0.142 857 142 857 142 867 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.142 857 142 857 142 867(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
0.142 857 142 857 142 867(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: 0.142 857 142 857 142 867.

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) 0.142 857 142 857 142 867 × 2 = 0 + 0.285 714 285 714 285 734;
  • 2) 0.285 714 285 714 285 734 × 2 = 0 + 0.571 428 571 428 571 468;
  • 3) 0.571 428 571 428 571 468 × 2 = 1 + 0.142 857 142 857 142 936;
  • 4) 0.142 857 142 857 142 936 × 2 = 0 + 0.285 714 285 714 285 872;
  • 5) 0.285 714 285 714 285 872 × 2 = 0 + 0.571 428 571 428 571 744;
  • 6) 0.571 428 571 428 571 744 × 2 = 1 + 0.142 857 142 857 143 488;
  • 7) 0.142 857 142 857 143 488 × 2 = 0 + 0.285 714 285 714 286 976;
  • 8) 0.285 714 285 714 286 976 × 2 = 0 + 0.571 428 571 428 573 952;
  • 9) 0.571 428 571 428 573 952 × 2 = 1 + 0.142 857 142 857 147 904;
  • 10) 0.142 857 142 857 147 904 × 2 = 0 + 0.285 714 285 714 295 808;
  • 11) 0.285 714 285 714 295 808 × 2 = 0 + 0.571 428 571 428 591 616;
  • 12) 0.571 428 571 428 591 616 × 2 = 1 + 0.142 857 142 857 183 232;
  • 13) 0.142 857 142 857 183 232 × 2 = 0 + 0.285 714 285 714 366 464;
  • 14) 0.285 714 285 714 366 464 × 2 = 0 + 0.571 428 571 428 732 928;
  • 15) 0.571 428 571 428 732 928 × 2 = 1 + 0.142 857 142 857 465 856;
  • 16) 0.142 857 142 857 465 856 × 2 = 0 + 0.285 714 285 714 931 712;
  • 17) 0.285 714 285 714 931 712 × 2 = 0 + 0.571 428 571 429 863 424;
  • 18) 0.571 428 571 429 863 424 × 2 = 1 + 0.142 857 142 859 726 848;
  • 19) 0.142 857 142 859 726 848 × 2 = 0 + 0.285 714 285 719 453 696;
  • 20) 0.285 714 285 719 453 696 × 2 = 0 + 0.571 428 571 438 907 392;
  • 21) 0.571 428 571 438 907 392 × 2 = 1 + 0.142 857 142 877 814 784;
  • 22) 0.142 857 142 877 814 784 × 2 = 0 + 0.285 714 285 755 629 568;
  • 23) 0.285 714 285 755 629 568 × 2 = 0 + 0.571 428 571 511 259 136;
  • 24) 0.571 428 571 511 259 136 × 2 = 1 + 0.142 857 143 022 518 272;
  • 25) 0.142 857 143 022 518 272 × 2 = 0 + 0.285 714 286 045 036 544;
  • 26) 0.285 714 286 045 036 544 × 2 = 0 + 0.571 428 572 090 073 088;
  • 27) 0.571 428 572 090 073 088 × 2 = 1 + 0.142 857 144 180 146 176;
  • 28) 0.142 857 144 180 146 176 × 2 = 0 + 0.285 714 288 360 292 352;
  • 29) 0.285 714 288 360 292 352 × 2 = 0 + 0.571 428 576 720 584 704;
  • 30) 0.571 428 576 720 584 704 × 2 = 1 + 0.142 857 153 441 169 408;
  • 31) 0.142 857 153 441 169 408 × 2 = 0 + 0.285 714 306 882 338 816;
  • 32) 0.285 714 306 882 338 816 × 2 = 0 + 0.571 428 613 764 677 632;
  • 33) 0.571 428 613 764 677 632 × 2 = 1 + 0.142 857 227 529 355 264;
  • 34) 0.142 857 227 529 355 264 × 2 = 0 + 0.285 714 455 058 710 528;
  • 35) 0.285 714 455 058 710 528 × 2 = 0 + 0.571 428 910 117 421 056;
  • 36) 0.571 428 910 117 421 056 × 2 = 1 + 0.142 857 820 234 842 112;
  • 37) 0.142 857 820 234 842 112 × 2 = 0 + 0.285 715 640 469 684 224;
  • 38) 0.285 715 640 469 684 224 × 2 = 0 + 0.571 431 280 939 368 448;
  • 39) 0.571 431 280 939 368 448 × 2 = 1 + 0.142 862 561 878 736 896;
  • 40) 0.142 862 561 878 736 896 × 2 = 0 + 0.285 725 123 757 473 792;
  • 41) 0.285 725 123 757 473 792 × 2 = 0 + 0.571 450 247 514 947 584;
  • 42) 0.571 450 247 514 947 584 × 2 = 1 + 0.142 900 495 029 895 168;
  • 43) 0.142 900 495 029 895 168 × 2 = 0 + 0.285 800 990 059 790 336;
  • 44) 0.285 800 990 059 790 336 × 2 = 0 + 0.571 601 980 119 580 672;
  • 45) 0.571 601 980 119 580 672 × 2 = 1 + 0.143 203 960 239 161 344;
  • 46) 0.143 203 960 239 161 344 × 2 = 0 + 0.286 407 920 478 322 688;
  • 47) 0.286 407 920 478 322 688 × 2 = 0 + 0.572 815 840 956 645 376;
  • 48) 0.572 815 840 956 645 376 × 2 = 1 + 0.145 631 681 913 290 752;
  • 49) 0.145 631 681 913 290 752 × 2 = 0 + 0.291 263 363 826 581 504;
  • 50) 0.291 263 363 826 581 504 × 2 = 0 + 0.582 526 727 653 163 008;
  • 51) 0.582 526 727 653 163 008 × 2 = 1 + 0.165 053 455 306 326 016;
  • 52) 0.165 053 455 306 326 016 × 2 = 0 + 0.330 106 910 612 652 032;
  • 53) 0.330 106 910 612 652 032 × 2 = 0 + 0.660 213 821 225 304 064;
  • 54) 0.660 213 821 225 304 064 × 2 = 1 + 0.320 427 642 450 608 128;
  • 55) 0.320 427 642 450 608 128 × 2 = 0 + 0.640 855 284 901 216 256;

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:


0.142 857 142 857 142 867(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(2)

5. Positive number before normalization:

0.142 857 142 857 142 867(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.142 857 142 857 142 867(10) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(2) =

0.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 010(2) × 20 =

1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010(2) × 2-3


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): -3

Mantissa (not normalized):
1.0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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) =

1020(10) =

011 1111 1100(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. 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 =

0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 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 1100

Mantissa (52 bits) =
0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010


Decimal number 0.142 857 142 857 142 867 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1100 - 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 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