64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1.637 395 701 684 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1.637 395 701 684(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. First, convert to binary (in base 2) the integer part: 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;
  • 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.


1(10) =


1(2)


3. Convert to binary (base 2) the fractional part: 0.637 395 701 684.

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) 0.637 395 701 684 × 2 = 1 + 0.274 791 403 368;
  • 2) 0.274 791 403 368 × 2 = 0 + 0.549 582 806 736;
  • 3) 0.549 582 806 736 × 2 = 1 + 0.099 165 613 472;
  • 4) 0.099 165 613 472 × 2 = 0 + 0.198 331 226 944;
  • 5) 0.198 331 226 944 × 2 = 0 + 0.396 662 453 888;
  • 6) 0.396 662 453 888 × 2 = 0 + 0.793 324 907 776;
  • 7) 0.793 324 907 776 × 2 = 1 + 0.586 649 815 552;
  • 8) 0.586 649 815 552 × 2 = 1 + 0.173 299 631 104;
  • 9) 0.173 299 631 104 × 2 = 0 + 0.346 599 262 208;
  • 10) 0.346 599 262 208 × 2 = 0 + 0.693 198 524 416;
  • 11) 0.693 198 524 416 × 2 = 1 + 0.386 397 048 832;
  • 12) 0.386 397 048 832 × 2 = 0 + 0.772 794 097 664;
  • 13) 0.772 794 097 664 × 2 = 1 + 0.545 588 195 328;
  • 14) 0.545 588 195 328 × 2 = 1 + 0.091 176 390 656;
  • 15) 0.091 176 390 656 × 2 = 0 + 0.182 352 781 312;
  • 16) 0.182 352 781 312 × 2 = 0 + 0.364 705 562 624;
  • 17) 0.364 705 562 624 × 2 = 0 + 0.729 411 125 248;
  • 18) 0.729 411 125 248 × 2 = 1 + 0.458 822 250 496;
  • 19) 0.458 822 250 496 × 2 = 0 + 0.917 644 500 992;
  • 20) 0.917 644 500 992 × 2 = 1 + 0.835 289 001 984;
  • 21) 0.835 289 001 984 × 2 = 1 + 0.670 578 003 968;
  • 22) 0.670 578 003 968 × 2 = 1 + 0.341 156 007 936;
  • 23) 0.341 156 007 936 × 2 = 0 + 0.682 312 015 872;
  • 24) 0.682 312 015 872 × 2 = 1 + 0.364 624 031 744;
  • 25) 0.364 624 031 744 × 2 = 0 + 0.729 248 063 488;
  • 26) 0.729 248 063 488 × 2 = 1 + 0.458 496 126 976;
  • 27) 0.458 496 126 976 × 2 = 0 + 0.916 992 253 952;
  • 28) 0.916 992 253 952 × 2 = 1 + 0.833 984 507 904;
  • 29) 0.833 984 507 904 × 2 = 1 + 0.667 969 015 808;
  • 30) 0.667 969 015 808 × 2 = 1 + 0.335 938 031 616;
  • 31) 0.335 938 031 616 × 2 = 0 + 0.671 876 063 232;
  • 32) 0.671 876 063 232 × 2 = 1 + 0.343 752 126 464;
  • 33) 0.343 752 126 464 × 2 = 0 + 0.687 504 252 928;
  • 34) 0.687 504 252 928 × 2 = 1 + 0.375 008 505 856;
  • 35) 0.375 008 505 856 × 2 = 0 + 0.750 017 011 712;
  • 36) 0.750 017 011 712 × 2 = 1 + 0.500 034 023 424;
  • 37) 0.500 034 023 424 × 2 = 1 + 0.000 068 046 848;
  • 38) 0.000 068 046 848 × 2 = 0 + 0.000 136 093 696;
  • 39) 0.000 136 093 696 × 2 = 0 + 0.000 272 187 392;
  • 40) 0.000 272 187 392 × 2 = 0 + 0.000 544 374 784;
  • 41) 0.000 544 374 784 × 2 = 0 + 0.001 088 749 568;
  • 42) 0.001 088 749 568 × 2 = 0 + 0.002 177 499 136;
  • 43) 0.002 177 499 136 × 2 = 0 + 0.004 354 998 272;
  • 44) 0.004 354 998 272 × 2 = 0 + 0.008 709 996 544;
  • 45) 0.008 709 996 544 × 2 = 0 + 0.017 419 993 088;
  • 46) 0.017 419 993 088 × 2 = 0 + 0.034 839 986 176;
  • 47) 0.034 839 986 176 × 2 = 0 + 0.069 679 972 352;
  • 48) 0.069 679 972 352 × 2 = 0 + 0.139 359 944 704;
  • 49) 0.139 359 944 704 × 2 = 0 + 0.278 719 889 408;
  • 50) 0.278 719 889 408 × 2 = 0 + 0.557 439 778 816;
  • 51) 0.557 439 778 816 × 2 = 1 + 0.114 879 557 632;
  • 52) 0.114 879 557 632 × 2 = 0 + 0.229 759 115 264;
  • 53) 0.229 759 115 264 × 2 = 0 + 0.459 518 230 528;

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...)


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:


0.637 395 701 684(10) =


0.1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010 0(2)


5. Positive number before normalization:

1.637 395 701 684(10) =


1.1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.637 395 701 684(10) =


1.1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010 0(2) =


1.1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010 0(2) × 20


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): 0


Mantissa (not normalized):
1.1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


0 + 2(11-1) - 1 =


(0 + 1 023)(10) =


1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =


1023(10) =


011 1111 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, 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. 1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010 0 =


1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010


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 1111


Mantissa (52 bits) =
1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010


The base ten decimal number 1.637 395 701 684 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1111 - 1010 0011 0010 1100 0101 1101 0101 1101 0101 1000 0000 0000 0010

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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