6.441 148 781 594 706 389 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.441 148 781 594 706 389 17(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
6.441 148 781 594 706 389 17(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: 0.441 148 781 594 706 389 17.

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) 0.441 148 781 594 706 389 17 × 2 = 0 + 0.882 297 563 189 412 778 34;
  • 2) 0.882 297 563 189 412 778 34 × 2 = 1 + 0.764 595 126 378 825 556 68;
  • 3) 0.764 595 126 378 825 556 68 × 2 = 1 + 0.529 190 252 757 651 113 36;
  • 4) 0.529 190 252 757 651 113 36 × 2 = 1 + 0.058 380 505 515 302 226 72;
  • 5) 0.058 380 505 515 302 226 72 × 2 = 0 + 0.116 761 011 030 604 453 44;
  • 6) 0.116 761 011 030 604 453 44 × 2 = 0 + 0.233 522 022 061 208 906 88;
  • 7) 0.233 522 022 061 208 906 88 × 2 = 0 + 0.467 044 044 122 417 813 76;
  • 8) 0.467 044 044 122 417 813 76 × 2 = 0 + 0.934 088 088 244 835 627 52;
  • 9) 0.934 088 088 244 835 627 52 × 2 = 1 + 0.868 176 176 489 671 255 04;
  • 10) 0.868 176 176 489 671 255 04 × 2 = 1 + 0.736 352 352 979 342 510 08;
  • 11) 0.736 352 352 979 342 510 08 × 2 = 1 + 0.472 704 705 958 685 020 16;
  • 12) 0.472 704 705 958 685 020 16 × 2 = 0 + 0.945 409 411 917 370 040 32;
  • 13) 0.945 409 411 917 370 040 32 × 2 = 1 + 0.890 818 823 834 740 080 64;
  • 14) 0.890 818 823 834 740 080 64 × 2 = 1 + 0.781 637 647 669 480 161 28;
  • 15) 0.781 637 647 669 480 161 28 × 2 = 1 + 0.563 275 295 338 960 322 56;
  • 16) 0.563 275 295 338 960 322 56 × 2 = 1 + 0.126 550 590 677 920 645 12;
  • 17) 0.126 550 590 677 920 645 12 × 2 = 0 + 0.253 101 181 355 841 290 24;
  • 18) 0.253 101 181 355 841 290 24 × 2 = 0 + 0.506 202 362 711 682 580 48;
  • 19) 0.506 202 362 711 682 580 48 × 2 = 1 + 0.012 404 725 423 365 160 96;
  • 20) 0.012 404 725 423 365 160 96 × 2 = 0 + 0.024 809 450 846 730 321 92;
  • 21) 0.024 809 450 846 730 321 92 × 2 = 0 + 0.049 618 901 693 460 643 84;
  • 22) 0.049 618 901 693 460 643 84 × 2 = 0 + 0.099 237 803 386 921 287 68;
  • 23) 0.099 237 803 386 921 287 68 × 2 = 0 + 0.198 475 606 773 842 575 36;
  • 24) 0.198 475 606 773 842 575 36 × 2 = 0 + 0.396 951 213 547 685 150 72;
  • 25) 0.396 951 213 547 685 150 72 × 2 = 0 + 0.793 902 427 095 370 301 44;
  • 26) 0.793 902 427 095 370 301 44 × 2 = 1 + 0.587 804 854 190 740 602 88;
  • 27) 0.587 804 854 190 740 602 88 × 2 = 1 + 0.175 609 708 381 481 205 76;
  • 28) 0.175 609 708 381 481 205 76 × 2 = 0 + 0.351 219 416 762 962 411 52;
  • 29) 0.351 219 416 762 962 411 52 × 2 = 0 + 0.702 438 833 525 924 823 04;
  • 30) 0.702 438 833 525 924 823 04 × 2 = 1 + 0.404 877 667 051 849 646 08;
  • 31) 0.404 877 667 051 849 646 08 × 2 = 0 + 0.809 755 334 103 699 292 16;
  • 32) 0.809 755 334 103 699 292 16 × 2 = 1 + 0.619 510 668 207 398 584 32;
  • 33) 0.619 510 668 207 398 584 32 × 2 = 1 + 0.239 021 336 414 797 168 64;
  • 34) 0.239 021 336 414 797 168 64 × 2 = 0 + 0.478 042 672 829 594 337 28;
  • 35) 0.478 042 672 829 594 337 28 × 2 = 0 + 0.956 085 345 659 188 674 56;
  • 36) 0.956 085 345 659 188 674 56 × 2 = 1 + 0.912 170 691 318 377 349 12;
  • 37) 0.912 170 691 318 377 349 12 × 2 = 1 + 0.824 341 382 636 754 698 24;
  • 38) 0.824 341 382 636 754 698 24 × 2 = 1 + 0.648 682 765 273 509 396 48;
  • 39) 0.648 682 765 273 509 396 48 × 2 = 1 + 0.297 365 530 547 018 792 96;
  • 40) 0.297 365 530 547 018 792 96 × 2 = 0 + 0.594 731 061 094 037 585 92;
  • 41) 0.594 731 061 094 037 585 92 × 2 = 1 + 0.189 462 122 188 075 171 84;
  • 42) 0.189 462 122 188 075 171 84 × 2 = 0 + 0.378 924 244 376 150 343 68;
  • 43) 0.378 924 244 376 150 343 68 × 2 = 0 + 0.757 848 488 752 300 687 36;
  • 44) 0.757 848 488 752 300 687 36 × 2 = 1 + 0.515 696 977 504 601 374 72;
  • 45) 0.515 696 977 504 601 374 72 × 2 = 1 + 0.031 393 955 009 202 749 44;
  • 46) 0.031 393 955 009 202 749 44 × 2 = 0 + 0.062 787 910 018 405 498 88;
  • 47) 0.062 787 910 018 405 498 88 × 2 = 0 + 0.125 575 820 036 810 997 76;
  • 48) 0.125 575 820 036 810 997 76 × 2 = 0 + 0.251 151 640 073 621 995 52;
  • 49) 0.251 151 640 073 621 995 52 × 2 = 0 + 0.502 303 280 147 243 991 04;
  • 50) 0.502 303 280 147 243 991 04 × 2 = 1 + 0.004 606 560 294 487 982 08;
  • 51) 0.004 606 560 294 487 982 08 × 2 = 0 + 0.009 213 120 588 975 964 16;
  • 52) 0.009 213 120 588 975 964 16 × 2 = 0 + 0.018 426 241 177 951 928 32;
  • 53) 0.018 426 241 177 951 928 32 × 2 = 0 + 0.036 852 482 355 903 856 64;

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:


0.441 148 781 594 706 389 17(10) =

0.0111 0000 1110 1111 0010 0000 0110 0101 1001 1110 1001 1000 0100 0(2)

5. Positive number before normalization:

6.441 148 781 594 706 389 17(10) =

110.0111 0000 1110 1111 0010 0000 0110 0101 1001 1110 1001 1000 0100 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:


6.441 148 781 594 706 389 17(10) =

110.0111 0000 1110 1111 0010 0000 0110 0101 1001 1110 1001 1000 0100 0(2) =

110.0111 0000 1110 1111 0010 0000 0110 0101 1001 1110 1001 1000 0100 0(2) × 20 =

1.1001 1100 0011 1011 1100 1000 0001 1001 0110 0111 1010 0110 0001 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 1100 0011 1011 1100 1000 0001 1001 0110 0111 1010 0110 0001 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 1100 0011 1011 1100 1000 0001 1001 0110 0111 1010 0110 0001 000 =

1001 1100 0011 1011 1100 1000 0001 1001 0110 0111 1010 0110 0001


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 1100 0011 1011 1100 1000 0001 1001 0110 0111 1010 0110 0001


Decimal number 6.441 148 781 594 706 389 17 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1001 1100 0011 1011 1100 1000 0001 1001 0110 0111 1010 0110 0001

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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