-0.000 282 037 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 037 4(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
-0.000 282 037 4(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:

|-0.000 282 037 4| = 0.000 282 037 4


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: 0.000 282 037 4.

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) 0.000 282 037 4 × 2 = 0 + 0.000 564 074 8;
  • 2) 0.000 564 074 8 × 2 = 0 + 0.001 128 149 6;
  • 3) 0.001 128 149 6 × 2 = 0 + 0.002 256 299 2;
  • 4) 0.002 256 299 2 × 2 = 0 + 0.004 512 598 4;
  • 5) 0.004 512 598 4 × 2 = 0 + 0.009 025 196 8;
  • 6) 0.009 025 196 8 × 2 = 0 + 0.018 050 393 6;
  • 7) 0.018 050 393 6 × 2 = 0 + 0.036 100 787 2;
  • 8) 0.036 100 787 2 × 2 = 0 + 0.072 201 574 4;
  • 9) 0.072 201 574 4 × 2 = 0 + 0.144 403 148 8;
  • 10) 0.144 403 148 8 × 2 = 0 + 0.288 806 297 6;
  • 11) 0.288 806 297 6 × 2 = 0 + 0.577 612 595 2;
  • 12) 0.577 612 595 2 × 2 = 1 + 0.155 225 190 4;
  • 13) 0.155 225 190 4 × 2 = 0 + 0.310 450 380 8;
  • 14) 0.310 450 380 8 × 2 = 0 + 0.620 900 761 6;
  • 15) 0.620 900 761 6 × 2 = 1 + 0.241 801 523 2;
  • 16) 0.241 801 523 2 × 2 = 0 + 0.483 603 046 4;
  • 17) 0.483 603 046 4 × 2 = 0 + 0.967 206 092 8;
  • 18) 0.967 206 092 8 × 2 = 1 + 0.934 412 185 6;
  • 19) 0.934 412 185 6 × 2 = 1 + 0.868 824 371 2;
  • 20) 0.868 824 371 2 × 2 = 1 + 0.737 648 742 4;
  • 21) 0.737 648 742 4 × 2 = 1 + 0.475 297 484 8;
  • 22) 0.475 297 484 8 × 2 = 0 + 0.950 594 969 6;
  • 23) 0.950 594 969 6 × 2 = 1 + 0.901 189 939 2;
  • 24) 0.901 189 939 2 × 2 = 1 + 0.802 379 878 4;
  • 25) 0.802 379 878 4 × 2 = 1 + 0.604 759 756 8;
  • 26) 0.604 759 756 8 × 2 = 1 + 0.209 519 513 6;
  • 27) 0.209 519 513 6 × 2 = 0 + 0.419 039 027 2;
  • 28) 0.419 039 027 2 × 2 = 0 + 0.838 078 054 4;
  • 29) 0.838 078 054 4 × 2 = 1 + 0.676 156 108 8;
  • 30) 0.676 156 108 8 × 2 = 1 + 0.352 312 217 6;
  • 31) 0.352 312 217 6 × 2 = 0 + 0.704 624 435 2;
  • 32) 0.704 624 435 2 × 2 = 1 + 0.409 248 870 4;
  • 33) 0.409 248 870 4 × 2 = 0 + 0.818 497 740 8;
  • 34) 0.818 497 740 8 × 2 = 1 + 0.636 995 481 6;
  • 35) 0.636 995 481 6 × 2 = 1 + 0.273 990 963 2;
  • 36) 0.273 990 963 2 × 2 = 0 + 0.547 981 926 4;
  • 37) 0.547 981 926 4 × 2 = 1 + 0.095 963 852 8;
  • 38) 0.095 963 852 8 × 2 = 0 + 0.191 927 705 6;
  • 39) 0.191 927 705 6 × 2 = 0 + 0.383 855 411 2;
  • 40) 0.383 855 411 2 × 2 = 0 + 0.767 710 822 4;
  • 41) 0.767 710 822 4 × 2 = 1 + 0.535 421 644 8;
  • 42) 0.535 421 644 8 × 2 = 1 + 0.070 843 289 6;
  • 43) 0.070 843 289 6 × 2 = 0 + 0.141 686 579 2;
  • 44) 0.141 686 579 2 × 2 = 0 + 0.283 373 158 4;
  • 45) 0.283 373 158 4 × 2 = 0 + 0.566 746 316 8;
  • 46) 0.566 746 316 8 × 2 = 1 + 0.133 492 633 6;
  • 47) 0.133 492 633 6 × 2 = 0 + 0.266 985 267 2;
  • 48) 0.266 985 267 2 × 2 = 0 + 0.533 970 534 4;
  • 49) 0.533 970 534 4 × 2 = 1 + 0.067 941 068 8;
  • 50) 0.067 941 068 8 × 2 = 0 + 0.135 882 137 6;
  • 51) 0.135 882 137 6 × 2 = 0 + 0.271 764 275 2;
  • 52) 0.271 764 275 2 × 2 = 0 + 0.543 528 550 4;
  • 53) 0.543 528 550 4 × 2 = 1 + 0.087 057 100 8;
  • 54) 0.087 057 100 8 × 2 = 0 + 0.174 114 201 6;
  • 55) 0.174 114 201 6 × 2 = 0 + 0.348 228 403 2;
  • 56) 0.348 228 403 2 × 2 = 0 + 0.696 456 806 4;
  • 57) 0.696 456 806 4 × 2 = 1 + 0.392 913 612 8;
  • 58) 0.392 913 612 8 × 2 = 0 + 0.785 827 225 6;
  • 59) 0.785 827 225 6 × 2 = 1 + 0.571 654 451 2;
  • 60) 0.571 654 451 2 × 2 = 1 + 0.143 308 902 4;
  • 61) 0.143 308 902 4 × 2 = 0 + 0.286 617 804 8;
  • 62) 0.286 617 804 8 × 2 = 0 + 0.573 235 609 6;
  • 63) 0.573 235 609 6 × 2 = 1 + 0.146 471 219 2;
  • 64) 0.146 471 219 2 × 2 = 0 + 0.292 942 438 4;

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:


0.000 282 037 4(10) =

0.0000 0000 0001 0010 0111 1011 1100 1101 0110 1000 1100 0100 1000 1000 1011 0010(2)

6. Positive number before normalization:

0.000 282 037 4(10) =

0.0000 0000 0001 0010 0111 1011 1100 1101 0110 1000 1100 0100 1000 1000 1011 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:


0.000 282 037 4(10) =

0.0000 0000 0001 0010 0111 1011 1100 1101 0110 1000 1100 0100 1000 1000 1011 0010(2) =

0.0000 0000 0001 0010 0111 1011 1100 1101 0110 1000 1100 0100 1000 1000 1011 0010(2) × 20 =

1.0010 0111 1011 1100 1101 0110 1000 1100 0100 1000 1000 1011 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 1100 1101 0110 1000 1100 0100 1000 1000 1011 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 1100 1101 0110 1000 1100 0100 1000 1000 1011 0010 =

0010 0111 1011 1100 1101 0110 1000 1100 0100 1000 1000 1011 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 1100 1101 0110 1000 1100 0100 1000 1000 1011 0010


Decimal number -0.000 282 037 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1100 1101 0110 1000 1100 0100 1000 1000 1011 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