2.560 879 601 235 235 325 098 774 26 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.560 879 601 235 235 325 098 774 26(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
2.560 879 601 235 235 325 098 774 26(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.560 879 601 235 235 325 098 774 26.

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) 0.560 879 601 235 235 325 098 774 26 × 2 = 1 + 0.121 759 202 470 470 650 197 548 52;
  • 2) 0.121 759 202 470 470 650 197 548 52 × 2 = 0 + 0.243 518 404 940 941 300 395 097 04;
  • 3) 0.243 518 404 940 941 300 395 097 04 × 2 = 0 + 0.487 036 809 881 882 600 790 194 08;
  • 4) 0.487 036 809 881 882 600 790 194 08 × 2 = 0 + 0.974 073 619 763 765 201 580 388 16;
  • 5) 0.974 073 619 763 765 201 580 388 16 × 2 = 1 + 0.948 147 239 527 530 403 160 776 32;
  • 6) 0.948 147 239 527 530 403 160 776 32 × 2 = 1 + 0.896 294 479 055 060 806 321 552 64;
  • 7) 0.896 294 479 055 060 806 321 552 64 × 2 = 1 + 0.792 588 958 110 121 612 643 105 28;
  • 8) 0.792 588 958 110 121 612 643 105 28 × 2 = 1 + 0.585 177 916 220 243 225 286 210 56;
  • 9) 0.585 177 916 220 243 225 286 210 56 × 2 = 1 + 0.170 355 832 440 486 450 572 421 12;
  • 10) 0.170 355 832 440 486 450 572 421 12 × 2 = 0 + 0.340 711 664 880 972 901 144 842 24;
  • 11) 0.340 711 664 880 972 901 144 842 24 × 2 = 0 + 0.681 423 329 761 945 802 289 684 48;
  • 12) 0.681 423 329 761 945 802 289 684 48 × 2 = 1 + 0.362 846 659 523 891 604 579 368 96;
  • 13) 0.362 846 659 523 891 604 579 368 96 × 2 = 0 + 0.725 693 319 047 783 209 158 737 92;
  • 14) 0.725 693 319 047 783 209 158 737 92 × 2 = 1 + 0.451 386 638 095 566 418 317 475 84;
  • 15) 0.451 386 638 095 566 418 317 475 84 × 2 = 0 + 0.902 773 276 191 132 836 634 951 68;
  • 16) 0.902 773 276 191 132 836 634 951 68 × 2 = 1 + 0.805 546 552 382 265 673 269 903 36;
  • 17) 0.805 546 552 382 265 673 269 903 36 × 2 = 1 + 0.611 093 104 764 531 346 539 806 72;
  • 18) 0.611 093 104 764 531 346 539 806 72 × 2 = 1 + 0.222 186 209 529 062 693 079 613 44;
  • 19) 0.222 186 209 529 062 693 079 613 44 × 2 = 0 + 0.444 372 419 058 125 386 159 226 88;
  • 20) 0.444 372 419 058 125 386 159 226 88 × 2 = 0 + 0.888 744 838 116 250 772 318 453 76;
  • 21) 0.888 744 838 116 250 772 318 453 76 × 2 = 1 + 0.777 489 676 232 501 544 636 907 52;
  • 22) 0.777 489 676 232 501 544 636 907 52 × 2 = 1 + 0.554 979 352 465 003 089 273 815 04;
  • 23) 0.554 979 352 465 003 089 273 815 04 × 2 = 1 + 0.109 958 704 930 006 178 547 630 08;
  • 24) 0.109 958 704 930 006 178 547 630 08 × 2 = 0 + 0.219 917 409 860 012 357 095 260 16;
  • 25) 0.219 917 409 860 012 357 095 260 16 × 2 = 0 + 0.439 834 819 720 024 714 190 520 32;
  • 26) 0.439 834 819 720 024 714 190 520 32 × 2 = 0 + 0.879 669 639 440 049 428 381 040 64;
  • 27) 0.879 669 639 440 049 428 381 040 64 × 2 = 1 + 0.759 339 278 880 098 856 762 081 28;
  • 28) 0.759 339 278 880 098 856 762 081 28 × 2 = 1 + 0.518 678 557 760 197 713 524 162 56;
  • 29) 0.518 678 557 760 197 713 524 162 56 × 2 = 1 + 0.037 357 115 520 395 427 048 325 12;
  • 30) 0.037 357 115 520 395 427 048 325 12 × 2 = 0 + 0.074 714 231 040 790 854 096 650 24;
  • 31) 0.074 714 231 040 790 854 096 650 24 × 2 = 0 + 0.149 428 462 081 581 708 193 300 48;
  • 32) 0.149 428 462 081 581 708 193 300 48 × 2 = 0 + 0.298 856 924 163 163 416 386 600 96;
  • 33) 0.298 856 924 163 163 416 386 600 96 × 2 = 0 + 0.597 713 848 326 326 832 773 201 92;
  • 34) 0.597 713 848 326 326 832 773 201 92 × 2 = 1 + 0.195 427 696 652 653 665 546 403 84;
  • 35) 0.195 427 696 652 653 665 546 403 84 × 2 = 0 + 0.390 855 393 305 307 331 092 807 68;
  • 36) 0.390 855 393 305 307 331 092 807 68 × 2 = 0 + 0.781 710 786 610 614 662 185 615 36;
  • 37) 0.781 710 786 610 614 662 185 615 36 × 2 = 1 + 0.563 421 573 221 229 324 371 230 72;
  • 38) 0.563 421 573 221 229 324 371 230 72 × 2 = 1 + 0.126 843 146 442 458 648 742 461 44;
  • 39) 0.126 843 146 442 458 648 742 461 44 × 2 = 0 + 0.253 686 292 884 917 297 484 922 88;
  • 40) 0.253 686 292 884 917 297 484 922 88 × 2 = 0 + 0.507 372 585 769 834 594 969 845 76;
  • 41) 0.507 372 585 769 834 594 969 845 76 × 2 = 1 + 0.014 745 171 539 669 189 939 691 52;
  • 42) 0.014 745 171 539 669 189 939 691 52 × 2 = 0 + 0.029 490 343 079 338 379 879 383 04;
  • 43) 0.029 490 343 079 338 379 879 383 04 × 2 = 0 + 0.058 980 686 158 676 759 758 766 08;
  • 44) 0.058 980 686 158 676 759 758 766 08 × 2 = 0 + 0.117 961 372 317 353 519 517 532 16;
  • 45) 0.117 961 372 317 353 519 517 532 16 × 2 = 0 + 0.235 922 744 634 707 039 035 064 32;
  • 46) 0.235 922 744 634 707 039 035 064 32 × 2 = 0 + 0.471 845 489 269 414 078 070 128 64;
  • 47) 0.471 845 489 269 414 078 070 128 64 × 2 = 0 + 0.943 690 978 538 828 156 140 257 28;
  • 48) 0.943 690 978 538 828 156 140 257 28 × 2 = 1 + 0.887 381 957 077 656 312 280 514 56;
  • 49) 0.887 381 957 077 656 312 280 514 56 × 2 = 1 + 0.774 763 914 155 312 624 561 029 12;
  • 50) 0.774 763 914 155 312 624 561 029 12 × 2 = 1 + 0.549 527 828 310 625 249 122 058 24;
  • 51) 0.549 527 828 310 625 249 122 058 24 × 2 = 1 + 0.099 055 656 621 250 498 244 116 48;
  • 52) 0.099 055 656 621 250 498 244 116 48 × 2 = 0 + 0.198 111 313 242 500 996 488 232 96;
  • 53) 0.198 111 313 242 500 996 488 232 96 × 2 = 0 + 0.396 222 626 485 001 992 976 465 92;

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:


0.560 879 601 235 235 325 098 774 26(10) =

0.1000 1111 1001 0101 1100 1110 0011 1000 0100 1100 1000 0001 1110 0(2)

5. Positive number before normalization:

2.560 879 601 235 235 325 098 774 26(10) =

10.1000 1111 1001 0101 1100 1110 0011 1000 0100 1100 1000 0001 1110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.560 879 601 235 235 325 098 774 26(10) =

10.1000 1111 1001 0101 1100 1110 0011 1000 0100 1100 1000 0001 1110 0(2) =

10.1000 1111 1001 0101 1100 1110 0011 1000 0100 1100 1000 0001 1110 0(2) × 20 =

1.0100 0111 1100 1010 1110 0111 0001 1100 0010 0110 0100 0000 1111 00(2) × 21


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): 1

Mantissa (not normalized):
1.0100 0111 1100 1010 1110 0111 0001 1100 0010 0110 0100 0000 1111 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0100 0111 1100 1010 1110 0111 0001 1100 0010 0110 0100 0000 1111 00 =

0100 0111 1100 1010 1110 0111 0001 1100 0010 0110 0100 0000 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) =
100 0000 0000

Mantissa (52 bits) =
0100 0111 1100 1010 1110 0111 0001 1100 0010 0110 0100 0000 1111


Decimal number 2.560 879 601 235 235 325 098 774 26 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0100 0111 1100 1010 1110 0111 0001 1100 0010 0110 0100 0000 1111

» Convert decimal 2.560 879 601 235 235 325 098 774 72 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