Decimal to 64 Bit IEEE 754 Binary: Convert Number 9 439 544 819 036 717 377 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 9 439 544 819 036 717 377(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. 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;
  • 9 439 544 819 036 717 377 ÷ 2 = 4 719 772 409 518 358 688 + 1;
  • 4 719 772 409 518 358 688 ÷ 2 = 2 359 886 204 759 179 344 + 0;
  • 2 359 886 204 759 179 344 ÷ 2 = 1 179 943 102 379 589 672 + 0;
  • 1 179 943 102 379 589 672 ÷ 2 = 589 971 551 189 794 836 + 0;
  • 589 971 551 189 794 836 ÷ 2 = 294 985 775 594 897 418 + 0;
  • 294 985 775 594 897 418 ÷ 2 = 147 492 887 797 448 709 + 0;
  • 147 492 887 797 448 709 ÷ 2 = 73 746 443 898 724 354 + 1;
  • 73 746 443 898 724 354 ÷ 2 = 36 873 221 949 362 177 + 0;
  • 36 873 221 949 362 177 ÷ 2 = 18 436 610 974 681 088 + 1;
  • 18 436 610 974 681 088 ÷ 2 = 9 218 305 487 340 544 + 0;
  • 9 218 305 487 340 544 ÷ 2 = 4 609 152 743 670 272 + 0;
  • 4 609 152 743 670 272 ÷ 2 = 2 304 576 371 835 136 + 0;
  • 2 304 576 371 835 136 ÷ 2 = 1 152 288 185 917 568 + 0;
  • 1 152 288 185 917 568 ÷ 2 = 576 144 092 958 784 + 0;
  • 576 144 092 958 784 ÷ 2 = 288 072 046 479 392 + 0;
  • 288 072 046 479 392 ÷ 2 = 144 036 023 239 696 + 0;
  • 144 036 023 239 696 ÷ 2 = 72 018 011 619 848 + 0;
  • 72 018 011 619 848 ÷ 2 = 36 009 005 809 924 + 0;
  • 36 009 005 809 924 ÷ 2 = 18 004 502 904 962 + 0;
  • 18 004 502 904 962 ÷ 2 = 9 002 251 452 481 + 0;
  • 9 002 251 452 481 ÷ 2 = 4 501 125 726 240 + 1;
  • 4 501 125 726 240 ÷ 2 = 2 250 562 863 120 + 0;
  • 2 250 562 863 120 ÷ 2 = 1 125 281 431 560 + 0;
  • 1 125 281 431 560 ÷ 2 = 562 640 715 780 + 0;
  • 562 640 715 780 ÷ 2 = 281 320 357 890 + 0;
  • 281 320 357 890 ÷ 2 = 140 660 178 945 + 0;
  • 140 660 178 945 ÷ 2 = 70 330 089 472 + 1;
  • 70 330 089 472 ÷ 2 = 35 165 044 736 + 0;
  • 35 165 044 736 ÷ 2 = 17 582 522 368 + 0;
  • 17 582 522 368 ÷ 2 = 8 791 261 184 + 0;
  • 8 791 261 184 ÷ 2 = 4 395 630 592 + 0;
  • 4 395 630 592 ÷ 2 = 2 197 815 296 + 0;
  • 2 197 815 296 ÷ 2 = 1 098 907 648 + 0;
  • 1 098 907 648 ÷ 2 = 549 453 824 + 0;
  • 549 453 824 ÷ 2 = 274 726 912 + 0;
  • 274 726 912 ÷ 2 = 137 363 456 + 0;
  • 137 363 456 ÷ 2 = 68 681 728 + 0;
  • 68 681 728 ÷ 2 = 34 340 864 + 0;
  • 34 340 864 ÷ 2 = 17 170 432 + 0;
  • 17 170 432 ÷ 2 = 8 585 216 + 0;
  • 8 585 216 ÷ 2 = 4 292 608 + 0;
  • 4 292 608 ÷ 2 = 2 146 304 + 0;
  • 2 146 304 ÷ 2 = 1 073 152 + 0;
  • 1 073 152 ÷ 2 = 536 576 + 0;
  • 536 576 ÷ 2 = 268 288 + 0;
  • 268 288 ÷ 2 = 134 144 + 0;
  • 134 144 ÷ 2 = 67 072 + 0;
  • 67 072 ÷ 2 = 33 536 + 0;
  • 33 536 ÷ 2 = 16 768 + 0;
  • 16 768 ÷ 2 = 8 384 + 0;
  • 8 384 ÷ 2 = 4 192 + 0;
  • 4 192 ÷ 2 = 2 096 + 0;
  • 2 096 ÷ 2 = 1 048 + 0;
  • 1 048 ÷ 2 = 524 + 0;
  • 524 ÷ 2 = 262 + 0;
  • 262 ÷ 2 = 131 + 0;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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

9 439 544 819 036 717 377(10) =


1000 0011 0000 0000 0000 0000 0000 0000 0000 0100 0001 0000 0000 0001 0100 0001(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the left, so that only one non zero digit remains to the left of it:


9 439 544 819 036 717 377(10) =


1000 0011 0000 0000 0000 0000 0000 0000 0000 0100 0001 0000 0000 0001 0100 0001(2) =


1000 0011 0000 0000 0000 0000 0000 0000 0000 0100 0001 0000 0000 0001 0100 0001(2) × 20 =


1.0000 0110 0000 0000 0000 0000 0000 0000 0000 1000 0010 0000 0000 0010 1000 001(2) × 263


4. 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): 63


Mantissa (not normalized):
1.0000 0110 0000 0000 0000 0000 0000 0000 0000 1000 0010 0000 0000 0010 1000 001


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


63 + 2(11-1) - 1 =


(63 + 1 023)(10) =


1 086(10)


6. 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;
  • 1 086 ÷ 2 = 543 + 0;
  • 543 ÷ 2 = 271 + 1;
  • 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;

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

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


Exponent (adjusted) =


1086(10) =


100 0011 1110(2)


8. 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. 0000 0110 0000 0000 0000 0000 0000 0000 0000 1000 0010 0000 0000 001 0100 0001 =


0000 0110 0000 0000 0000 0000 0000 0000 0000 1000 0010 0000 0000


9. 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 0011 1110


Mantissa (52 bits) =
0000 0110 0000 0000 0000 0000 0000 0000 0000 1000 0010 0000 0000


The base ten decimal number 9 439 544 819 036 717 377 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0011 1110 - 0000 0110 0000 0000 0000 0000 0000 0000 0000 1000 0010 0000 0000

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