-0.381 966 011 250 104 21 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 250 104 21(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
-0.381 966 011 250 104 21(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:

|-0.381 966 011 250 104 21| = 0.381 966 011 250 104 21


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

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.381 966 011 250 104 21.

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) 0.381 966 011 250 104 21 × 2 = 0 + 0.763 932 022 500 208 42;
  • 2) 0.763 932 022 500 208 42 × 2 = 1 + 0.527 864 045 000 416 84;
  • 3) 0.527 864 045 000 416 84 × 2 = 1 + 0.055 728 090 000 833 68;
  • 4) 0.055 728 090 000 833 68 × 2 = 0 + 0.111 456 180 001 667 36;
  • 5) 0.111 456 180 001 667 36 × 2 = 0 + 0.222 912 360 003 334 72;
  • 6) 0.222 912 360 003 334 72 × 2 = 0 + 0.445 824 720 006 669 44;
  • 7) 0.445 824 720 006 669 44 × 2 = 0 + 0.891 649 440 013 338 88;
  • 8) 0.891 649 440 013 338 88 × 2 = 1 + 0.783 298 880 026 677 76;
  • 9) 0.783 298 880 026 677 76 × 2 = 1 + 0.566 597 760 053 355 52;
  • 10) 0.566 597 760 053 355 52 × 2 = 1 + 0.133 195 520 106 711 04;
  • 11) 0.133 195 520 106 711 04 × 2 = 0 + 0.266 391 040 213 422 08;
  • 12) 0.266 391 040 213 422 08 × 2 = 0 + 0.532 782 080 426 844 16;
  • 13) 0.532 782 080 426 844 16 × 2 = 1 + 0.065 564 160 853 688 32;
  • 14) 0.065 564 160 853 688 32 × 2 = 0 + 0.131 128 321 707 376 64;
  • 15) 0.131 128 321 707 376 64 × 2 = 0 + 0.262 256 643 414 753 28;
  • 16) 0.262 256 643 414 753 28 × 2 = 0 + 0.524 513 286 829 506 56;
  • 17) 0.524 513 286 829 506 56 × 2 = 1 + 0.049 026 573 659 013 12;
  • 18) 0.049 026 573 659 013 12 × 2 = 0 + 0.098 053 147 318 026 24;
  • 19) 0.098 053 147 318 026 24 × 2 = 0 + 0.196 106 294 636 052 48;
  • 20) 0.196 106 294 636 052 48 × 2 = 0 + 0.392 212 589 272 104 96;
  • 21) 0.392 212 589 272 104 96 × 2 = 0 + 0.784 425 178 544 209 92;
  • 22) 0.784 425 178 544 209 92 × 2 = 1 + 0.568 850 357 088 419 84;
  • 23) 0.568 850 357 088 419 84 × 2 = 1 + 0.137 700 714 176 839 68;
  • 24) 0.137 700 714 176 839 68 × 2 = 0 + 0.275 401 428 353 679 36;
  • 25) 0.275 401 428 353 679 36 × 2 = 0 + 0.550 802 856 707 358 72;
  • 26) 0.550 802 856 707 358 72 × 2 = 1 + 0.101 605 713 414 717 44;
  • 27) 0.101 605 713 414 717 44 × 2 = 0 + 0.203 211 426 829 434 88;
  • 28) 0.203 211 426 829 434 88 × 2 = 0 + 0.406 422 853 658 869 76;
  • 29) 0.406 422 853 658 869 76 × 2 = 0 + 0.812 845 707 317 739 52;
  • 30) 0.812 845 707 317 739 52 × 2 = 1 + 0.625 691 414 635 479 04;
  • 31) 0.625 691 414 635 479 04 × 2 = 1 + 0.251 382 829 270 958 08;
  • 32) 0.251 382 829 270 958 08 × 2 = 0 + 0.502 765 658 541 916 16;
  • 33) 0.502 765 658 541 916 16 × 2 = 1 + 0.005 531 317 083 832 32;
  • 34) 0.005 531 317 083 832 32 × 2 = 0 + 0.011 062 634 167 664 64;
  • 35) 0.011 062 634 167 664 64 × 2 = 0 + 0.022 125 268 335 329 28;
  • 36) 0.022 125 268 335 329 28 × 2 = 0 + 0.044 250 536 670 658 56;
  • 37) 0.044 250 536 670 658 56 × 2 = 0 + 0.088 501 073 341 317 12;
  • 38) 0.088 501 073 341 317 12 × 2 = 0 + 0.177 002 146 682 634 24;
  • 39) 0.177 002 146 682 634 24 × 2 = 0 + 0.354 004 293 365 268 48;
  • 40) 0.354 004 293 365 268 48 × 2 = 0 + 0.708 008 586 730 536 96;
  • 41) 0.708 008 586 730 536 96 × 2 = 1 + 0.416 017 173 461 073 92;
  • 42) 0.416 017 173 461 073 92 × 2 = 0 + 0.832 034 346 922 147 84;
  • 43) 0.832 034 346 922 147 84 × 2 = 1 + 0.664 068 693 844 295 68;
  • 44) 0.664 068 693 844 295 68 × 2 = 1 + 0.328 137 387 688 591 36;
  • 45) 0.328 137 387 688 591 36 × 2 = 0 + 0.656 274 775 377 182 72;
  • 46) 0.656 274 775 377 182 72 × 2 = 1 + 0.312 549 550 754 365 44;
  • 47) 0.312 549 550 754 365 44 × 2 = 0 + 0.625 099 101 508 730 88;
  • 48) 0.625 099 101 508 730 88 × 2 = 1 + 0.250 198 203 017 461 76;
  • 49) 0.250 198 203 017 461 76 × 2 = 0 + 0.500 396 406 034 923 52;
  • 50) 0.500 396 406 034 923 52 × 2 = 1 + 0.000 792 812 069 847 04;
  • 51) 0.000 792 812 069 847 04 × 2 = 0 + 0.001 585 624 139 694 08;
  • 52) 0.001 585 624 139 694 08 × 2 = 0 + 0.003 171 248 279 388 16;
  • 53) 0.003 171 248 279 388 16 × 2 = 0 + 0.006 342 496 558 776 32;
  • 54) 0.006 342 496 558 776 32 × 2 = 0 + 0.012 684 993 117 552 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).


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:


0.381 966 011 250 104 21(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0100 00(2)

6. Positive number before normalization:

0.381 966 011 250 104 21(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0100 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.381 966 011 250 104 21(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0100 00(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0000 1011 0101 0100 00(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 0000(2) × 2-2


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): -2

Mantissa (not normalized):
1.1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 0000


9. 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) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1021(10) =

011 1111 1101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 0000 =

1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 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) =
011 1111 1101

Mantissa (52 bits) =
1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 0000


Decimal number -0.381 966 011 250 104 21 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 1010 0000 0010 1101 0101 0000

» Convert decimal -0.381 966 011 250 103 99 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