-0.000 282 005 917 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 917 12(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
-0.000 282 005 917 12(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:

|-0.000 282 005 917 12| = 0.000 282 005 917 12


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: 0.000 282 005 917 12.

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) 0.000 282 005 917 12 × 2 = 0 + 0.000 564 011 834 24;
  • 2) 0.000 564 011 834 24 × 2 = 0 + 0.001 128 023 668 48;
  • 3) 0.001 128 023 668 48 × 2 = 0 + 0.002 256 047 336 96;
  • 4) 0.002 256 047 336 96 × 2 = 0 + 0.004 512 094 673 92;
  • 5) 0.004 512 094 673 92 × 2 = 0 + 0.009 024 189 347 84;
  • 6) 0.009 024 189 347 84 × 2 = 0 + 0.018 048 378 695 68;
  • 7) 0.018 048 378 695 68 × 2 = 0 + 0.036 096 757 391 36;
  • 8) 0.036 096 757 391 36 × 2 = 0 + 0.072 193 514 782 72;
  • 9) 0.072 193 514 782 72 × 2 = 0 + 0.144 387 029 565 44;
  • 10) 0.144 387 029 565 44 × 2 = 0 + 0.288 774 059 130 88;
  • 11) 0.288 774 059 130 88 × 2 = 0 + 0.577 548 118 261 76;
  • 12) 0.577 548 118 261 76 × 2 = 1 + 0.155 096 236 523 52;
  • 13) 0.155 096 236 523 52 × 2 = 0 + 0.310 192 473 047 04;
  • 14) 0.310 192 473 047 04 × 2 = 0 + 0.620 384 946 094 08;
  • 15) 0.620 384 946 094 08 × 2 = 1 + 0.240 769 892 188 16;
  • 16) 0.240 769 892 188 16 × 2 = 0 + 0.481 539 784 376 32;
  • 17) 0.481 539 784 376 32 × 2 = 0 + 0.963 079 568 752 64;
  • 18) 0.963 079 568 752 64 × 2 = 1 + 0.926 159 137 505 28;
  • 19) 0.926 159 137 505 28 × 2 = 1 + 0.852 318 275 010 56;
  • 20) 0.852 318 275 010 56 × 2 = 1 + 0.704 636 550 021 12;
  • 21) 0.704 636 550 021 12 × 2 = 1 + 0.409 273 100 042 24;
  • 22) 0.409 273 100 042 24 × 2 = 0 + 0.818 546 200 084 48;
  • 23) 0.818 546 200 084 48 × 2 = 1 + 0.637 092 400 168 96;
  • 24) 0.637 092 400 168 96 × 2 = 1 + 0.274 184 800 337 92;
  • 25) 0.274 184 800 337 92 × 2 = 0 + 0.548 369 600 675 84;
  • 26) 0.548 369 600 675 84 × 2 = 1 + 0.096 739 201 351 68;
  • 27) 0.096 739 201 351 68 × 2 = 0 + 0.193 478 402 703 36;
  • 28) 0.193 478 402 703 36 × 2 = 0 + 0.386 956 805 406 72;
  • 29) 0.386 956 805 406 72 × 2 = 0 + 0.773 913 610 813 44;
  • 30) 0.773 913 610 813 44 × 2 = 1 + 0.547 827 221 626 88;
  • 31) 0.547 827 221 626 88 × 2 = 1 + 0.095 654 443 253 76;
  • 32) 0.095 654 443 253 76 × 2 = 0 + 0.191 308 886 507 52;
  • 33) 0.191 308 886 507 52 × 2 = 0 + 0.382 617 773 015 04;
  • 34) 0.382 617 773 015 04 × 2 = 0 + 0.765 235 546 030 08;
  • 35) 0.765 235 546 030 08 × 2 = 1 + 0.530 471 092 060 16;
  • 36) 0.530 471 092 060 16 × 2 = 1 + 0.060 942 184 120 32;
  • 37) 0.060 942 184 120 32 × 2 = 0 + 0.121 884 368 240 64;
  • 38) 0.121 884 368 240 64 × 2 = 0 + 0.243 768 736 481 28;
  • 39) 0.243 768 736 481 28 × 2 = 0 + 0.487 537 472 962 56;
  • 40) 0.487 537 472 962 56 × 2 = 0 + 0.975 074 945 925 12;
  • 41) 0.975 074 945 925 12 × 2 = 1 + 0.950 149 891 850 24;
  • 42) 0.950 149 891 850 24 × 2 = 1 + 0.900 299 783 700 48;
  • 43) 0.900 299 783 700 48 × 2 = 1 + 0.800 599 567 400 96;
  • 44) 0.800 599 567 400 96 × 2 = 1 + 0.601 199 134 801 92;
  • 45) 0.601 199 134 801 92 × 2 = 1 + 0.202 398 269 603 84;
  • 46) 0.202 398 269 603 84 × 2 = 0 + 0.404 796 539 207 68;
  • 47) 0.404 796 539 207 68 × 2 = 0 + 0.809 593 078 415 36;
  • 48) 0.809 593 078 415 36 × 2 = 1 + 0.619 186 156 830 72;
  • 49) 0.619 186 156 830 72 × 2 = 1 + 0.238 372 313 661 44;
  • 50) 0.238 372 313 661 44 × 2 = 0 + 0.476 744 627 322 88;
  • 51) 0.476 744 627 322 88 × 2 = 0 + 0.953 489 254 645 76;
  • 52) 0.953 489 254 645 76 × 2 = 1 + 0.906 978 509 291 52;
  • 53) 0.906 978 509 291 52 × 2 = 1 + 0.813 957 018 583 04;
  • 54) 0.813 957 018 583 04 × 2 = 1 + 0.627 914 037 166 08;
  • 55) 0.627 914 037 166 08 × 2 = 1 + 0.255 828 074 332 16;
  • 56) 0.255 828 074 332 16 × 2 = 0 + 0.511 656 148 664 32;
  • 57) 0.511 656 148 664 32 × 2 = 1 + 0.023 312 297 328 64;
  • 58) 0.023 312 297 328 64 × 2 = 0 + 0.046 624 594 657 28;
  • 59) 0.046 624 594 657 28 × 2 = 0 + 0.093 249 189 314 56;
  • 60) 0.093 249 189 314 56 × 2 = 0 + 0.186 498 378 629 12;
  • 61) 0.186 498 378 629 12 × 2 = 0 + 0.372 996 757 258 24;
  • 62) 0.372 996 757 258 24 × 2 = 0 + 0.745 993 514 516 48;
  • 63) 0.745 993 514 516 48 × 2 = 1 + 0.491 987 029 032 96;
  • 64) 0.491 987 029 032 96 × 2 = 0 + 0.983 974 058 065 92;

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:


0.000 282 005 917 12(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1111 1001 1001 1110 1000 0010(2)

6. Positive number before normalization:

0.000 282 005 917 12(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1111 1001 1001 1110 1000 0010(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:


0.000 282 005 917 12(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1111 1001 1001 1110 1000 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 1111 1001 1001 1110 1000 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0000 1111 1001 1001 1110 1000 0010(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 0011 0000 1111 1001 1001 1110 1000 0010


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 0011 0000 1111 1001 1001 1110 1000 0010 =

0010 0111 1011 0100 0110 0011 0000 1111 1001 1001 1110 1000 0010


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 0011 0000 1111 1001 1001 1110 1000 0010


Decimal number -0.000 282 005 917 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0000 1111 1001 1001 1110 1000 0010

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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