-0.000 282 557 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 557(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
-0.000 282 557(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:

|-0.000 282 557| = 0.000 282 557


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: 0.000 282 557.

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) 0.000 282 557 × 2 = 0 + 0.000 565 114;
  • 2) 0.000 565 114 × 2 = 0 + 0.001 130 228;
  • 3) 0.001 130 228 × 2 = 0 + 0.002 260 456;
  • 4) 0.002 260 456 × 2 = 0 + 0.004 520 912;
  • 5) 0.004 520 912 × 2 = 0 + 0.009 041 824;
  • 6) 0.009 041 824 × 2 = 0 + 0.018 083 648;
  • 7) 0.018 083 648 × 2 = 0 + 0.036 167 296;
  • 8) 0.036 167 296 × 2 = 0 + 0.072 334 592;
  • 9) 0.072 334 592 × 2 = 0 + 0.144 669 184;
  • 10) 0.144 669 184 × 2 = 0 + 0.289 338 368;
  • 11) 0.289 338 368 × 2 = 0 + 0.578 676 736;
  • 12) 0.578 676 736 × 2 = 1 + 0.157 353 472;
  • 13) 0.157 353 472 × 2 = 0 + 0.314 706 944;
  • 14) 0.314 706 944 × 2 = 0 + 0.629 413 888;
  • 15) 0.629 413 888 × 2 = 1 + 0.258 827 776;
  • 16) 0.258 827 776 × 2 = 0 + 0.517 655 552;
  • 17) 0.517 655 552 × 2 = 1 + 0.035 311 104;
  • 18) 0.035 311 104 × 2 = 0 + 0.070 622 208;
  • 19) 0.070 622 208 × 2 = 0 + 0.141 244 416;
  • 20) 0.141 244 416 × 2 = 0 + 0.282 488 832;
  • 21) 0.282 488 832 × 2 = 0 + 0.564 977 664;
  • 22) 0.564 977 664 × 2 = 1 + 0.129 955 328;
  • 23) 0.129 955 328 × 2 = 0 + 0.259 910 656;
  • 24) 0.259 910 656 × 2 = 0 + 0.519 821 312;
  • 25) 0.519 821 312 × 2 = 1 + 0.039 642 624;
  • 26) 0.039 642 624 × 2 = 0 + 0.079 285 248;
  • 27) 0.079 285 248 × 2 = 0 + 0.158 570 496;
  • 28) 0.158 570 496 × 2 = 0 + 0.317 140 992;
  • 29) 0.317 140 992 × 2 = 0 + 0.634 281 984;
  • 30) 0.634 281 984 × 2 = 1 + 0.268 563 968;
  • 31) 0.268 563 968 × 2 = 0 + 0.537 127 936;
  • 32) 0.537 127 936 × 2 = 1 + 0.074 255 872;
  • 33) 0.074 255 872 × 2 = 0 + 0.148 511 744;
  • 34) 0.148 511 744 × 2 = 0 + 0.297 023 488;
  • 35) 0.297 023 488 × 2 = 0 + 0.594 046 976;
  • 36) 0.594 046 976 × 2 = 1 + 0.188 093 952;
  • 37) 0.188 093 952 × 2 = 0 + 0.376 187 904;
  • 38) 0.376 187 904 × 2 = 0 + 0.752 375 808;
  • 39) 0.752 375 808 × 2 = 1 + 0.504 751 616;
  • 40) 0.504 751 616 × 2 = 1 + 0.009 503 232;
  • 41) 0.009 503 232 × 2 = 0 + 0.019 006 464;
  • 42) 0.019 006 464 × 2 = 0 + 0.038 012 928;
  • 43) 0.038 012 928 × 2 = 0 + 0.076 025 856;
  • 44) 0.076 025 856 × 2 = 0 + 0.152 051 712;
  • 45) 0.152 051 712 × 2 = 0 + 0.304 103 424;
  • 46) 0.304 103 424 × 2 = 0 + 0.608 206 848;
  • 47) 0.608 206 848 × 2 = 1 + 0.216 413 696;
  • 48) 0.216 413 696 × 2 = 0 + 0.432 827 392;
  • 49) 0.432 827 392 × 2 = 0 + 0.865 654 784;
  • 50) 0.865 654 784 × 2 = 1 + 0.731 309 568;
  • 51) 0.731 309 568 × 2 = 1 + 0.462 619 136;
  • 52) 0.462 619 136 × 2 = 0 + 0.925 238 272;
  • 53) 0.925 238 272 × 2 = 1 + 0.850 476 544;
  • 54) 0.850 476 544 × 2 = 1 + 0.700 953 088;
  • 55) 0.700 953 088 × 2 = 1 + 0.401 906 176;
  • 56) 0.401 906 176 × 2 = 0 + 0.803 812 352;
  • 57) 0.803 812 352 × 2 = 1 + 0.607 624 704;
  • 58) 0.607 624 704 × 2 = 1 + 0.215 249 408;
  • 59) 0.215 249 408 × 2 = 0 + 0.430 498 816;
  • 60) 0.430 498 816 × 2 = 0 + 0.860 997 632;
  • 61) 0.860 997 632 × 2 = 1 + 0.721 995 264;
  • 62) 0.721 995 264 × 2 = 1 + 0.443 990 528;
  • 63) 0.443 990 528 × 2 = 0 + 0.887 981 056;
  • 64) 0.887 981 056 × 2 = 1 + 0.775 962 112;

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:


0.000 282 557(10) =

0.0000 0000 0001 0010 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101(2)

6. Positive number before normalization:

0.000 282 557(10) =

0.0000 0000 0001 0010 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101(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:


0.000 282 557(10) =

0.0000 0000 0001 0010 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101(2) =

0.0000 0000 0001 0010 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101(2) × 20 =

1.0010 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101(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 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101


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 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101 =

0010 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101


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 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101


Decimal number -0.000 282 557 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0100 1000 0101 0001 0011 0000 0010 0110 1110 1100 1101

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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