-0.000 282 003 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 003 59(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 59(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 59| = 0.000 282 003 59


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 59.

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 59 × 2 = 0 + 0.000 564 007 18;
  • 2) 0.000 564 007 18 × 2 = 0 + 0.001 128 014 36;
  • 3) 0.001 128 014 36 × 2 = 0 + 0.002 256 028 72;
  • 4) 0.002 256 028 72 × 2 = 0 + 0.004 512 057 44;
  • 5) 0.004 512 057 44 × 2 = 0 + 0.009 024 114 88;
  • 6) 0.009 024 114 88 × 2 = 0 + 0.018 048 229 76;
  • 7) 0.018 048 229 76 × 2 = 0 + 0.036 096 459 52;
  • 8) 0.036 096 459 52 × 2 = 0 + 0.072 192 919 04;
  • 9) 0.072 192 919 04 × 2 = 0 + 0.144 385 838 08;
  • 10) 0.144 385 838 08 × 2 = 0 + 0.288 771 676 16;
  • 11) 0.288 771 676 16 × 2 = 0 + 0.577 543 352 32;
  • 12) 0.577 543 352 32 × 2 = 1 + 0.155 086 704 64;
  • 13) 0.155 086 704 64 × 2 = 0 + 0.310 173 409 28;
  • 14) 0.310 173 409 28 × 2 = 0 + 0.620 346 818 56;
  • 15) 0.620 346 818 56 × 2 = 1 + 0.240 693 637 12;
  • 16) 0.240 693 637 12 × 2 = 0 + 0.481 387 274 24;
  • 17) 0.481 387 274 24 × 2 = 0 + 0.962 774 548 48;
  • 18) 0.962 774 548 48 × 2 = 1 + 0.925 549 096 96;
  • 19) 0.925 549 096 96 × 2 = 1 + 0.851 098 193 92;
  • 20) 0.851 098 193 92 × 2 = 1 + 0.702 196 387 84;
  • 21) 0.702 196 387 84 × 2 = 1 + 0.404 392 775 68;
  • 22) 0.404 392 775 68 × 2 = 0 + 0.808 785 551 36;
  • 23) 0.808 785 551 36 × 2 = 1 + 0.617 571 102 72;
  • 24) 0.617 571 102 72 × 2 = 1 + 0.235 142 205 44;
  • 25) 0.235 142 205 44 × 2 = 0 + 0.470 284 410 88;
  • 26) 0.470 284 410 88 × 2 = 0 + 0.940 568 821 76;
  • 27) 0.940 568 821 76 × 2 = 1 + 0.881 137 643 52;
  • 28) 0.881 137 643 52 × 2 = 1 + 0.762 275 287 04;
  • 29) 0.762 275 287 04 × 2 = 1 + 0.524 550 574 08;
  • 30) 0.524 550 574 08 × 2 = 1 + 0.049 101 148 16;
  • 31) 0.049 101 148 16 × 2 = 0 + 0.098 202 296 32;
  • 32) 0.098 202 296 32 × 2 = 0 + 0.196 404 592 64;
  • 33) 0.196 404 592 64 × 2 = 0 + 0.392 809 185 28;
  • 34) 0.392 809 185 28 × 2 = 0 + 0.785 618 370 56;
  • 35) 0.785 618 370 56 × 2 = 1 + 0.571 236 741 12;
  • 36) 0.571 236 741 12 × 2 = 1 + 0.142 473 482 24;
  • 37) 0.142 473 482 24 × 2 = 0 + 0.284 946 964 48;
  • 38) 0.284 946 964 48 × 2 = 0 + 0.569 893 928 96;
  • 39) 0.569 893 928 96 × 2 = 1 + 0.139 787 857 92;
  • 40) 0.139 787 857 92 × 2 = 0 + 0.279 575 715 84;
  • 41) 0.279 575 715 84 × 2 = 0 + 0.559 151 431 68;
  • 42) 0.559 151 431 68 × 2 = 1 + 0.118 302 863 36;
  • 43) 0.118 302 863 36 × 2 = 0 + 0.236 605 726 72;
  • 44) 0.236 605 726 72 × 2 = 0 + 0.473 211 453 44;
  • 45) 0.473 211 453 44 × 2 = 0 + 0.946 422 906 88;
  • 46) 0.946 422 906 88 × 2 = 1 + 0.892 845 813 76;
  • 47) 0.892 845 813 76 × 2 = 1 + 0.785 691 627 52;
  • 48) 0.785 691 627 52 × 2 = 1 + 0.571 383 255 04;
  • 49) 0.571 383 255 04 × 2 = 1 + 0.142 766 510 08;
  • 50) 0.142 766 510 08 × 2 = 0 + 0.285 533 020 16;
  • 51) 0.285 533 020 16 × 2 = 0 + 0.571 066 040 32;
  • 52) 0.571 066 040 32 × 2 = 1 + 0.142 132 080 64;
  • 53) 0.142 132 080 64 × 2 = 0 + 0.284 264 161 28;
  • 54) 0.284 264 161 28 × 2 = 0 + 0.568 528 322 56;
  • 55) 0.568 528 322 56 × 2 = 1 + 0.137 056 645 12;
  • 56) 0.137 056 645 12 × 2 = 0 + 0.274 113 290 24;
  • 57) 0.274 113 290 24 × 2 = 0 + 0.548 226 580 48;
  • 58) 0.548 226 580 48 × 2 = 1 + 0.096 453 160 96;
  • 59) 0.096 453 160 96 × 2 = 0 + 0.192 906 321 92;
  • 60) 0.192 906 321 92 × 2 = 0 + 0.385 812 643 84;
  • 61) 0.385 812 643 84 × 2 = 0 + 0.771 625 287 68;
  • 62) 0.771 625 287 68 × 2 = 1 + 0.543 250 575 36;
  • 63) 0.543 250 575 36 × 2 = 1 + 0.086 501 150 72;
  • 64) 0.086 501 150 72 × 2 = 0 + 0.173 002 301 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:


0.000 282 003 59(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0011 0010 0100 0111 1001 0010 0100 0110(2)

6. Positive number before normalization:

0.000 282 003 59(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0011 0010 0100 0111 1001 0010 0100 0110(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 59(10) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0011 0010 0100 0111 1001 0010 0100 0110(2) =

0.0000 0000 0001 0010 0111 1011 0011 1100 0011 0010 0100 0111 1001 0010 0100 0110(2) × 20 =

1.0010 0111 1011 0011 1100 0011 0010 0100 0111 1001 0010 0100 0110(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 1100 0011 0010 0100 0111 1001 0010 0100 0110


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 1100 0011 0010 0100 0111 1001 0010 0100 0110 =

0010 0111 1011 0011 1100 0011 0010 0100 0111 1001 0010 0100 0110


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 1100 0011 0010 0100 0111 1001 0010 0100 0110


Decimal number -0.000 282 003 59 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1100 0011 0010 0100 0111 1001 0010 0100 0110

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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