 72) 0.628 236 010 409 164 8 × 2 = 1 + 0.256 472 020 818 329 6;
  • 73) 0.256 472 020 818 329 6 × 2 = 0 + 0.512 944 041 636 659 2;
  • 74) 0.512 944 041 636 659 2 × 2 = 1 + 0.025 888 083 273 318 4;
  • 75) 0.025 888 083 273 318 4 × 2 = 0 + 0.051 776 166 546 636 8;
  • 76) 0.051 776 166 546 636 8 × 2 = 0 + 0.103 552 333 093 273 6;
  • 77) 0.103 552 333 093 273 6 × 2 = 0 + 0.207 104 666 186 547 2;
  • 78) 0.207 104 666 186 547 2 × 2 = 0 + 0.414 209 332 373 094 4;
  • 79) 0.414 209 332 373 094 4 × 2 = 0 + 0.828 418 664 746 188 8;
  • 80) 0.828 418 664 746 188 8 × 2 = 1 + 0.656 837 329 492 377 6;
  • 81) 0.656 837 329 492 377 6 × 2 = 1 + 0.313 674 658 984 755 2;
  • 82) 0.313 674 658 984 755 2 × 2 = 0 + 0.627 349 317 969 510 4;
  • 83) 0.627 349 317 969 510 4 × 2 = 1 + 0.254 698 635 939 020 8;
  • 84) 0.254 698 635 939 020 8 × 2 = 0 + 0.509 397 271 878 041 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:


0.000 000 000 436 572 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010(2)

5. Positive number before normalization:

0.000 000 000 436 572 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 436 572 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010(2) × 20 =

1.1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010(2) × 2-32


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): -32

Mantissa (not normalized):
1.1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(11-1) - 1 =

(-32 + 1 023)(10) =

991(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;
  • 991 ÷ 2 = 495 + 1;
  • 495 ÷ 2 = 247 + 1;
  • 247 ÷ 2 = 123 + 1;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

991(10) =

011 1101 1111(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 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010 =

1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 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 1101 1111

Mantissa (52 bits) =
1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 1010


Decimal number 0.000 000 000 436 572 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1111 - 1110 0000 0000 0100 0100 0011 0010 1101 1100 1011 0100 0001 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