6.285 714 285 714 285 714 285 714 285 776 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 714 285 714 285 776(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
6.285 714 285 714 285 714 285 714 285 776(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: 0.285 714 285 714 285 714 285 714 285 776.

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) 0.285 714 285 714 285 714 285 714 285 776 × 2 = 0 + 0.571 428 571 428 571 428 571 428 571 552;
  • 2) 0.571 428 571 428 571 428 571 428 571 552 × 2 = 1 + 0.142 857 142 857 142 857 142 857 143 104;
  • 3) 0.142 857 142 857 142 857 142 857 143 104 × 2 = 0 + 0.285 714 285 714 285 714 285 714 286 208;
  • 4) 0.285 714 285 714 285 714 285 714 286 208 × 2 = 0 + 0.571 428 571 428 571 428 571 428 572 416;
  • 5) 0.571 428 571 428 571 428 571 428 572 416 × 2 = 1 + 0.142 857 142 857 142 857 142 857 144 832;
  • 6) 0.142 857 142 857 142 857 142 857 144 832 × 2 = 0 + 0.285 714 285 714 285 714 285 714 289 664;
  • 7) 0.285 714 285 714 285 714 285 714 289 664 × 2 = 0 + 0.571 428 571 428 571 428 571 428 579 328;
  • 8) 0.571 428 571 428 571 428 571 428 579 328 × 2 = 1 + 0.142 857 142 857 142 857 142 857 158 656;
  • 9) 0.142 857 142 857 142 857 142 857 158 656 × 2 = 0 + 0.285 714 285 714 285 714 285 714 317 312;
  • 10) 0.285 714 285 714 285 714 285 714 317 312 × 2 = 0 + 0.571 428 571 428 571 428 571 428 634 624;
  • 11) 0.571 428 571 428 571 428 571 428 634 624 × 2 = 1 + 0.142 857 142 857 142 857 142 857 269 248;
  • 12) 0.142 857 142 857 142 857 142 857 269 248 × 2 = 0 + 0.285 714 285 714 285 714 285 714 538 496;
  • 13) 0.285 714 285 714 285 714 285 714 538 496 × 2 = 0 + 0.571 428 571 428 571 428 571 429 076 992;
  • 14) 0.571 428 571 428 571 428 571 429 076 992 × 2 = 1 + 0.142 857 142 857 142 857 142 858 153 984;
  • 15) 0.142 857 142 857 142 857 142 858 153 984 × 2 = 0 + 0.285 714 285 714 285 714 285 716 307 968;
  • 16) 0.285 714 285 714 285 714 285 716 307 968 × 2 = 0 + 0.571 428 571 428 571 428 571 432 615 936;
  • 17) 0.571 428 571 428 571 428 571 432 615 936 × 2 = 1 + 0.142 857 142 857 142 857 142 865 231 872;
  • 18) 0.142 857 142 857 142 857 142 865 231 872 × 2 = 0 + 0.285 714 285 714 285 714 285 730 463 744;
  • 19) 0.285 714 285 714 285 714 285 730 463 744 × 2 = 0 + 0.571 428 571 428 571 428 571 460 927 488;
  • 20) 0.571 428 571 428 571 428 571 460 927 488 × 2 = 1 + 0.142 857 142 857 142 857 142 921 854 976;
  • 21) 0.142 857 142 857 142 857 142 921 854 976 × 2 = 0 + 0.285 714 285 714 285 714 285 843 709 952;
  • 22) 0.285 714 285 714 285 714 285 843 709 952 × 2 = 0 + 0.571 428 571 428 571 428 571 687 419 904;
  • 23) 0.571 428 571 428 571 428 571 687 419 904 × 2 = 1 + 0.142 857 142 857 142 857 143 374 839 808;
  • 24) 0.142 857 142 857 142 857 143 374 839 808 × 2 = 0 + 0.285 714 285 714 285 714 286 749 679 616;
  • 25) 0.285 714 285 714 285 714 286 749 679 616 × 2 = 0 + 0.571 428 571 428 571 428 573 499 359 232;
  • 26) 0.571 428 571 428 571 428 573 499 359 232 × 2 = 1 + 0.142 857 142 857 142 857 146 998 718 464;
  • 27) 0.142 857 142 857 142 857 146 998 718 464 × 2 = 0 + 0.285 714 285 714 285 714 293 997 436 928;
  • 28) 0.285 714 285 714 285 714 293 997 436 928 × 2 = 0 + 0.571 428 571 428 571 428 587 994 873 856;
  • 29) 0.571 428 571 428 571 428 587 994 873 856 × 2 = 1 + 0.142 857 142 857 142 857 175 989 747 712;
  • 30) 0.142 857 142 857 142 857 175 989 747 712 × 2 = 0 + 0.285 714 285 714 285 714 351 979 495 424;
  • 31) 0.285 714 285 714 285 714 351 979 495 424 × 2 = 0 + 0.571 428 571 428 571 428 703 958 990 848;
  • 32) 0.571 428 571 428 571 428 703 958 990 848 × 2 = 1 + 0.142 857 142 857 142 857 407 917 981 696;
  • 33) 0.142 857 142 857 142 857 407 917 981 696 × 2 = 0 + 0.285 714 285 714 285 714 815 835 963 392;
  • 34) 0.285 714 285 714 285 714 815 835 963 392 × 2 = 0 + 0.571 428 571 428 571 429 631 671 926 784;
  • 35) 0.571 428 571 428 571 429 631 671 926 784 × 2 = 1 + 0.142 857 142 857 142 859 263 343 853 568;
  • 36) 0.142 857 142 857 142 859 263 343 853 568 × 2 = 0 + 0.285 714 285 714 285 718 526 687 707 136;
  • 37) 0.285 714 285 714 285 718 526 687 707 136 × 2 = 0 + 0.571 428 571 428 571 437 053 375 414 272;
  • 38) 0.571 428 571 428 571 437 053 375 414 272 × 2 = 1 + 0.142 857 142 857 142 874 106 750 828 544;
  • 39) 0.142 857 142 857 142 874 106 750 828 544 × 2 = 0 + 0.285 714 285 714 285 748 213 501 657 088;
  • 40) 0.285 714 285 714 285 748 213 501 657 088 × 2 = 0 + 0.571 428 571 428 571 496 427 003 314 176;
  • 41) 0.571 428 571 428 571 496 427 003 314 176 × 2 = 1 + 0.142 857 142 857 142 992 854 006 628 352;
  • 42) 0.142 857 142 857 142 992 854 006 628 352 × 2 = 0 + 0.285 714 285 714 285 985 708 013 256 704;
  • 43) 0.285 714 285 714 285 985 708 013 256 704 × 2 = 0 + 0.571 428 571 428 571 971 416 026 513 408;
  • 44) 0.571 428 571 428 571 971 416 026 513 408 × 2 = 1 + 0.142 857 142 857 143 942 832 053 026 816;
  • 45) 0.142 857 142 857 143 942 832 053 026 816 × 2 = 0 + 0.285 714 285 714 287 885 664 106 053 632;
  • 46) 0.285 714 285 714 287 885 664 106 053 632 × 2 = 0 + 0.571 428 571 428 575 771 328 212 107 264;
  • 47) 0.571 428 571 428 575 771 328 212 107 264 × 2 = 1 + 0.142 857 142 857 151 542 656 424 214 528;
  • 48) 0.142 857 142 857 151 542 656 424 214 528 × 2 = 0 + 0.285 714 285 714 303 085 312 848 429 056;
  • 49) 0.285 714 285 714 303 085 312 848 429 056 × 2 = 0 + 0.571 428 571 428 606 170 625 696 858 112;
  • 50) 0.571 428 571 428 606 170 625 696 858 112 × 2 = 1 + 0.142 857 142 857 212 341 251 393 716 224;
  • 51) 0.142 857 142 857 212 341 251 393 716 224 × 2 = 0 + 0.285 714 285 714 424 682 502 787 432 448;
  • 52) 0.285 714 285 714 424 682 502 787 432 448 × 2 = 0 + 0.571 428 571 428 849 365 005 574 864 896;
  • 53) 0.571 428 571 428 849 365 005 574 864 896 × 2 = 1 + 0.142 857 142 857 698 730 011 149 729 792;

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:


0.285 714 285 714 285 714 285 714 285 776(10) =

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

5. Positive number before normalization:

6.285 714 285 714 285 714 285 714 285 776(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:


6.285 714 285 714 285 714 285 714 285 776(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 6.285 714 285 714 285 714 285 714 285 776 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