-0.000 282 009 44 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 44(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
-0.000 282 009 44(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:

|-0.000 282 009 44| = 0.000 282 009 44


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: 0.000 282 009 44.

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) 0.000 282 009 44 × 2 = 0 + 0.000 564 018 88;
  • 2) 0.000 564 018 88 × 2 = 0 + 0.001 128 037 76;
  • 3) 0.001 128 037 76 × 2 = 0 + 0.002 256 075 52;
  • 4) 0.002 256 075 52 × 2 = 0 + 0.004 512 151 04;
  • 5) 0.004 512 151 04 × 2 = 0 + 0.009 024 302 08;
  • 6) 0.009 024 302 08 × 2 = 0 + 0.018 048 604 16;
  • 7) 0.018 048 604 16 × 2 = 0 + 0.036 097 208 32;
  • 8) 0.036 097 208 32 × 2 = 0 + 0.072 194 416 64;
  • 9) 0.072 194 416 64 × 2 = 0 + 0.144 388 833 28;
  • 10) 0.144 388 833 28 × 2 = 0 + 0.288 777 666 56;
  • 11) 0.288 777 666 56 × 2 = 0 + 0.577 555 333 12;
  • 12) 0.577 555 333 12 × 2 = 1 + 0.155 110 666 24;
  • 13) 0.155 110 666 24 × 2 = 0 + 0.310 221 332 48;
  • 14) 0.310 221 332 48 × 2 = 0 + 0.620 442 664 96;
  • 15) 0.620 442 664 96 × 2 = 1 + 0.240 885 329 92;
  • 16) 0.240 885 329 92 × 2 = 0 + 0.481 770 659 84;
  • 17) 0.481 770 659 84 × 2 = 0 + 0.963 541 319 68;
  • 18) 0.963 541 319 68 × 2 = 1 + 0.927 082 639 36;
  • 19) 0.927 082 639 36 × 2 = 1 + 0.854 165 278 72;
  • 20) 0.854 165 278 72 × 2 = 1 + 0.708 330 557 44;
  • 21) 0.708 330 557 44 × 2 = 1 + 0.416 661 114 88;
  • 22) 0.416 661 114 88 × 2 = 0 + 0.833 322 229 76;
  • 23) 0.833 322 229 76 × 2 = 1 + 0.666 644 459 52;
  • 24) 0.666 644 459 52 × 2 = 1 + 0.333 288 919 04;
  • 25) 0.333 288 919 04 × 2 = 0 + 0.666 577 838 08;
  • 26) 0.666 577 838 08 × 2 = 1 + 0.333 155 676 16;
  • 27) 0.333 155 676 16 × 2 = 0 + 0.666 311 352 32;
  • 28) 0.666 311 352 32 × 2 = 1 + 0.332 622 704 64;
  • 29) 0.332 622 704 64 × 2 = 0 + 0.665 245 409 28;
  • 30) 0.665 245 409 28 × 2 = 1 + 0.330 490 818 56;
  • 31) 0.330 490 818 56 × 2 = 0 + 0.660 981 637 12;
  • 32) 0.660 981 637 12 × 2 = 1 + 0.321 963 274 24;
  • 33) 0.321 963 274 24 × 2 = 0 + 0.643 926 548 48;
  • 34) 0.643 926 548 48 × 2 = 1 + 0.287 853 096 96;
  • 35) 0.287 853 096 96 × 2 = 0 + 0.575 706 193 92;
  • 36) 0.575 706 193 92 × 2 = 1 + 0.151 412 387 84;
  • 37) 0.151 412 387 84 × 2 = 0 + 0.302 824 775 68;
  • 38) 0.302 824 775 68 × 2 = 0 + 0.605 649 551 36;
  • 39) 0.605 649 551 36 × 2 = 1 + 0.211 299 102 72;
  • 40) 0.211 299 102 72 × 2 = 0 + 0.422 598 205 44;
  • 41) 0.422 598 205 44 × 2 = 0 + 0.845 196 410 88;
  • 42) 0.845 196 410 88 × 2 = 1 + 0.690 392 821 76;
  • 43) 0.690 392 821 76 × 2 = 1 + 0.380 785 643 52;
  • 44) 0.380 785 643 52 × 2 = 0 + 0.761 571 287 04;
  • 45) 0.761 571 287 04 × 2 = 1 + 0.523 142 574 08;
  • 46) 0.523 142 574 08 × 2 = 1 + 0.046 285 148 16;
  • 47) 0.046 285 148 16 × 2 = 0 + 0.092 570 296 32;
  • 48) 0.092 570 296 32 × 2 = 0 + 0.185 140 592 64;
  • 49) 0.185 140 592 64 × 2 = 0 + 0.370 281 185 28;
  • 50) 0.370 281 185 28 × 2 = 0 + 0.740 562 370 56;
  • 51) 0.740 562 370 56 × 2 = 1 + 0.481 124 741 12;
  • 52) 0.481 124 741 12 × 2 = 0 + 0.962 249 482 24;
  • 53) 0.962 249 482 24 × 2 = 1 + 0.924 498 964 48;
  • 54) 0.924 498 964 48 × 2 = 1 + 0.848 997 928 96;
  • 55) 0.848 997 928 96 × 2 = 1 + 0.697 995 857 92;
  • 56) 0.697 995 857 92 × 2 = 1 + 0.395 991 715 84;
  • 57) 0.395 991 715 84 × 2 = 0 + 0.791 983 431 68;
  • 58) 0.791 983 431 68 × 2 = 1 + 0.583 966 863 36;
  • 59) 0.583 966 863 36 × 2 = 1 + 0.167 933 726 72;
  • 60) 0.167 933 726 72 × 2 = 0 + 0.335 867 453 44;
  • 61) 0.335 867 453 44 × 2 = 0 + 0.671 734 906 88;
  • 62) 0.671 734 906 88 × 2 = 1 + 0.343 469 813 76;
  • 63) 0.343 469 813 76 × 2 = 0 + 0.686 939 627 52;
  • 64) 0.686 939 627 52 × 2 = 1 + 0.373 879 255 04;

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:


0.000 282 009 44(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0101 0010 0110 1100 0010 1111 0110 0101(2)

6. Positive number before normalization:

0.000 282 009 44(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0101 0010 0110 1100 0010 1111 0110 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:


0.000 282 009 44(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0101 0010 0110 1100 0010 1111 0110 0101(2) =

0.0000 0000 0001 0010 0111 1011 0101 0101 0101 0010 0110 1100 0010 1111 0110 0101(2) × 20 =

1.0010 0111 1011 0101 0101 0101 0010 0110 1100 0010 1111 0110 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 0101 0101 0101 0010 0110 1100 0010 1111 0110 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 0101 0101 0101 0010 0110 1100 0010 1111 0110 0101 =

0010 0111 1011 0101 0101 0101 0010 0110 1100 0010 1111 0110 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 0101 0101 0101 0010 0110 1100 0010 1111 0110 0101


Decimal number -0.000 282 009 44 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0101 0101 0010 0110 1100 0010 1111 0110 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