6.285 714 285 714 285 699 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 699 5(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
6.285 714 285 714 285 699 5(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: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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.

6(10) =

110(2)


3. Convert to binary (base 2) the fractional part: 0.285 714 285 714 285 699 5.

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) 0.285 714 285 714 285 699 5 × 2 = 0 + 0.571 428 571 428 571 399;
  • 2) 0.571 428 571 428 571 399 × 2 = 1 + 0.142 857 142 857 142 798;
  • 3) 0.142 857 142 857 142 798 × 2 = 0 + 0.285 714 285 714 285 596;
  • 4) 0.285 714 285 714 285 596 × 2 = 0 + 0.571 428 571 428 571 192;
  • 5) 0.571 428 571 428 571 192 × 2 = 1 + 0.142 857 142 857 142 384;
  • 6) 0.142 857 142 857 142 384 × 2 = 0 + 0.285 714 285 714 284 768;
  • 7) 0.285 714 285 714 284 768 × 2 = 0 + 0.571 428 571 428 569 536;
  • 8) 0.571 428 571 428 569 536 × 2 = 1 + 0.142 857 142 857 139 072;
  • 9) 0.142 857 142 857 139 072 × 2 = 0 + 0.285 714 285 714 278 144;
  • 10) 0.285 714 285 714 278 144 × 2 = 0 + 0.571 428 571 428 556 288;
  • 11) 0.571 428 571 428 556 288 × 2 = 1 + 0.142 857 142 857 112 576;
  • 12) 0.142 857 142 857 112 576 × 2 = 0 + 0.285 714 285 714 225 152;
  • 13) 0.285 714 285 714 225 152 × 2 = 0 + 0.571 428 571 428 450 304;
  • 14) 0.571 428 571 428 450 304 × 2 = 1 + 0.142 857 142 856 900 608;
  • 15) 0.142 857 142 856 900 608 × 2 = 0 + 0.285 714 285 713 801 216;
  • 16) 0.285 714 285 713 801 216 × 2 = 0 + 0.571 428 571 427 602 432;
  • 17) 0.571 428 571 427 602 432 × 2 = 1 + 0.142 857 142 855 204 864;
  • 18) 0.142 857 142 855 204 864 × 2 = 0 + 0.285 714 285 710 409 728;
  • 19) 0.285 714 285 710 409 728 × 2 = 0 + 0.571 428 571 420 819 456;
  • 20) 0.571 428 571 420 819 456 × 2 = 1 + 0.142 857 142 841 638 912;
  • 21) 0.142 857 142 841 638 912 × 2 = 0 + 0.285 714 285 683 277 824;
  • 22) 0.285 714 285 683 277 824 × 2 = 0 + 0.571 428 571 366 555 648;
  • 23) 0.571 428 571 366 555 648 × 2 = 1 + 0.142 857 142 733 111 296;
  • 24) 0.142 857 142 733 111 296 × 2 = 0 + 0.285 714 285 466 222 592;
  • 25) 0.285 714 285 466 222 592 × 2 = 0 + 0.571 428 570 932 445 184;
  • 26) 0.571 428 570 932 445 184 × 2 = 1 + 0.142 857 141 864 890 368;
  • 27) 0.142 857 141 864 890 368 × 2 = 0 + 0.285 714 283 729 780 736;
  • 28) 0.285 714 283 729 780 736 × 2 = 0 + 0.571 428 567 459 561 472;
  • 29) 0.571 428 567 459 561 472 × 2 = 1 + 0.142 857 134 919 122 944;
  • 30) 0.142 857 134 919 122 944 × 2 = 0 + 0.285 714 269 838 245 888;
  • 31) 0.285 714 269 838 245 888 × 2 = 0 + 0.571 428 539 676 491 776;
  • 32) 0.571 428 539 676 491 776 × 2 = 1 + 0.142 857 079 352 983 552;
  • 33) 0.142 857 079 352 983 552 × 2 = 0 + 0.285 714 158 705 967 104;
  • 34) 0.285 714 158 705 967 104 × 2 = 0 + 0.571 428 317 411 934 208;
  • 35) 0.571 428 317 411 934 208 × 2 = 1 + 0.142 856 634 823 868 416;
  • 36) 0.142 856 634 823 868 416 × 2 = 0 + 0.285 713 269 647 736 832;
  • 37) 0.285 713 269 647 736 832 × 2 = 0 + 0.571 426 539 295 473 664;
  • 38) 0.571 426 539 295 473 664 × 2 = 1 + 0.142 853 078 590 947 328;
  • 39) 0.142 853 078 590 947 328 × 2 = 0 + 0.285 706 157 181 894 656;
  • 40) 0.285 706 157 181 894 656 × 2 = 0 + 0.571 412 314 363 789 312;
  • 41) 0.571 412 314 363 789 312 × 2 = 1 + 0.142 824 628 727 578 624;
  • 42) 0.142 824 628 727 578 624 × 2 = 0 + 0.285 649 257 455 157 248;
  • 43) 0.285 649 257 455 157 248 × 2 = 0 + 0.571 298 514 910 314 496;
  • 44) 0.571 298 514 910 314 496 × 2 = 1 + 0.142 597 029 820 628 992;
  • 45) 0.142 597 029 820 628 992 × 2 = 0 + 0.285 194 059 641 257 984;
  • 46) 0.285 194 059 641 257 984 × 2 = 0 + 0.570 388 119 282 515 968;
  • 47) 0.570 388 119 282 515 968 × 2 = 1 + 0.140 776 238 565 031 936;
  • 48) 0.140 776 238 565 031 936 × 2 = 0 + 0.281 552 477 130 063 872;
  • 49) 0.281 552 477 130 063 872 × 2 = 0 + 0.563 104 954 260 127 744;
  • 50) 0.563 104 954 260 127 744 × 2 = 1 + 0.126 209 908 520 255 488;
  • 51) 0.126 209 908 520 255 488 × 2 = 0 + 0.252 419 817 040 510 976;
  • 52) 0.252 419 817 040 510 976 × 2 = 0 + 0.504 839 634 081 021 952;
  • 53) 0.504 839 634 081 021 952 × 2 = 1 + 0.009 679 268 162 043 904;

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:


0.285 714 285 714 285 699 5(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2)

5. Positive number before normalization:

6.285 714 285 714 285 699 5(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


6.285 714 285 714 285 699 5(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 001(2) × 22


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): 2

Mantissa (not normalized):
1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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) =

1025(10) =

100 0000 0001(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. 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 001 =

1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


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 0001

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


Decimal number 6.285 714 285 714 285 699 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001

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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