0.523 598 775 598 298 873 078 72 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.523 598 775 598 298 873 078 72(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
0.523 598 775 598 298 873 078 72(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: 0.523 598 775 598 298 873 078 72.

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) 0.523 598 775 598 298 873 078 72 × 2 = 1 + 0.047 197 551 196 597 746 157 44;
  • 2) 0.047 197 551 196 597 746 157 44 × 2 = 0 + 0.094 395 102 393 195 492 314 88;
  • 3) 0.094 395 102 393 195 492 314 88 × 2 = 0 + 0.188 790 204 786 390 984 629 76;
  • 4) 0.188 790 204 786 390 984 629 76 × 2 = 0 + 0.377 580 409 572 781 969 259 52;
  • 5) 0.377 580 409 572 781 969 259 52 × 2 = 0 + 0.755 160 819 145 563 938 519 04;
  • 6) 0.755 160 819 145 563 938 519 04 × 2 = 1 + 0.510 321 638 291 127 877 038 08;
  • 7) 0.510 321 638 291 127 877 038 08 × 2 = 1 + 0.020 643 276 582 255 754 076 16;
  • 8) 0.020 643 276 582 255 754 076 16 × 2 = 0 + 0.041 286 553 164 511 508 152 32;
  • 9) 0.041 286 553 164 511 508 152 32 × 2 = 0 + 0.082 573 106 329 023 016 304 64;
  • 10) 0.082 573 106 329 023 016 304 64 × 2 = 0 + 0.165 146 212 658 046 032 609 28;
  • 11) 0.165 146 212 658 046 032 609 28 × 2 = 0 + 0.330 292 425 316 092 065 218 56;
  • 12) 0.330 292 425 316 092 065 218 56 × 2 = 0 + 0.660 584 850 632 184 130 437 12;
  • 13) 0.660 584 850 632 184 130 437 12 × 2 = 1 + 0.321 169 701 264 368 260 874 24;
  • 14) 0.321 169 701 264 368 260 874 24 × 2 = 0 + 0.642 339 402 528 736 521 748 48;
  • 15) 0.642 339 402 528 736 521 748 48 × 2 = 1 + 0.284 678 805 057 473 043 496 96;
  • 16) 0.284 678 805 057 473 043 496 96 × 2 = 0 + 0.569 357 610 114 946 086 993 92;
  • 17) 0.569 357 610 114 946 086 993 92 × 2 = 1 + 0.138 715 220 229 892 173 987 84;
  • 18) 0.138 715 220 229 892 173 987 84 × 2 = 0 + 0.277 430 440 459 784 347 975 68;
  • 19) 0.277 430 440 459 784 347 975 68 × 2 = 0 + 0.554 860 880 919 568 695 951 36;
  • 20) 0.554 860 880 919 568 695 951 36 × 2 = 1 + 0.109 721 761 839 137 391 902 72;
  • 21) 0.109 721 761 839 137 391 902 72 × 2 = 0 + 0.219 443 523 678 274 783 805 44;
  • 22) 0.219 443 523 678 274 783 805 44 × 2 = 0 + 0.438 887 047 356 549 567 610 88;
  • 23) 0.438 887 047 356 549 567 610 88 × 2 = 0 + 0.877 774 094 713 099 135 221 76;
  • 24) 0.877 774 094 713 099 135 221 76 × 2 = 1 + 0.755 548 189 426 198 270 443 52;
  • 25) 0.755 548 189 426 198 270 443 52 × 2 = 1 + 0.511 096 378 852 396 540 887 04;
  • 26) 0.511 096 378 852 396 540 887 04 × 2 = 1 + 0.022 192 757 704 793 081 774 08;
  • 27) 0.022 192 757 704 793 081 774 08 × 2 = 0 + 0.044 385 515 409 586 163 548 16;
  • 28) 0.044 385 515 409 586 163 548 16 × 2 = 0 + 0.088 771 030 819 172 327 096 32;
  • 29) 0.088 771 030 819 172 327 096 32 × 2 = 0 + 0.177 542 061 638 344 654 192 64;
  • 30) 0.177 542 061 638 344 654 192 64 × 2 = 0 + 0.355 084 123 276 689 308 385 28;
  • 31) 0.355 084 123 276 689 308 385 28 × 2 = 0 + 0.710 168 246 553 378 616 770 56;
  • 32) 0.710 168 246 553 378 616 770 56 × 2 = 1 + 0.420 336 493 106 757 233 541 12;
  • 33) 0.420 336 493 106 757 233 541 12 × 2 = 0 + 0.840 672 986 213 514 467 082 24;
  • 34) 0.840 672 986 213 514 467 082 24 × 2 = 1 + 0.681 345 972 427 028 934 164 48;
  • 35) 0.681 345 972 427 028 934 164 48 × 2 = 1 + 0.362 691 944 854 057 868 328 96;
  • 36) 0.362 691 944 854 057 868 328 96 × 2 = 0 + 0.725 383 889 708 115 736 657 92;
  • 37) 0.725 383 889 708 115 736 657 92 × 2 = 1 + 0.450 767 779 416 231 473 315 84;
  • 38) 0.450 767 779 416 231 473 315 84 × 2 = 0 + 0.901 535 558 832 462 946 631 68;
  • 39) 0.901 535 558 832 462 946 631 68 × 2 = 1 + 0.803 071 117 664 925 893 263 36;
  • 40) 0.803 071 117 664 925 893 263 36 × 2 = 1 + 0.606 142 235 329 851 786 526 72;
  • 41) 0.606 142 235 329 851 786 526 72 × 2 = 1 + 0.212 284 470 659 703 573 053 44;
  • 42) 0.212 284 470 659 703 573 053 44 × 2 = 0 + 0.424 568 941 319 407 146 106 88;
  • 43) 0.424 568 941 319 407 146 106 88 × 2 = 0 + 0.849 137 882 638 814 292 213 76;
  • 44) 0.849 137 882 638 814 292 213 76 × 2 = 1 + 0.698 275 765 277 628 584 427 52;
  • 45) 0.698 275 765 277 628 584 427 52 × 2 = 1 + 0.396 551 530 555 257 168 855 04;
  • 46) 0.396 551 530 555 257 168 855 04 × 2 = 0 + 0.793 103 061 110 514 337 710 08;
  • 47) 0.793 103 061 110 514 337 710 08 × 2 = 1 + 0.586 206 122 221 028 675 420 16;
  • 48) 0.586 206 122 221 028 675 420 16 × 2 = 1 + 0.172 412 244 442 057 350 840 32;
  • 49) 0.172 412 244 442 057 350 840 32 × 2 = 0 + 0.344 824 488 884 114 701 680 64;
  • 50) 0.344 824 488 884 114 701 680 64 × 2 = 0 + 0.689 648 977 768 229 403 361 28;
  • 51) 0.689 648 977 768 229 403 361 28 × 2 = 1 + 0.379 297 955 536 458 806 722 56;
  • 52) 0.379 297 955 536 458 806 722 56 × 2 = 0 + 0.758 595 911 072 917 613 445 12;
  • 53) 0.758 595 911 072 917 613 445 12 × 2 = 1 + 0.517 191 822 145 835 226 890 24;

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:


0.523 598 775 598 298 873 078 72(10) =

0.1000 0110 0000 1010 1001 0001 1100 0001 0110 1011 1001 1011 0010 1(2)

5. Positive number before normalization:

0.523 598 775 598 298 873 078 72(10) =

0.1000 0110 0000 1010 1001 0001 1100 0001 0110 1011 1001 1011 0010 1(2)

6. Normalize the binary representation of the number.

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


0.523 598 775 598 298 873 078 72(10) =

0.1000 0110 0000 1010 1001 0001 1100 0001 0110 1011 1001 1011 0010 1(2) =

0.1000 0110 0000 1010 1001 0001 1100 0001 0110 1011 1001 1011 0010 1(2) × 20 =

1.0000 1100 0001 0101 0010 0011 1000 0010 1101 0111 0011 0110 0101(2) × 2-1


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):
1.0000 1100 0001 0101 0010 0011 1000 0010 1101 0111 0011 0110 0101


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) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 0000 1100 0001 0101 0010 0011 1000 0010 1101 0111 0011 0110 0101 =

0000 1100 0001 0101 0010 0011 1000 0010 1101 0111 0011 0110 0101


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 1110

Mantissa (52 bits) =
0000 1100 0001 0101 0010 0011 1000 0010 1101 0111 0011 0110 0101


Decimal number 0.523 598 775 598 298 873 078 72 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0000 1100 0001 0101 0010 0011 1000 0010 1101 0111 0011 0110 0101

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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