-0.000 282 005 914 396 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 396 7(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
-0.000 282 005 914 396 7(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:

|-0.000 282 005 914 396 7| = 0.000 282 005 914 396 7


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: 0.000 282 005 914 396 7.

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) 0.000 282 005 914 396 7 × 2 = 0 + 0.000 564 011 828 793 4;
  • 2) 0.000 564 011 828 793 4 × 2 = 0 + 0.001 128 023 657 586 8;
  • 3) 0.001 128 023 657 586 8 × 2 = 0 + 0.002 256 047 315 173 6;
  • 4) 0.002 256 047 315 173 6 × 2 = 0 + 0.004 512 094 630 347 2;
  • 5) 0.004 512 094 630 347 2 × 2 = 0 + 0.009 024 189 260 694 4;
  • 6) 0.009 024 189 260 694 4 × 2 = 0 + 0.018 048 378 521 388 8;
  • 7) 0.018 048 378 521 388 8 × 2 = 0 + 0.036 096 757 042 777 6;
  • 8) 0.036 096 757 042 777 6 × 2 = 0 + 0.072 193 514 085 555 2;
  • 9) 0.072 193 514 085 555 2 × 2 = 0 + 0.144 387 028 171 110 4;
  • 10) 0.144 387 028 171 110 4 × 2 = 0 + 0.288 774 056 342 220 8;
  • 11) 0.288 774 056 342 220 8 × 2 = 0 + 0.577 548 112 684 441 6;
  • 12) 0.577 548 112 684 441 6 × 2 = 1 + 0.155 096 225 368 883 2;
  • 13) 0.155 096 225 368 883 2 × 2 = 0 + 0.310 192 450 737 766 4;
  • 14) 0.310 192 450 737 766 4 × 2 = 0 + 0.620 384 901 475 532 8;
  • 15) 0.620 384 901 475 532 8 × 2 = 1 + 0.240 769 802 951 065 6;
  • 16) 0.240 769 802 951 065 6 × 2 = 0 + 0.481 539 605 902 131 2;
  • 17) 0.481 539 605 902 131 2 × 2 = 0 + 0.963 079 211 804 262 4;
  • 18) 0.963 079 211 804 262 4 × 2 = 1 + 0.926 158 423 608 524 8;
  • 19) 0.926 158 423 608 524 8 × 2 = 1 + 0.852 316 847 217 049 6;
  • 20) 0.852 316 847 217 049 6 × 2 = 1 + 0.704 633 694 434 099 2;
  • 21) 0.704 633 694 434 099 2 × 2 = 1 + 0.409 267 388 868 198 4;
  • 22) 0.409 267 388 868 198 4 × 2 = 0 + 0.818 534 777 736 396 8;
  • 23) 0.818 534 777 736 396 8 × 2 = 1 + 0.637 069 555 472 793 6;
  • 24) 0.637 069 555 472 793 6 × 2 = 1 + 0.274 139 110 945 587 2;
  • 25) 0.274 139 110 945 587 2 × 2 = 0 + 0.548 278 221 891 174 4;
  • 26) 0.548 278 221 891 174 4 × 2 = 1 + 0.096 556 443 782 348 8;
  • 27) 0.096 556 443 782 348 8 × 2 = 0 + 0.193 112 887 564 697 6;
  • 28) 0.193 112 887 564 697 6 × 2 = 0 + 0.386 225 775 129 395 2;
  • 29) 0.386 225 775 129 395 2 × 2 = 0 + 0.772 451 550 258 790 4;
  • 30) 0.772 451 550 258 790 4 × 2 = 1 + 0.544 903 100 517 580 8;
  • 31) 0.544 903 100 517 580 8 × 2 = 1 + 0.089 806 201 035 161 6;
  • 32) 0.089 806 201 035 161 6 × 2 = 0 + 0.179 612 402 070 323 2;
  • 33) 0.179 612 402 070 323 2 × 2 = 0 + 0.359 224 804 140 646 4;
  • 34) 0.359 224 804 140 646 4 × 2 = 0 + 0.718 449 608 281 292 8;
  • 35) 0.718 449 608 281 292 8 × 2 = 1 + 0.436 899 216 562 585 6;
  • 36) 0.436 899 216 562 585 6 × 2 = 0 + 0.873 798 433 125 171 2;
  • 37) 0.873 798 433 125 171 2 × 2 = 1 + 0.747 596 866 250 342 4;
  • 38) 0.747 596 866 250 342 4 × 2 = 1 + 0.495 193 732 500 684 8;
  • 39) 0.495 193 732 500 684 8 × 2 = 0 + 0.990 387 465 001 369 6;
  • 40) 0.990 387 465 001 369 6 × 2 = 1 + 0.980 774 930 002 739 2;
  • 41) 0.980 774 930 002 739 2 × 2 = 1 + 0.961 549 860 005 478 4;
  • 42) 0.961 549 860 005 478 4 × 2 = 1 + 0.923 099 720 010 956 8;
  • 43) 0.923 099 720 010 956 8 × 2 = 1 + 0.846 199 440 021 913 6;
  • 44) 0.846 199 440 021 913 6 × 2 = 1 + 0.692 398 880 043 827 2;
  • 45) 0.692 398 880 043 827 2 × 2 = 1 + 0.384 797 760 087 654 4;
  • 46) 0.384 797 760 087 654 4 × 2 = 0 + 0.769 595 520 175 308 8;
  • 47) 0.769 595 520 175 308 8 × 2 = 1 + 0.539 191 040 350 617 6;
  • 48) 0.539 191 040 350 617 6 × 2 = 1 + 0.078 382 080 701 235 2;
  • 49) 0.078 382 080 701 235 2 × 2 = 0 + 0.156 764 161 402 470 4;
  • 50) 0.156 764 161 402 470 4 × 2 = 0 + 0.313 528 322 804 940 8;
  • 51) 0.313 528 322 804 940 8 × 2 = 0 + 0.627 056 645 609 881 6;
  • 52) 0.627 056 645 609 881 6 × 2 = 1 + 0.254 113 291 219 763 2;
  • 53) 0.254 113 291 219 763 2 × 2 = 0 + 0.508 226 582 439 526 4;
  • 54) 0.508 226 582 439 526 4 × 2 = 1 + 0.016 453 164 879 052 8;
  • 55) 0.016 453 164 879 052 8 × 2 = 0 + 0.032 906 329 758 105 6;
  • 56) 0.032 906 329 758 105 6 × 2 = 0 + 0.065 812 659 516 211 2;
  • 57) 0.065 812 659 516 211 2 × 2 = 0 + 0.131 625 319 032 422 4;
  • 58) 0.131 625 319 032 422 4 × 2 = 0 + 0.263 250 638 064 844 8;
  • 59) 0.263 250 638 064 844 8 × 2 = 0 + 0.526 501 276 129 689 6;
  • 60) 0.526 501 276 129 689 6 × 2 = 1 + 0.053 002 552 259 379 2;
  • 61) 0.053 002 552 259 379 2 × 2 = 0 + 0.106 005 104 518 758 4;
  • 62) 0.106 005 104 518 758 4 × 2 = 0 + 0.212 010 209 037 516 8;
  • 63) 0.212 010 209 037 516 8 × 2 = 0 + 0.424 020 418 075 033 6;
  • 64) 0.424 020 418 075 033 6 × 2 = 0 + 0.848 040 836 150 067 2;

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:


0.000 282 005 914 396 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000(2)

6. Positive number before normalization:

0.000 282 005 914 396 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 914 396 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000 =

0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000


Decimal number -0.000 282 005 914 396 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1011 0001 0100 0001 0000

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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