64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -3 528 711 045 533 138 969 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -3 528 711 045 533 138 969(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-3 528 711 045 533 138 969| = 3 528 711 045 533 138 969

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;
  • 3 528 711 045 533 138 969 ÷ 2 = 1 764 355 522 766 569 484 + 1;
  • 1 764 355 522 766 569 484 ÷ 2 = 882 177 761 383 284 742 + 0;
  • 882 177 761 383 284 742 ÷ 2 = 441 088 880 691 642 371 + 0;
  • 441 088 880 691 642 371 ÷ 2 = 220 544 440 345 821 185 + 1;
  • 220 544 440 345 821 185 ÷ 2 = 110 272 220 172 910 592 + 1;
  • 110 272 220 172 910 592 ÷ 2 = 55 136 110 086 455 296 + 0;
  • 55 136 110 086 455 296 ÷ 2 = 27 568 055 043 227 648 + 0;
  • 27 568 055 043 227 648 ÷ 2 = 13 784 027 521 613 824 + 0;
  • 13 784 027 521 613 824 ÷ 2 = 6 892 013 760 806 912 + 0;
  • 6 892 013 760 806 912 ÷ 2 = 3 446 006 880 403 456 + 0;
  • 3 446 006 880 403 456 ÷ 2 = 1 723 003 440 201 728 + 0;
  • 1 723 003 440 201 728 ÷ 2 = 861 501 720 100 864 + 0;
  • 861 501 720 100 864 ÷ 2 = 430 750 860 050 432 + 0;
  • 430 750 860 050 432 ÷ 2 = 215 375 430 025 216 + 0;
  • 215 375 430 025 216 ÷ 2 = 107 687 715 012 608 + 0;
  • 107 687 715 012 608 ÷ 2 = 53 843 857 506 304 + 0;
  • 53 843 857 506 304 ÷ 2 = 26 921 928 753 152 + 0;
  • 26 921 928 753 152 ÷ 2 = 13 460 964 376 576 + 0;
  • 13 460 964 376 576 ÷ 2 = 6 730 482 188 288 + 0;
  • 6 730 482 188 288 ÷ 2 = 3 365 241 094 144 + 0;
  • 3 365 241 094 144 ÷ 2 = 1 682 620 547 072 + 0;
  • 1 682 620 547 072 ÷ 2 = 841 310 273 536 + 0;
  • 841 310 273 536 ÷ 2 = 420 655 136 768 + 0;
  • 420 655 136 768 ÷ 2 = 210 327 568 384 + 0;
  • 210 327 568 384 ÷ 2 = 105 163 784 192 + 0;
  • 105 163 784 192 ÷ 2 = 52 581 892 096 + 0;
  • 52 581 892 096 ÷ 2 = 26 290 946 048 + 0;
  • 26 290 946 048 ÷ 2 = 13 145 473 024 + 0;
  • 13 145 473 024 ÷ 2 = 6 572 736 512 + 0;
  • 6 572 736 512 ÷ 2 = 3 286 368 256 + 0;
  • 3 286 368 256 ÷ 2 = 1 643 184 128 + 0;
  • 1 643 184 128 ÷ 2 = 821 592 064 + 0;
  • 821 592 064 ÷ 2 = 410 796 032 + 0;
  • 410 796 032 ÷ 2 = 205 398 016 + 0;
  • 205 398 016 ÷ 2 = 102 699 008 + 0;
  • 102 699 008 ÷ 2 = 51 349 504 + 0;
  • 51 349 504 ÷ 2 = 25 674 752 + 0;
  • 25 674 752 ÷ 2 = 12 837 376 + 0;
  • 12 837 376 ÷ 2 = 6 418 688 + 0;
  • 6 418 688 ÷ 2 = 3 209 344 + 0;
  • 3 209 344 ÷ 2 = 1 604 672 + 0;
  • 1 604 672 ÷ 2 = 802 336 + 0;
  • 802 336 ÷ 2 = 401 168 + 0;
  • 401 168 ÷ 2 = 200 584 + 0;
  • 200 584 ÷ 2 = 100 292 + 0;
  • 100 292 ÷ 2 = 50 146 + 0;
  • 50 146 ÷ 2 = 25 073 + 0;
  • 25 073 ÷ 2 = 12 536 + 1;
  • 12 536 ÷ 2 = 6 268 + 0;
  • 6 268 ÷ 2 = 3 134 + 0;
  • 3 134 ÷ 2 = 1 567 + 0;
  • 1 567 ÷ 2 = 783 + 1;
  • 783 ÷ 2 = 391 + 1;
  • 391 ÷ 2 = 195 + 1;
  • 195 ÷ 2 = 97 + 1;
  • 97 ÷ 2 = 48 + 1;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.


3 528 711 045 533 138 969(10) =


11 0000 1111 1000 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1001(2)


4. Normalize the binary representation of the number.

Shift the decimal mark 61 positions to the left, so that only one non zero digit remains to the left of it:


3 528 711 045 533 138 969(10) =


11 0000 1111 1000 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1001(2) =


11 0000 1111 1000 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1001(2) × 20 =


1.1000 0111 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 1(2) × 261


5. 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): 61


Mantissa (not normalized):
1.1000 0111 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 1


6. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


61 + 2(11-1) - 1 =


(61 + 1 023)(10) =


1 084(10)


7. 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;
  • 1 084 ÷ 2 = 542 + 0;
  • 542 ÷ 2 = 271 + 0;
  • 271 ÷ 2 = 135 + 1;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

8. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1084(10) =


100 0011 1100(2)


9. 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. 1000 0111 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0 0001 1001 =


1000 0111 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000


10. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
100 0011 1100


Mantissa (52 bits) =
1000 0111 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000


The base ten decimal number -3 528 711 045 533 138 969 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0011 1100 - 1000 0111 1100 0100 0000 0000 0000 0000 0000 0000 0000 0000 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