-0.000 282 003 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 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
-0.000 282 003 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:

|-0.000 282 003 12| = 0.000 282 003 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: 0.000 282 003 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) 0.000 282 003 12 × 2 = 0 + 0.000 564 006 24;
  • 2) 0.000 564 006 24 × 2 = 0 + 0.001 128 012 48;
  • 3) 0.001 128 012 48 × 2 = 0 + 0.002 256 024 96;
  • 4) 0.002 256 024 96 × 2 = 0 + 0.004 512 049 92;
  • 5) 0.004 512 049 92 × 2 = 0 + 0.009 024 099 84;
  • 6) 0.009 024 099 84 × 2 = 0 + 0.018 048 199 68;
  • 7) 0.018 048 199 68 × 2 = 0 + 0.036 096 399 36;
  • 8) 0.036 096 399 36 × 2 = 0 + 0.072 192 798 72;
  • 9) 0.072 192 798 72 × 2 = 0 + 0.144 385 597 44;
  • 10) 0.144 385 597 44 × 2 = 0 + 0.288 771 194 88;
  • 11) 0.288 771 194 88 × 2 = 0 + 0.577 542 389 76;
  • 12) 0.577 542 389 76 × 2 = 1 + 0.155 084 779 52;
  • 13) 0.155 084 779 52 × 2 = 0 + 0.310 169 559 04;
  • 14) 0.310 169 559 04 × 2 = 0 + 0.620 339 118 08;
  • 15) 0.620 339 118 08 × 2 = 1 + 0.240 678 236 16;
  • 16) 0.240 678 236 16 × 2 = 0 + 0.481 356 472 32;
  • 17) 0.481 356 472 32 × 2 = 0 + 0.962 712 944 64;
  • 18) 0.962 712 944 64 × 2 = 1 + 0.925 425 889 28;
  • 19) 0.925 425 889 28 × 2 = 1 + 0.850 851 778 56;
  • 20) 0.850 851 778 56 × 2 = 1 + 0.701 703 557 12;
  • 21) 0.701 703 557 12 × 2 = 1 + 0.403 407 114 24;
  • 22) 0.403 407 114 24 × 2 = 0 + 0.806 814 228 48;
  • 23) 0.806 814 228 48 × 2 = 1 + 0.613 628 456 96;
  • 24) 0.613 628 456 96 × 2 = 1 + 0.227 256 913 92;
  • 25) 0.227 256 913 92 × 2 = 0 + 0.454 513 827 84;
  • 26) 0.454 513 827 84 × 2 = 0 + 0.909 027 655 68;
  • 27) 0.909 027 655 68 × 2 = 1 + 0.818 055 311 36;
  • 28) 0.818 055 311 36 × 2 = 1 + 0.636 110 622 72;
  • 29) 0.636 110 622 72 × 2 = 1 + 0.272 221 245 44;
  • 30) 0.272 221 245 44 × 2 = 0 + 0.544 442 490 88;
  • 31) 0.544 442 490 88 × 2 = 1 + 0.088 884 981 76;
  • 32) 0.088 884 981 76 × 2 = 0 + 0.177 769 963 52;
  • 33) 0.177 769 963 52 × 2 = 0 + 0.355 539 927 04;
  • 34) 0.355 539 927 04 × 2 = 0 + 0.711 079 854 08;
  • 35) 0.711 079 854 08 × 2 = 1 + 0.422 159 708 16;
  • 36) 0.422 159 708 16 × 2 = 0 + 0.844 319 416 32;
  • 37) 0.844 319 416 32 × 2 = 1 + 0.688 638 832 64;
  • 38) 0.688 638 832 64 × 2 = 1 + 0.377 277 665 28;
  • 39) 0.377 277 665 28 × 2 = 0 + 0.754 555 330 56;
  • 40) 0.754 555 330 56 × 2 = 1 + 0.509 110 661 12;
  • 41) 0.509 110 661 12 × 2 = 1 + 0.018 221 322 24;
  • 42) 0.018 221 322 24 × 2 = 0 + 0.036 442 644 48;
  • 43) 0.036 442 644 48 × 2 = 0 + 0.072 885 288 96;
  • 44) 0.072 885 288 96 × 2 = 0 + 0.145 770 577 92;
  • 45) 0.145 770 577 92 × 2 = 0 + 0.291 541 155 84;
  • 46) 0.291 541 155 84 × 2 = 0 + 0.583 082 311 68;
  • 47) 0.583 082 311 68 × 2 = 1 + 0.166 164 623 36;
  • 48) 0.166 164 623 36 × 2 = 0 + 0.332 329 246 72;
  • 49) 0.332 329 246 72 × 2 = 0 + 0.664 658 493 44;
  • 50) 0.664 658 493 44 × 2 = 1 + 0.329 316 986 88;
  • 51) 0.329 316 986 88 × 2 = 0 + 0.658 633 973 76;
  • 52) 0.658 633 973 76 × 2 = 1 + 0.317 267 947 52;
  • 53) 0.317 267 947 52 × 2 = 0 + 0.634 535 895 04;
  • 54) 0.634 535 895 04 × 2 = 1 + 0.269 071 790 08;
  • 55) 0.269 071 790 08 × 2 = 0 + 0.538 143 580 16;
  • 56) 0.538 143 580 16 × 2 = 1 + 0.076 287 160 32;
  • 57) 0.076 287 160 32 × 2 = 0 + 0.152 574 320 64;
  • 58) 0.152 574 320 64 × 2 = 0 + 0.305 148 641 28;
  • 59) 0.305 148 641 28 × 2 = 0 + 0.610 297 282 56;
  • 60) 0.610 297 282 56 × 2 = 1 + 0.220 594 565 12;
  • 61) 0.220 594 565 12 × 2 = 0 + 0.441 189 130 24;
  • 62) 0.441 189 130 24 × 2 = 0 + 0.882 378 260 48;
  • 63) 0.882 378 260 48 × 2 = 1 + 0.764 756 520 96;
  • 64) 0.764 756 520 96 × 2 = 1 + 0.529 513 041 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:


0.000 282 003 12(10) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011(2)

6. Positive number before normalization:

0.000 282 003 12(10) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011(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:


0.000 282 003 12(10) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011(2) =

0.0000 0000 0001 0010 0111 1011 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011(2) × 20 =

1.0010 0111 1011 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011(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 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011


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 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011 =

0010 0111 1011 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011


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 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011


Decimal number -0.000 282 003 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1010 0010 1101 1000 0010 0101 0101 0001 0011

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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