0.000 700 457 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 700 457 59(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
0.000 700 457 59(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: 0.000 700 457 59.

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) 0.000 700 457 59 × 2 = 0 + 0.001 400 915 18;
  • 2) 0.001 400 915 18 × 2 = 0 + 0.002 801 830 36;
  • 3) 0.002 801 830 36 × 2 = 0 + 0.005 603 660 72;
  • 4) 0.005 603 660 72 × 2 = 0 + 0.011 207 321 44;
  • 5) 0.011 207 321 44 × 2 = 0 + 0.022 414 642 88;
  • 6) 0.022 414 642 88 × 2 = 0 + 0.044 829 285 76;
  • 7) 0.044 829 285 76 × 2 = 0 + 0.089 658 571 52;
  • 8) 0.089 658 571 52 × 2 = 0 + 0.179 317 143 04;
  • 9) 0.179 317 143 04 × 2 = 0 + 0.358 634 286 08;
  • 10) 0.358 634 286 08 × 2 = 0 + 0.717 268 572 16;
  • 11) 0.717 268 572 16 × 2 = 1 + 0.434 537 144 32;
  • 12) 0.434 537 144 32 × 2 = 0 + 0.869 074 288 64;
  • 13) 0.869 074 288 64 × 2 = 1 + 0.738 148 577 28;
  • 14) 0.738 148 577 28 × 2 = 1 + 0.476 297 154 56;
  • 15) 0.476 297 154 56 × 2 = 0 + 0.952 594 309 12;
  • 16) 0.952 594 309 12 × 2 = 1 + 0.905 188 618 24;
  • 17) 0.905 188 618 24 × 2 = 1 + 0.810 377 236 48;
  • 18) 0.810 377 236 48 × 2 = 1 + 0.620 754 472 96;
  • 19) 0.620 754 472 96 × 2 = 1 + 0.241 508 945 92;
  • 20) 0.241 508 945 92 × 2 = 0 + 0.483 017 891 84;
  • 21) 0.483 017 891 84 × 2 = 0 + 0.966 035 783 68;
  • 22) 0.966 035 783 68 × 2 = 1 + 0.932 071 567 36;
  • 23) 0.932 071 567 36 × 2 = 1 + 0.864 143 134 72;
  • 24) 0.864 143 134 72 × 2 = 1 + 0.728 286 269 44;
  • 25) 0.728 286 269 44 × 2 = 1 + 0.456 572 538 88;
  • 26) 0.456 572 538 88 × 2 = 0 + 0.913 145 077 76;
  • 27) 0.913 145 077 76 × 2 = 1 + 0.826 290 155 52;
  • 28) 0.826 290 155 52 × 2 = 1 + 0.652 580 311 04;
  • 29) 0.652 580 311 04 × 2 = 1 + 0.305 160 622 08;
  • 30) 0.305 160 622 08 × 2 = 0 + 0.610 321 244 16;
  • 31) 0.610 321 244 16 × 2 = 1 + 0.220 642 488 32;
  • 32) 0.220 642 488 32 × 2 = 0 + 0.441 284 976 64;
  • 33) 0.441 284 976 64 × 2 = 0 + 0.882 569 953 28;
  • 34) 0.882 569 953 28 × 2 = 1 + 0.765 139 906 56;
  • 35) 0.765 139 906 56 × 2 = 1 + 0.530 279 813 12;
  • 36) 0.530 279 813 12 × 2 = 1 + 0.060 559 626 24;
  • 37) 0.060 559 626 24 × 2 = 0 + 0.121 119 252 48;
  • 38) 0.121 119 252 48 × 2 = 0 + 0.242 238 504 96;
  • 39) 0.242 238 504 96 × 2 = 0 + 0.484 477 009 92;
  • 40) 0.484 477 009 92 × 2 = 0 + 0.968 954 019 84;
  • 41) 0.968 954 019 84 × 2 = 1 + 0.937 908 039 68;
  • 42) 0.937 908 039 68 × 2 = 1 + 0.875 816 079 36;
  • 43) 0.875 816 079 36 × 2 = 1 + 0.751 632 158 72;
  • 44) 0.751 632 158 72 × 2 = 1 + 0.503 264 317 44;
  • 45) 0.503 264 317 44 × 2 = 1 + 0.006 528 634 88;
  • 46) 0.006 528 634 88 × 2 = 0 + 0.013 057 269 76;
  • 47) 0.013 057 269 76 × 2 = 0 + 0.026 114 539 52;
  • 48) 0.026 114 539 52 × 2 = 0 + 0.052 229 079 04;
  • 49) 0.052 229 079 04 × 2 = 0 + 0.104 458 158 08;
  • 50) 0.104 458 158 08 × 2 = 0 + 0.208 916 316 16;
  • 51) 0.208 916 316 16 × 2 = 0 + 0.417 832 632 32;
  • 52) 0.417 832 632 32 × 2 = 0 + 0.835 665 264 64;
  • 53) 0.835 665 264 64 × 2 = 1 + 0.671 330 529 28;
  • 54) 0.671 330 529 28 × 2 = 1 + 0.342 661 058 56;
  • 55) 0.342 661 058 56 × 2 = 0 + 0.685 322 117 12;
  • 56) 0.685 322 117 12 × 2 = 1 + 0.370 644 234 24;
  • 57) 0.370 644 234 24 × 2 = 0 + 0.741 288 468 48;
  • 58) 0.741 288 468 48 × 2 = 1 + 0.482 576 936 96;
  • 59) 0.482 576 936 96 × 2 = 0 + 0.965 153 873 92;
  • 60) 0.965 153 873 92 × 2 = 1 + 0.930 307 747 84;
  • 61) 0.930 307 747 84 × 2 = 1 + 0.860 615 495 68;
  • 62) 0.860 615 495 68 × 2 = 1 + 0.721 230 991 36;
  • 63) 0.721 230 991 36 × 2 = 1 + 0.442 461 982 72;

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:


0.000 700 457 59(10) =

0.0000 0000 0010 1101 1110 0111 1011 1010 0111 0000 1111 1000 0000 1101 0101 111(2)

5. Positive number before normalization:

0.000 700 457 59(10) =

0.0000 0000 0010 1101 1110 0111 1011 1010 0111 0000 1111 1000 0000 1101 0101 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 700 457 59(10) =

0.0000 0000 0010 1101 1110 0111 1011 1010 0111 0000 1111 1000 0000 1101 0101 111(2) =

0.0000 0000 0010 1101 1110 0111 1011 1010 0111 0000 1111 1000 0000 1101 0101 111(2) × 20 =

1.0110 1111 0011 1101 1101 0011 1000 0111 1100 0000 0110 1010 1111(2) × 2-11


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): -11

Mantissa (not normalized):
1.0110 1111 0011 1101 1101 0011 1000 0111 1100 0000 0110 1010 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-11 + 1 023)(10) =

1 012(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;
  • 1 012 ÷ 2 = 506 + 0;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1012(10) =

011 1111 0100(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. 0110 1111 0011 1101 1101 0011 1000 0111 1100 0000 0110 1010 1111 =

0110 1111 0011 1101 1101 0011 1000 0111 1100 0000 0110 1010 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) =
011 1111 0100

Mantissa (52 bits) =
0110 1111 0011 1101 1101 0011 1000 0111 1100 0000 0110 1010 1111


Decimal number 0.000 700 457 59 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0100 - 0110 1111 0011 1101 1101 0011 1000 0111 1100 0000 0110 1010 1111

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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