6.316 668 999 999 999 201 122 591 338 39 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.316 668 999 999 999 201 122 591 338 39(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
6.316 668 999 999 999 201 122 591 338 39(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: 0.316 668 999 999 999 201 122 591 338 39.

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) 0.316 668 999 999 999 201 122 591 338 39 × 2 = 0 + 0.633 337 999 999 998 402 245 182 676 78;
  • 2) 0.633 337 999 999 998 402 245 182 676 78 × 2 = 1 + 0.266 675 999 999 996 804 490 365 353 56;
  • 3) 0.266 675 999 999 996 804 490 365 353 56 × 2 = 0 + 0.533 351 999 999 993 608 980 730 707 12;
  • 4) 0.533 351 999 999 993 608 980 730 707 12 × 2 = 1 + 0.066 703 999 999 987 217 961 461 414 24;
  • 5) 0.066 703 999 999 987 217 961 461 414 24 × 2 = 0 + 0.133 407 999 999 974 435 922 922 828 48;
  • 6) 0.133 407 999 999 974 435 922 922 828 48 × 2 = 0 + 0.266 815 999 999 948 871 845 845 656 96;
  • 7) 0.266 815 999 999 948 871 845 845 656 96 × 2 = 0 + 0.533 631 999 999 897 743 691 691 313 92;
  • 8) 0.533 631 999 999 897 743 691 691 313 92 × 2 = 1 + 0.067 263 999 999 795 487 383 382 627 84;
  • 9) 0.067 263 999 999 795 487 383 382 627 84 × 2 = 0 + 0.134 527 999 999 590 974 766 765 255 68;
  • 10) 0.134 527 999 999 590 974 766 765 255 68 × 2 = 0 + 0.269 055 999 999 181 949 533 530 511 36;
  • 11) 0.269 055 999 999 181 949 533 530 511 36 × 2 = 0 + 0.538 111 999 998 363 899 067 061 022 72;
  • 12) 0.538 111 999 998 363 899 067 061 022 72 × 2 = 1 + 0.076 223 999 996 727 798 134 122 045 44;
  • 13) 0.076 223 999 996 727 798 134 122 045 44 × 2 = 0 + 0.152 447 999 993 455 596 268 244 090 88;
  • 14) 0.152 447 999 993 455 596 268 244 090 88 × 2 = 0 + 0.304 895 999 986 911 192 536 488 181 76;
  • 15) 0.304 895 999 986 911 192 536 488 181 76 × 2 = 0 + 0.609 791 999 973 822 385 072 976 363 52;
  • 16) 0.609 791 999 973 822 385 072 976 363 52 × 2 = 1 + 0.219 583 999 947 644 770 145 952 727 04;
  • 17) 0.219 583 999 947 644 770 145 952 727 04 × 2 = 0 + 0.439 167 999 895 289 540 291 905 454 08;
  • 18) 0.439 167 999 895 289 540 291 905 454 08 × 2 = 0 + 0.878 335 999 790 579 080 583 810 908 16;
  • 19) 0.878 335 999 790 579 080 583 810 908 16 × 2 = 1 + 0.756 671 999 581 158 161 167 621 816 32;
  • 20) 0.756 671 999 581 158 161 167 621 816 32 × 2 = 1 + 0.513 343 999 162 316 322 335 243 632 64;
  • 21) 0.513 343 999 162 316 322 335 243 632 64 × 2 = 1 + 0.026 687 998 324 632 644 670 487 265 28;
  • 22) 0.026 687 998 324 632 644 670 487 265 28 × 2 = 0 + 0.053 375 996 649 265 289 340 974 530 56;
  • 23) 0.053 375 996 649 265 289 340 974 530 56 × 2 = 0 + 0.106 751 993 298 530 578 681 949 061 12;
  • 24) 0.106 751 993 298 530 578 681 949 061 12 × 2 = 0 + 0.213 503 986 597 061 157 363 898 122 24;
  • 25) 0.213 503 986 597 061 157 363 898 122 24 × 2 = 0 + 0.427 007 973 194 122 314 727 796 244 48;
  • 26) 0.427 007 973 194 122 314 727 796 244 48 × 2 = 0 + 0.854 015 946 388 244 629 455 592 488 96;
  • 27) 0.854 015 946 388 244 629 455 592 488 96 × 2 = 1 + 0.708 031 892 776 489 258 911 184 977 92;
  • 28) 0.708 031 892 776 489 258 911 184 977 92 × 2 = 1 + 0.416 063 785 552 978 517 822 369 955 84;
  • 29) 0.416 063 785 552 978 517 822 369 955 84 × 2 = 0 + 0.832 127 571 105 957 035 644 739 911 68;
  • 30) 0.832 127 571 105 957 035 644 739 911 68 × 2 = 1 + 0.664 255 142 211 914 071 289 479 823 36;
  • 31) 0.664 255 142 211 914 071 289 479 823 36 × 2 = 1 + 0.328 510 284 423 828 142 578 959 646 72;
  • 32) 0.328 510 284 423 828 142 578 959 646 72 × 2 = 0 + 0.657 020 568 847 656 285 157 919 293 44;
  • 33) 0.657 020 568 847 656 285 157 919 293 44 × 2 = 1 + 0.314 041 137 695 312 570 315 838 586 88;
  • 34) 0.314 041 137 695 312 570 315 838 586 88 × 2 = 0 + 0.628 082 275 390 625 140 631 677 173 76;
  • 35) 0.628 082 275 390 625 140 631 677 173 76 × 2 = 1 + 0.256 164 550 781 250 281 263 354 347 52;
  • 36) 0.256 164 550 781 250 281 263 354 347 52 × 2 = 0 + 0.512 329 101 562 500 562 526 708 695 04;
  • 37) 0.512 329 101 562 500 562 526 708 695 04 × 2 = 1 + 0.024 658 203 125 001 125 053 417 390 08;
  • 38) 0.024 658 203 125 001 125 053 417 390 08 × 2 = 0 + 0.049 316 406 250 002 250 106 834 780 16;
  • 39) 0.049 316 406 250 002 250 106 834 780 16 × 2 = 0 + 0.098 632 812 500 004 500 213 669 560 32;
  • 40) 0.098 632 812 500 004 500 213 669 560 32 × 2 = 0 + 0.197 265 625 000 009 000 427 339 120 64;
  • 41) 0.197 265 625 000 009 000 427 339 120 64 × 2 = 0 + 0.394 531 250 000 018 000 854 678 241 28;
  • 42) 0.394 531 250 000 018 000 854 678 241 28 × 2 = 0 + 0.789 062 500 000 036 001 709 356 482 56;
  • 43) 0.789 062 500 000 036 001 709 356 482 56 × 2 = 1 + 0.578 125 000 000 072 003 418 712 965 12;
  • 44) 0.578 125 000 000 072 003 418 712 965 12 × 2 = 1 + 0.156 250 000 000 144 006 837 425 930 24;
  • 45) 0.156 250 000 000 144 006 837 425 930 24 × 2 = 0 + 0.312 500 000 000 288 013 674 851 860 48;
  • 46) 0.312 500 000 000 288 013 674 851 860 48 × 2 = 0 + 0.625 000 000 000 576 027 349 703 720 96;
  • 47) 0.625 000 000 000 576 027 349 703 720 96 × 2 = 1 + 0.250 000 000 001 152 054 699 407 441 92;
  • 48) 0.250 000 000 001 152 054 699 407 441 92 × 2 = 0 + 0.500 000 000 002 304 109 398 814 883 84;
  • 49) 0.500 000 000 002 304 109 398 814 883 84 × 2 = 1 + 0.000 000 000 004 608 218 797 629 767 68;
  • 50) 0.000 000 000 004 608 218 797 629 767 68 × 2 = 0 + 0.000 000 000 009 216 437 595 259 535 36;
  • 51) 0.000 000 000 009 216 437 595 259 535 36 × 2 = 0 + 0.000 000 000 018 432 875 190 519 070 72;
  • 52) 0.000 000 000 018 432 875 190 519 070 72 × 2 = 0 + 0.000 000 000 036 865 750 381 038 141 44;
  • 53) 0.000 000 000 036 865 750 381 038 141 44 × 2 = 0 + 0.000 000 000 073 731 500 762 076 282 88;

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:


0.316 668 999 999 999 201 122 591 338 39(10) =

0.0101 0001 0001 0001 0011 1000 0011 0110 1010 1000 0011 0010 1000 0(2)

5. Positive number before normalization:

6.316 668 999 999 999 201 122 591 338 39(10) =

110.0101 0001 0001 0001 0011 1000 0011 0110 1010 1000 0011 0010 1000 0(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:


6.316 668 999 999 999 201 122 591 338 39(10) =

110.0101 0001 0001 0001 0011 1000 0011 0110 1010 1000 0011 0010 1000 0(2) =

110.0101 0001 0001 0001 0011 1000 0011 0110 1010 1000 0011 0010 1000 0(2) × 20 =

1.1001 0100 0100 0100 0100 1110 0000 1101 1010 1010 0000 1100 1010 000(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 0100 0100 0100 0100 1110 0000 1101 1010 1010 0000 1100 1010 000


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 0100 0100 0100 0100 1110 0000 1101 1010 1010 0000 1100 1010 000 =

1001 0100 0100 0100 0100 1110 0000 1101 1010 1010 0000 1100 1010


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 0100 0100 0100 0100 1110 0000 1101 1010 1010 0000 1100 1010


Decimal number 6.316 668 999 999 999 201 122 591 338 39 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1001 0100 0100 0100 0100 1110 0000 1101 1010 1010 0000 1100 1010

» Convert decimal 6.316 668 999 999 999 201 122 591 338 57 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