-2.211 829 052 383 358 304 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.211 829 052 383 358 304 4(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
-2.211 829 052 383 358 304 4(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. Start with the positive version of the number:

|-2.211 829 052 383 358 304 4| = 2.211 829 052 383 358 304 4


2. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.

2(10) =

10(2)


4. Convert to binary (base 2) the fractional part: 0.211 829 052 383 358 304 4.

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) 0.211 829 052 383 358 304 4 × 2 = 0 + 0.423 658 104 766 716 608 8;
  • 2) 0.423 658 104 766 716 608 8 × 2 = 0 + 0.847 316 209 533 433 217 6;
  • 3) 0.847 316 209 533 433 217 6 × 2 = 1 + 0.694 632 419 066 866 435 2;
  • 4) 0.694 632 419 066 866 435 2 × 2 = 1 + 0.389 264 838 133 732 870 4;
  • 5) 0.389 264 838 133 732 870 4 × 2 = 0 + 0.778 529 676 267 465 740 8;
  • 6) 0.778 529 676 267 465 740 8 × 2 = 1 + 0.557 059 352 534 931 481 6;
  • 7) 0.557 059 352 534 931 481 6 × 2 = 1 + 0.114 118 705 069 862 963 2;
  • 8) 0.114 118 705 069 862 963 2 × 2 = 0 + 0.228 237 410 139 725 926 4;
  • 9) 0.228 237 410 139 725 926 4 × 2 = 0 + 0.456 474 820 279 451 852 8;
  • 10) 0.456 474 820 279 451 852 8 × 2 = 0 + 0.912 949 640 558 903 705 6;
  • 11) 0.912 949 640 558 903 705 6 × 2 = 1 + 0.825 899 281 117 807 411 2;
  • 12) 0.825 899 281 117 807 411 2 × 2 = 1 + 0.651 798 562 235 614 822 4;
  • 13) 0.651 798 562 235 614 822 4 × 2 = 1 + 0.303 597 124 471 229 644 8;
  • 14) 0.303 597 124 471 229 644 8 × 2 = 0 + 0.607 194 248 942 459 289 6;
  • 15) 0.607 194 248 942 459 289 6 × 2 = 1 + 0.214 388 497 884 918 579 2;
  • 16) 0.214 388 497 884 918 579 2 × 2 = 0 + 0.428 776 995 769 837 158 4;
  • 17) 0.428 776 995 769 837 158 4 × 2 = 0 + 0.857 553 991 539 674 316 8;
  • 18) 0.857 553 991 539 674 316 8 × 2 = 1 + 0.715 107 983 079 348 633 6;
  • 19) 0.715 107 983 079 348 633 6 × 2 = 1 + 0.430 215 966 158 697 267 2;
  • 20) 0.430 215 966 158 697 267 2 × 2 = 0 + 0.860 431 932 317 394 534 4;
  • 21) 0.860 431 932 317 394 534 4 × 2 = 1 + 0.720 863 864 634 789 068 8;
  • 22) 0.720 863 864 634 789 068 8 × 2 = 1 + 0.441 727 729 269 578 137 6;
  • 23) 0.441 727 729 269 578 137 6 × 2 = 0 + 0.883 455 458 539 156 275 2;
  • 24) 0.883 455 458 539 156 275 2 × 2 = 1 + 0.766 910 917 078 312 550 4;
  • 25) 0.766 910 917 078 312 550 4 × 2 = 1 + 0.533 821 834 156 625 100 8;
  • 26) 0.533 821 834 156 625 100 8 × 2 = 1 + 0.067 643 668 313 250 201 6;
  • 27) 0.067 643 668 313 250 201 6 × 2 = 0 + 0.135 287 336 626 500 403 2;
  • 28) 0.135 287 336 626 500 403 2 × 2 = 0 + 0.270 574 673 253 000 806 4;
  • 29) 0.270 574 673 253 000 806 4 × 2 = 0 + 0.541 149 346 506 001 612 8;
  • 30) 0.541 149 346 506 001 612 8 × 2 = 1 + 0.082 298 693 012 003 225 6;
  • 31) 0.082 298 693 012 003 225 6 × 2 = 0 + 0.164 597 386 024 006 451 2;
  • 32) 0.164 597 386 024 006 451 2 × 2 = 0 + 0.329 194 772 048 012 902 4;
  • 33) 0.329 194 772 048 012 902 4 × 2 = 0 + 0.658 389 544 096 025 804 8;
  • 34) 0.658 389 544 096 025 804 8 × 2 = 1 + 0.316 779 088 192 051 609 6;
  • 35) 0.316 779 088 192 051 609 6 × 2 = 0 + 0.633 558 176 384 103 219 2;
  • 36) 0.633 558 176 384 103 219 2 × 2 = 1 + 0.267 116 352 768 206 438 4;
  • 37) 0.267 116 352 768 206 438 4 × 2 = 0 + 0.534 232 705 536 412 876 8;
  • 38) 0.534 232 705 536 412 876 8 × 2 = 1 + 0.068 465 411 072 825 753 6;
  • 39) 0.068 465 411 072 825 753 6 × 2 = 0 + 0.136 930 822 145 651 507 2;
  • 40) 0.136 930 822 145 651 507 2 × 2 = 0 + 0.273 861 644 291 303 014 4;
  • 41) 0.273 861 644 291 303 014 4 × 2 = 0 + 0.547 723 288 582 606 028 8;
  • 42) 0.547 723 288 582 606 028 8 × 2 = 1 + 0.095 446 577 165 212 057 6;
  • 43) 0.095 446 577 165 212 057 6 × 2 = 0 + 0.190 893 154 330 424 115 2;
  • 44) 0.190 893 154 330 424 115 2 × 2 = 0 + 0.381 786 308 660 848 230 4;
  • 45) 0.381 786 308 660 848 230 4 × 2 = 0 + 0.763 572 617 321 696 460 8;
  • 46) 0.763 572 617 321 696 460 8 × 2 = 1 + 0.527 145 234 643 392 921 6;
  • 47) 0.527 145 234 643 392 921 6 × 2 = 1 + 0.054 290 469 286 785 843 2;
  • 48) 0.054 290 469 286 785 843 2 × 2 = 0 + 0.108 580 938 573 571 686 4;
  • 49) 0.108 580 938 573 571 686 4 × 2 = 0 + 0.217 161 877 147 143 372 8;
  • 50) 0.217 161 877 147 143 372 8 × 2 = 0 + 0.434 323 754 294 286 745 6;
  • 51) 0.434 323 754 294 286 745 6 × 2 = 0 + 0.868 647 508 588 573 491 2;
  • 52) 0.868 647 508 588 573 491 2 × 2 = 1 + 0.737 295 017 177 146 982 4;
  • 53) 0.737 295 017 177 146 982 4 × 2 = 1 + 0.474 590 034 354 293 964 8;

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).


5. 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:


0.211 829 052 383 358 304 4(10) =

0.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2)

6. Positive number before normalization:

2.211 829 052 383 358 304 4(10) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.211 829 052 383 358 304 4(10) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) =

10.0011 0110 0011 1010 0110 1101 1100 0100 0101 0100 0100 0110 0001 1(2) × 20 =

1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11(2) × 21


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): 1

Mantissa (not normalized):
1.0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


10. 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;
  • 1 024 ÷ 2 = 512 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1024(10) =

100 0000 0000(2)


12. 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. 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000 11 =

0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000


Decimal number -2.211 829 052 383 358 304 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 0001 1011 0001 1101 0011 0110 1110 0010 0010 1010 0010 0011 0000

» Convert decimal -2.211 829 052 383 358 296 7 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