6.285 714 285 714 285 714 285 714 047 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 714 285 714 047(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
6.285 714 285 714 285 714 285 714 047(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: 0.285 714 285 714 285 714 285 714 047.

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) 0.285 714 285 714 285 714 285 714 047 × 2 = 0 + 0.571 428 571 428 571 428 571 428 094;
  • 2) 0.571 428 571 428 571 428 571 428 094 × 2 = 1 + 0.142 857 142 857 142 857 142 856 188;
  • 3) 0.142 857 142 857 142 857 142 856 188 × 2 = 0 + 0.285 714 285 714 285 714 285 712 376;
  • 4) 0.285 714 285 714 285 714 285 712 376 × 2 = 0 + 0.571 428 571 428 571 428 571 424 752;
  • 5) 0.571 428 571 428 571 428 571 424 752 × 2 = 1 + 0.142 857 142 857 142 857 142 849 504;
  • 6) 0.142 857 142 857 142 857 142 849 504 × 2 = 0 + 0.285 714 285 714 285 714 285 699 008;
  • 7) 0.285 714 285 714 285 714 285 699 008 × 2 = 0 + 0.571 428 571 428 571 428 571 398 016;
  • 8) 0.571 428 571 428 571 428 571 398 016 × 2 = 1 + 0.142 857 142 857 142 857 142 796 032;
  • 9) 0.142 857 142 857 142 857 142 796 032 × 2 = 0 + 0.285 714 285 714 285 714 285 592 064;
  • 10) 0.285 714 285 714 285 714 285 592 064 × 2 = 0 + 0.571 428 571 428 571 428 571 184 128;
  • 11) 0.571 428 571 428 571 428 571 184 128 × 2 = 1 + 0.142 857 142 857 142 857 142 368 256;
  • 12) 0.142 857 142 857 142 857 142 368 256 × 2 = 0 + 0.285 714 285 714 285 714 284 736 512;
  • 13) 0.285 714 285 714 285 714 284 736 512 × 2 = 0 + 0.571 428 571 428 571 428 569 473 024;
  • 14) 0.571 428 571 428 571 428 569 473 024 × 2 = 1 + 0.142 857 142 857 142 857 138 946 048;
  • 15) 0.142 857 142 857 142 857 138 946 048 × 2 = 0 + 0.285 714 285 714 285 714 277 892 096;
  • 16) 0.285 714 285 714 285 714 277 892 096 × 2 = 0 + 0.571 428 571 428 571 428 555 784 192;
  • 17) 0.571 428 571 428 571 428 555 784 192 × 2 = 1 + 0.142 857 142 857 142 857 111 568 384;
  • 18) 0.142 857 142 857 142 857 111 568 384 × 2 = 0 + 0.285 714 285 714 285 714 223 136 768;
  • 19) 0.285 714 285 714 285 714 223 136 768 × 2 = 0 + 0.571 428 571 428 571 428 446 273 536;
  • 20) 0.571 428 571 428 571 428 446 273 536 × 2 = 1 + 0.142 857 142 857 142 856 892 547 072;
  • 21) 0.142 857 142 857 142 856 892 547 072 × 2 = 0 + 0.285 714 285 714 285 713 785 094 144;
  • 22) 0.285 714 285 714 285 713 785 094 144 × 2 = 0 + 0.571 428 571 428 571 427 570 188 288;
  • 23) 0.571 428 571 428 571 427 570 188 288 × 2 = 1 + 0.142 857 142 857 142 855 140 376 576;
  • 24) 0.142 857 142 857 142 855 140 376 576 × 2 = 0 + 0.285 714 285 714 285 710 280 753 152;
  • 25) 0.285 714 285 714 285 710 280 753 152 × 2 = 0 + 0.571 428 571 428 571 420 561 506 304;
  • 26) 0.571 428 571 428 571 420 561 506 304 × 2 = 1 + 0.142 857 142 857 142 841 123 012 608;
  • 27) 0.142 857 142 857 142 841 123 012 608 × 2 = 0 + 0.285 714 285 714 285 682 246 025 216;
  • 28) 0.285 714 285 714 285 682 246 025 216 × 2 = 0 + 0.571 428 571 428 571 364 492 050 432;
  • 29) 0.571 428 571 428 571 364 492 050 432 × 2 = 1 + 0.142 857 142 857 142 728 984 100 864;
  • 30) 0.142 857 142 857 142 728 984 100 864 × 2 = 0 + 0.285 714 285 714 285 457 968 201 728;
  • 31) 0.285 714 285 714 285 457 968 201 728 × 2 = 0 + 0.571 428 571 428 570 915 936 403 456;
  • 32) 0.571 428 571 428 570 915 936 403 456 × 2 = 1 + 0.142 857 142 857 141 831 872 806 912;
  • 33) 0.142 857 142 857 141 831 872 806 912 × 2 = 0 + 0.285 714 285 714 283 663 745 613 824;
  • 34) 0.285 714 285 714 283 663 745 613 824 × 2 = 0 + 0.571 428 571 428 567 327 491 227 648;
  • 35) 0.571 428 571 428 567 327 491 227 648 × 2 = 1 + 0.142 857 142 857 134 654 982 455 296;
  • 36) 0.142 857 142 857 134 654 982 455 296 × 2 = 0 + 0.285 714 285 714 269 309 964 910 592;
  • 37) 0.285 714 285 714 269 309 964 910 592 × 2 = 0 + 0.571 428 571 428 538 619 929 821 184;
  • 38) 0.571 428 571 428 538 619 929 821 184 × 2 = 1 + 0.142 857 142 857 077 239 859 642 368;
  • 39) 0.142 857 142 857 077 239 859 642 368 × 2 = 0 + 0.285 714 285 714 154 479 719 284 736;
  • 40) 0.285 714 285 714 154 479 719 284 736 × 2 = 0 + 0.571 428 571 428 308 959 438 569 472;
  • 41) 0.571 428 571 428 308 959 438 569 472 × 2 = 1 + 0.142 857 142 856 617 918 877 138 944;
  • 42) 0.142 857 142 856 617 918 877 138 944 × 2 = 0 + 0.285 714 285 713 235 837 754 277 888;
  • 43) 0.285 714 285 713 235 837 754 277 888 × 2 = 0 + 0.571 428 571 426 471 675 508 555 776;
  • 44) 0.571 428 571 426 471 675 508 555 776 × 2 = 1 + 0.142 857 142 852 943 351 017 111 552;
  • 45) 0.142 857 142 852 943 351 017 111 552 × 2 = 0 + 0.285 714 285 705 886 702 034 223 104;
  • 46) 0.285 714 285 705 886 702 034 223 104 × 2 = 0 + 0.571 428 571 411 773 404 068 446 208;
  • 47) 0.571 428 571 411 773 404 068 446 208 × 2 = 1 + 0.142 857 142 823 546 808 136 892 416;
  • 48) 0.142 857 142 823 546 808 136 892 416 × 2 = 0 + 0.285 714 285 647 093 616 273 784 832;
  • 49) 0.285 714 285 647 093 616 273 784 832 × 2 = 0 + 0.571 428 571 294 187 232 547 569 664;
  • 50) 0.571 428 571 294 187 232 547 569 664 × 2 = 1 + 0.142 857 142 588 374 465 095 139 328;
  • 51) 0.142 857 142 588 374 465 095 139 328 × 2 = 0 + 0.285 714 285 176 748 930 190 278 656;
  • 52) 0.285 714 285 176 748 930 190 278 656 × 2 = 0 + 0.571 428 570 353 497 860 380 557 312;
  • 53) 0.571 428 570 353 497 860 380 557 312 × 2 = 1 + 0.142 857 140 706 995 720 761 114 624;

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:


0.285 714 285 714 285 714 285 714 047(10) =

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

5. Positive number before normalization:

6.285 714 285 714 285 714 285 714 047(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:


6.285 714 285 714 285 714 285 714 047(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 6.285 714 285 714 285 714 285 714 047 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

» Convert decimal 6.285 714 285 714 285 714 285 714 016 to 64 bit double precision IEEE 754 binary floating point representation standard

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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