2.714 285 714 157 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.714 285 714 157(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
2.714 285 714 157(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: 0.714 285 714 157.

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) 0.714 285 714 157 × 2 = 1 + 0.428 571 428 314;
  • 2) 0.428 571 428 314 × 2 = 0 + 0.857 142 856 628;
  • 3) 0.857 142 856 628 × 2 = 1 + 0.714 285 713 256;
  • 4) 0.714 285 713 256 × 2 = 1 + 0.428 571 426 512;
  • 5) 0.428 571 426 512 × 2 = 0 + 0.857 142 853 024;
  • 6) 0.857 142 853 024 × 2 = 1 + 0.714 285 706 048;
  • 7) 0.714 285 706 048 × 2 = 1 + 0.428 571 412 096;
  • 8) 0.428 571 412 096 × 2 = 0 + 0.857 142 824 192;
  • 9) 0.857 142 824 192 × 2 = 1 + 0.714 285 648 384;
  • 10) 0.714 285 648 384 × 2 = 1 + 0.428 571 296 768;
  • 11) 0.428 571 296 768 × 2 = 0 + 0.857 142 593 536;
  • 12) 0.857 142 593 536 × 2 = 1 + 0.714 285 187 072;
  • 13) 0.714 285 187 072 × 2 = 1 + 0.428 570 374 144;
  • 14) 0.428 570 374 144 × 2 = 0 + 0.857 140 748 288;
  • 15) 0.857 140 748 288 × 2 = 1 + 0.714 281 496 576;
  • 16) 0.714 281 496 576 × 2 = 1 + 0.428 562 993 152;
  • 17) 0.428 562 993 152 × 2 = 0 + 0.857 125 986 304;
  • 18) 0.857 125 986 304 × 2 = 1 + 0.714 251 972 608;
  • 19) 0.714 251 972 608 × 2 = 1 + 0.428 503 945 216;
  • 20) 0.428 503 945 216 × 2 = 0 + 0.857 007 890 432;
  • 21) 0.857 007 890 432 × 2 = 1 + 0.714 015 780 864;
  • 22) 0.714 015 780 864 × 2 = 1 + 0.428 031 561 728;
  • 23) 0.428 031 561 728 × 2 = 0 + 0.856 063 123 456;
  • 24) 0.856 063 123 456 × 2 = 1 + 0.712 126 246 912;
  • 25) 0.712 126 246 912 × 2 = 1 + 0.424 252 493 824;
  • 26) 0.424 252 493 824 × 2 = 0 + 0.848 504 987 648;
  • 27) 0.848 504 987 648 × 2 = 1 + 0.697 009 975 296;
  • 28) 0.697 009 975 296 × 2 = 1 + 0.394 019 950 592;
  • 29) 0.394 019 950 592 × 2 = 0 + 0.788 039 901 184;
  • 30) 0.788 039 901 184 × 2 = 1 + 0.576 079 802 368;
  • 31) 0.576 079 802 368 × 2 = 1 + 0.152 159 604 736;
  • 32) 0.152 159 604 736 × 2 = 0 + 0.304 319 209 472;
  • 33) 0.304 319 209 472 × 2 = 0 + 0.608 638 418 944;
  • 34) 0.608 638 418 944 × 2 = 1 + 0.217 276 837 888;
  • 35) 0.217 276 837 888 × 2 = 0 + 0.434 553 675 776;
  • 36) 0.434 553 675 776 × 2 = 0 + 0.869 107 351 552;
  • 37) 0.869 107 351 552 × 2 = 1 + 0.738 214 703 104;
  • 38) 0.738 214 703 104 × 2 = 1 + 0.476 429 406 208;
  • 39) 0.476 429 406 208 × 2 = 0 + 0.952 858 812 416;
  • 40) 0.952 858 812 416 × 2 = 1 + 0.905 717 624 832;
  • 41) 0.905 717 624 832 × 2 = 1 + 0.811 435 249 664;
  • 42) 0.811 435 249 664 × 2 = 1 + 0.622 870 499 328;
  • 43) 0.622 870 499 328 × 2 = 1 + 0.245 740 998 656;
  • 44) 0.245 740 998 656 × 2 = 0 + 0.491 481 997 312;
  • 45) 0.491 481 997 312 × 2 = 0 + 0.982 963 994 624;
  • 46) 0.982 963 994 624 × 2 = 1 + 0.965 927 989 248;
  • 47) 0.965 927 989 248 × 2 = 1 + 0.931 855 978 496;
  • 48) 0.931 855 978 496 × 2 = 1 + 0.863 711 956 992;
  • 49) 0.863 711 956 992 × 2 = 1 + 0.727 423 913 984;
  • 50) 0.727 423 913 984 × 2 = 1 + 0.454 847 827 968;
  • 51) 0.454 847 827 968 × 2 = 0 + 0.909 695 655 936;
  • 52) 0.909 695 655 936 × 2 = 1 + 0.819 391 311 872;
  • 53) 0.819 391 311 872 × 2 = 1 + 0.638 782 623 744;

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:


0.714 285 714 157(10) =

0.1011 0110 1101 1011 0110 1101 1011 0110 0100 1101 1110 0111 1101 1(2)

5. Positive number before normalization:

2.714 285 714 157(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 0100 1101 1110 0111 1101 1(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:


2.714 285 714 157(10) =

10.1011 0110 1101 1011 0110 1101 1011 0110 0100 1101 1110 0111 1101 1(2) =

10.1011 0110 1101 1011 0110 1101 1011 0110 0100 1101 1110 0111 1101 1(2) × 20 =

1.0101 1011 0110 1101 1011 0110 1101 1011 0010 0110 1111 0011 1110 11(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 1011 0110 1101 1011 0110 1101 1011 0010 0110 1111 0011 1110 11


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 1011 0110 1101 1011 0110 1101 1011 0010 0110 1111 0011 1110 11 =

0101 1011 0110 1101 1011 0110 1101 1011 0010 0110 1111 0011 1110


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 1011 0110 1101 1011 0110 1101 1011 0010 0110 1111 0011 1110


Decimal number 2.714 285 714 157 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 1011 0110 1101 1011 0110 1101 1011 0010 0110 1111 0011 1110

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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