-0.000 282 005 911 09 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 911 09(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
-0.000 282 005 911 09(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:

|-0.000 282 005 911 09| = 0.000 282 005 911 09


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: 0.000 282 005 911 09.

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) 0.000 282 005 911 09 × 2 = 0 + 0.000 564 011 822 18;
  • 2) 0.000 564 011 822 18 × 2 = 0 + 0.001 128 023 644 36;
  • 3) 0.001 128 023 644 36 × 2 = 0 + 0.002 256 047 288 72;
  • 4) 0.002 256 047 288 72 × 2 = 0 + 0.004 512 094 577 44;
  • 5) 0.004 512 094 577 44 × 2 = 0 + 0.009 024 189 154 88;
  • 6) 0.009 024 189 154 88 × 2 = 0 + 0.018 048 378 309 76;
  • 7) 0.018 048 378 309 76 × 2 = 0 + 0.036 096 756 619 52;
  • 8) 0.036 096 756 619 52 × 2 = 0 + 0.072 193 513 239 04;
  • 9) 0.072 193 513 239 04 × 2 = 0 + 0.144 387 026 478 08;
  • 10) 0.144 387 026 478 08 × 2 = 0 + 0.288 774 052 956 16;
  • 11) 0.288 774 052 956 16 × 2 = 0 + 0.577 548 105 912 32;
  • 12) 0.577 548 105 912 32 × 2 = 1 + 0.155 096 211 824 64;
  • 13) 0.155 096 211 824 64 × 2 = 0 + 0.310 192 423 649 28;
  • 14) 0.310 192 423 649 28 × 2 = 0 + 0.620 384 847 298 56;
  • 15) 0.620 384 847 298 56 × 2 = 1 + 0.240 769 694 597 12;
  • 16) 0.240 769 694 597 12 × 2 = 0 + 0.481 539 389 194 24;
  • 17) 0.481 539 389 194 24 × 2 = 0 + 0.963 078 778 388 48;
  • 18) 0.963 078 778 388 48 × 2 = 1 + 0.926 157 556 776 96;
  • 19) 0.926 157 556 776 96 × 2 = 1 + 0.852 315 113 553 92;
  • 20) 0.852 315 113 553 92 × 2 = 1 + 0.704 630 227 107 84;
  • 21) 0.704 630 227 107 84 × 2 = 1 + 0.409 260 454 215 68;
  • 22) 0.409 260 454 215 68 × 2 = 0 + 0.818 520 908 431 36;
  • 23) 0.818 520 908 431 36 × 2 = 1 + 0.637 041 816 862 72;
  • 24) 0.637 041 816 862 72 × 2 = 1 + 0.274 083 633 725 44;
  • 25) 0.274 083 633 725 44 × 2 = 0 + 0.548 167 267 450 88;
  • 26) 0.548 167 267 450 88 × 2 = 1 + 0.096 334 534 901 76;
  • 27) 0.096 334 534 901 76 × 2 = 0 + 0.192 669 069 803 52;
  • 28) 0.192 669 069 803 52 × 2 = 0 + 0.385 338 139 607 04;
  • 29) 0.385 338 139 607 04 × 2 = 0 + 0.770 676 279 214 08;
  • 30) 0.770 676 279 214 08 × 2 = 1 + 0.541 352 558 428 16;
  • 31) 0.541 352 558 428 16 × 2 = 1 + 0.082 705 116 856 32;
  • 32) 0.082 705 116 856 32 × 2 = 0 + 0.165 410 233 712 64;
  • 33) 0.165 410 233 712 64 × 2 = 0 + 0.330 820 467 425 28;
  • 34) 0.330 820 467 425 28 × 2 = 0 + 0.661 640 934 850 56;
  • 35) 0.661 640 934 850 56 × 2 = 1 + 0.323 281 869 701 12;
  • 36) 0.323 281 869 701 12 × 2 = 0 + 0.646 563 739 402 24;
  • 37) 0.646 563 739 402 24 × 2 = 1 + 0.293 127 478 804 48;
  • 38) 0.293 127 478 804 48 × 2 = 0 + 0.586 254 957 608 96;
  • 39) 0.586 254 957 608 96 × 2 = 1 + 0.172 509 915 217 92;
  • 40) 0.172 509 915 217 92 × 2 = 0 + 0.345 019 830 435 84;
  • 41) 0.345 019 830 435 84 × 2 = 0 + 0.690 039 660 871 68;
  • 42) 0.690 039 660 871 68 × 2 = 1 + 0.380 079 321 743 36;
  • 43) 0.380 079 321 743 36 × 2 = 0 + 0.760 158 643 486 72;
  • 44) 0.760 158 643 486 72 × 2 = 1 + 0.520 317 286 973 44;
  • 45) 0.520 317 286 973 44 × 2 = 1 + 0.040 634 573 946 88;
  • 46) 0.040 634 573 946 88 × 2 = 0 + 0.081 269 147 893 76;
  • 47) 0.081 269 147 893 76 × 2 = 0 + 0.162 538 295 787 52;
  • 48) 0.162 538 295 787 52 × 2 = 0 + 0.325 076 591 575 04;
  • 49) 0.325 076 591 575 04 × 2 = 0 + 0.650 153 183 150 08;
  • 50) 0.650 153 183 150 08 × 2 = 1 + 0.300 306 366 300 16;
  • 51) 0.300 306 366 300 16 × 2 = 0 + 0.600 612 732 600 32;
  • 52) 0.600 612 732 600 32 × 2 = 1 + 0.201 225 465 200 64;
  • 53) 0.201 225 465 200 64 × 2 = 0 + 0.402 450 930 401 28;
  • 54) 0.402 450 930 401 28 × 2 = 0 + 0.804 901 860 802 56;
  • 55) 0.804 901 860 802 56 × 2 = 1 + 0.609 803 721 605 12;
  • 56) 0.609 803 721 605 12 × 2 = 1 + 0.219 607 443 210 24;
  • 57) 0.219 607 443 210 24 × 2 = 0 + 0.439 214 886 420 48;
  • 58) 0.439 214 886 420 48 × 2 = 0 + 0.878 429 772 840 96;
  • 59) 0.878 429 772 840 96 × 2 = 1 + 0.756 859 545 681 92;
  • 60) 0.756 859 545 681 92 × 2 = 1 + 0.513 719 091 363 84;
  • 61) 0.513 719 091 363 84 × 2 = 1 + 0.027 438 182 727 68;
  • 62) 0.027 438 182 727 68 × 2 = 0 + 0.054 876 365 455 36;
  • 63) 0.054 876 365 455 36 × 2 = 0 + 0.109 752 730 910 72;
  • 64) 0.109 752 730 910 72 × 2 = 0 + 0.219 505 461 821 44;

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:


0.000 282 005 911 09(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 0101 1000 0101 0011 0011 1000(2)

6. Positive number before normalization:

0.000 282 005 911 09(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 0101 1000 0101 0011 0011 1000(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:


0.000 282 005 911 09(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 0101 1000 0101 0011 0011 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 0101 1000 0101 0011 0011 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1010 0101 1000 0101 0011 0011 1000(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 1010 0101 1000 0101 0011 0011 1000


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 1010 0101 1000 0101 0011 0011 1000 =

0010 0111 1011 0100 0110 0010 1010 0101 1000 0101 0011 0011 1000


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 1010 0101 1000 0101 0011 0011 1000


Decimal number -0.000 282 005 911 09 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1010 0101 1000 0101 0011 0011 1000

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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