2.629 632 384 924 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.629 632 384 924(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
2.629 632 384 924(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: 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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.629 632 384 924.

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) 0.629 632 384 924 × 2 = 1 + 0.259 264 769 848;
  • 2) 0.259 264 769 848 × 2 = 0 + 0.518 529 539 696;
  • 3) 0.518 529 539 696 × 2 = 1 + 0.037 059 079 392;
  • 4) 0.037 059 079 392 × 2 = 0 + 0.074 118 158 784;
  • 5) 0.074 118 158 784 × 2 = 0 + 0.148 236 317 568;
  • 6) 0.148 236 317 568 × 2 = 0 + 0.296 472 635 136;
  • 7) 0.296 472 635 136 × 2 = 0 + 0.592 945 270 272;
  • 8) 0.592 945 270 272 × 2 = 1 + 0.185 890 540 544;
  • 9) 0.185 890 540 544 × 2 = 0 + 0.371 781 081 088;
  • 10) 0.371 781 081 088 × 2 = 0 + 0.743 562 162 176;
  • 11) 0.743 562 162 176 × 2 = 1 + 0.487 124 324 352;
  • 12) 0.487 124 324 352 × 2 = 0 + 0.974 248 648 704;
  • 13) 0.974 248 648 704 × 2 = 1 + 0.948 497 297 408;
  • 14) 0.948 497 297 408 × 2 = 1 + 0.896 994 594 816;
  • 15) 0.896 994 594 816 × 2 = 1 + 0.793 989 189 632;
  • 16) 0.793 989 189 632 × 2 = 1 + 0.587 978 379 264;
  • 17) 0.587 978 379 264 × 2 = 1 + 0.175 956 758 528;
  • 18) 0.175 956 758 528 × 2 = 0 + 0.351 913 517 056;
  • 19) 0.351 913 517 056 × 2 = 0 + 0.703 827 034 112;
  • 20) 0.703 827 034 112 × 2 = 1 + 0.407 654 068 224;
  • 21) 0.407 654 068 224 × 2 = 0 + 0.815 308 136 448;
  • 22) 0.815 308 136 448 × 2 = 1 + 0.630 616 272 896;
  • 23) 0.630 616 272 896 × 2 = 1 + 0.261 232 545 792;
  • 24) 0.261 232 545 792 × 2 = 0 + 0.522 465 091 584;
  • 25) 0.522 465 091 584 × 2 = 1 + 0.044 930 183 168;
  • 26) 0.044 930 183 168 × 2 = 0 + 0.089 860 366 336;
  • 27) 0.089 860 366 336 × 2 = 0 + 0.179 720 732 672;
  • 28) 0.179 720 732 672 × 2 = 0 + 0.359 441 465 344;
  • 29) 0.359 441 465 344 × 2 = 0 + 0.718 882 930 688;
  • 30) 0.718 882 930 688 × 2 = 1 + 0.437 765 861 376;
  • 31) 0.437 765 861 376 × 2 = 0 + 0.875 531 722 752;
  • 32) 0.875 531 722 752 × 2 = 1 + 0.751 063 445 504;
  • 33) 0.751 063 445 504 × 2 = 1 + 0.502 126 891 008;
  • 34) 0.502 126 891 008 × 2 = 1 + 0.004 253 782 016;
  • 35) 0.004 253 782 016 × 2 = 0 + 0.008 507 564 032;
  • 36) 0.008 507 564 032 × 2 = 0 + 0.017 015 128 064;
  • 37) 0.017 015 128 064 × 2 = 0 + 0.034 030 256 128;
  • 38) 0.034 030 256 128 × 2 = 0 + 0.068 060 512 256;
  • 39) 0.068 060 512 256 × 2 = 0 + 0.136 121 024 512;
  • 40) 0.136 121 024 512 × 2 = 0 + 0.272 242 049 024;
  • 41) 0.272 242 049 024 × 2 = 0 + 0.544 484 098 048;
  • 42) 0.544 484 098 048 × 2 = 1 + 0.088 968 196 096;
  • 43) 0.088 968 196 096 × 2 = 0 + 0.177 936 392 192;
  • 44) 0.177 936 392 192 × 2 = 0 + 0.355 872 784 384;
  • 45) 0.355 872 784 384 × 2 = 0 + 0.711 745 568 768;
  • 46) 0.711 745 568 768 × 2 = 1 + 0.423 491 137 536;
  • 47) 0.423 491 137 536 × 2 = 0 + 0.846 982 275 072;
  • 48) 0.846 982 275 072 × 2 = 1 + 0.693 964 550 144;
  • 49) 0.693 964 550 144 × 2 = 1 + 0.387 929 100 288;
  • 50) 0.387 929 100 288 × 2 = 0 + 0.775 858 200 576;
  • 51) 0.775 858 200 576 × 2 = 1 + 0.551 716 401 152;
  • 52) 0.551 716 401 152 × 2 = 1 + 0.103 432 802 304;
  • 53) 0.103 432 802 304 × 2 = 0 + 0.206 865 604 608;

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:


0.629 632 384 924(10) =

0.1010 0001 0010 1111 1001 0110 1000 0101 1100 0000 0100 0101 1011 0(2)

5. Positive number before normalization:

2.629 632 384 924(10) =

10.1010 0001 0010 1111 1001 0110 1000 0101 1100 0000 0100 0101 1011 0(2)

6. Normalize the binary representation of the number.

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


2.629 632 384 924(10) =

10.1010 0001 0010 1111 1001 0110 1000 0101 1100 0000 0100 0101 1011 0(2) =

10.1010 0001 0010 1111 1001 0110 1000 0101 1100 0000 0100 0101 1011 0(2) × 20 =

1.0101 0000 1001 0111 1100 1011 0100 0010 1110 0000 0010 0010 1101 10(2) × 21


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):
1.0101 0000 1001 0111 1100 1011 0100 0010 1110 0000 0010 0010 1101 10


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0101 0000 1001 0111 1100 1011 0100 0010 1110 0000 0010 0010 1101 10 =

0101 0000 1001 0111 1100 1011 0100 0010 1110 0000 0010 0010 1101


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) =
100 0000 0000

Mantissa (52 bits) =
0101 0000 1001 0111 1100 1011 0100 0010 1110 0000 0010 0010 1101


Decimal number 2.629 632 384 924 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 0000 1001 0111 1100 1011 0100 0010 1110 0000 0010 0010 1101

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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