-0.000 282 006 18 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 18(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
-0.000 282 006 18(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:

|-0.000 282 006 18| = 0.000 282 006 18


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: 0.000 282 006 18.

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) 0.000 282 006 18 × 2 = 0 + 0.000 564 012 36;
  • 2) 0.000 564 012 36 × 2 = 0 + 0.001 128 024 72;
  • 3) 0.001 128 024 72 × 2 = 0 + 0.002 256 049 44;
  • 4) 0.002 256 049 44 × 2 = 0 + 0.004 512 098 88;
  • 5) 0.004 512 098 88 × 2 = 0 + 0.009 024 197 76;
  • 6) 0.009 024 197 76 × 2 = 0 + 0.018 048 395 52;
  • 7) 0.018 048 395 52 × 2 = 0 + 0.036 096 791 04;
  • 8) 0.036 096 791 04 × 2 = 0 + 0.072 193 582 08;
  • 9) 0.072 193 582 08 × 2 = 0 + 0.144 387 164 16;
  • 10) 0.144 387 164 16 × 2 = 0 + 0.288 774 328 32;
  • 11) 0.288 774 328 32 × 2 = 0 + 0.577 548 656 64;
  • 12) 0.577 548 656 64 × 2 = 1 + 0.155 097 313 28;
  • 13) 0.155 097 313 28 × 2 = 0 + 0.310 194 626 56;
  • 14) 0.310 194 626 56 × 2 = 0 + 0.620 389 253 12;
  • 15) 0.620 389 253 12 × 2 = 1 + 0.240 778 506 24;
  • 16) 0.240 778 506 24 × 2 = 0 + 0.481 557 012 48;
  • 17) 0.481 557 012 48 × 2 = 0 + 0.963 114 024 96;
  • 18) 0.963 114 024 96 × 2 = 1 + 0.926 228 049 92;
  • 19) 0.926 228 049 92 × 2 = 1 + 0.852 456 099 84;
  • 20) 0.852 456 099 84 × 2 = 1 + 0.704 912 199 68;
  • 21) 0.704 912 199 68 × 2 = 1 + 0.409 824 399 36;
  • 22) 0.409 824 399 36 × 2 = 0 + 0.819 648 798 72;
  • 23) 0.819 648 798 72 × 2 = 1 + 0.639 297 597 44;
  • 24) 0.639 297 597 44 × 2 = 1 + 0.278 595 194 88;
  • 25) 0.278 595 194 88 × 2 = 0 + 0.557 190 389 76;
  • 26) 0.557 190 389 76 × 2 = 1 + 0.114 380 779 52;
  • 27) 0.114 380 779 52 × 2 = 0 + 0.228 761 559 04;
  • 28) 0.228 761 559 04 × 2 = 0 + 0.457 523 118 08;
  • 29) 0.457 523 118 08 × 2 = 0 + 0.915 046 236 16;
  • 30) 0.915 046 236 16 × 2 = 1 + 0.830 092 472 32;
  • 31) 0.830 092 472 32 × 2 = 1 + 0.660 184 944 64;
  • 32) 0.660 184 944 64 × 2 = 1 + 0.320 369 889 28;
  • 33) 0.320 369 889 28 × 2 = 0 + 0.640 739 778 56;
  • 34) 0.640 739 778 56 × 2 = 1 + 0.281 479 557 12;
  • 35) 0.281 479 557 12 × 2 = 0 + 0.562 959 114 24;
  • 36) 0.562 959 114 24 × 2 = 1 + 0.125 918 228 48;
  • 37) 0.125 918 228 48 × 2 = 0 + 0.251 836 456 96;
  • 38) 0.251 836 456 96 × 2 = 0 + 0.503 672 913 92;
  • 39) 0.503 672 913 92 × 2 = 1 + 0.007 345 827 84;
  • 40) 0.007 345 827 84 × 2 = 0 + 0.014 691 655 68;
  • 41) 0.014 691 655 68 × 2 = 0 + 0.029 383 311 36;
  • 42) 0.029 383 311 36 × 2 = 0 + 0.058 766 622 72;
  • 43) 0.058 766 622 72 × 2 = 0 + 0.117 533 245 44;
  • 44) 0.117 533 245 44 × 2 = 0 + 0.235 066 490 88;
  • 45) 0.235 066 490 88 × 2 = 0 + 0.470 132 981 76;
  • 46) 0.470 132 981 76 × 2 = 0 + 0.940 265 963 52;
  • 47) 0.940 265 963 52 × 2 = 1 + 0.880 531 927 04;
  • 48) 0.880 531 927 04 × 2 = 1 + 0.761 063 854 08;
  • 49) 0.761 063 854 08 × 2 = 1 + 0.522 127 708 16;
  • 50) 0.522 127 708 16 × 2 = 1 + 0.044 255 416 32;
  • 51) 0.044 255 416 32 × 2 = 0 + 0.088 510 832 64;
  • 52) 0.088 510 832 64 × 2 = 0 + 0.177 021 665 28;
  • 53) 0.177 021 665 28 × 2 = 0 + 0.354 043 330 56;
  • 54) 0.354 043 330 56 × 2 = 0 + 0.708 086 661 12;
  • 55) 0.708 086 661 12 × 2 = 1 + 0.416 173 322 24;
  • 56) 0.416 173 322 24 × 2 = 0 + 0.832 346 644 48;
  • 57) 0.832 346 644 48 × 2 = 1 + 0.664 693 288 96;
  • 58) 0.664 693 288 96 × 2 = 1 + 0.329 386 577 92;
  • 59) 0.329 386 577 92 × 2 = 0 + 0.658 773 155 84;
  • 60) 0.658 773 155 84 × 2 = 1 + 0.317 546 311 68;
  • 61) 0.317 546 311 68 × 2 = 0 + 0.635 092 623 36;
  • 62) 0.635 092 623 36 × 2 = 1 + 0.270 185 246 72;
  • 63) 0.270 185 246 72 × 2 = 0 + 0.540 370 493 44;
  • 64) 0.540 370 493 44 × 2 = 1 + 0.080 740 986 88;

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:


0.000 282 006 18(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 0010 0000 0011 1100 0010 1101 0101(2)

6. Positive number before normalization:

0.000 282 006 18(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 0010 0000 0011 1100 0010 1101 0101(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:


0.000 282 006 18(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 0010 0000 0011 1100 0010 1101 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0101 0010 0000 0011 1100 0010 1101 0101(2) × 20 =

1.0010 0111 1011 0100 0111 0101 0010 0000 0011 1100 0010 1101 0101(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 0111 0101 0010 0000 0011 1100 0010 1101 0101


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 0111 0101 0010 0000 0011 1100 0010 1101 0101 =

0010 0111 1011 0100 0111 0101 0010 0000 0011 1100 0010 1101 0101


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 0111 0101 0010 0000 0011 1100 0010 1101 0101


Decimal number -0.000 282 006 18 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 0101 0010 0000 0011 1100 0010 1101 0101

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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