-0.000 282 059 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 059(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
-0.000 282 059(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:

|-0.000 282 059| = 0.000 282 059


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: 0.000 282 059.

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) 0.000 282 059 × 2 = 0 + 0.000 564 118;
  • 2) 0.000 564 118 × 2 = 0 + 0.001 128 236;
  • 3) 0.001 128 236 × 2 = 0 + 0.002 256 472;
  • 4) 0.002 256 472 × 2 = 0 + 0.004 512 944;
  • 5) 0.004 512 944 × 2 = 0 + 0.009 025 888;
  • 6) 0.009 025 888 × 2 = 0 + 0.018 051 776;
  • 7) 0.018 051 776 × 2 = 0 + 0.036 103 552;
  • 8) 0.036 103 552 × 2 = 0 + 0.072 207 104;
  • 9) 0.072 207 104 × 2 = 0 + 0.144 414 208;
  • 10) 0.144 414 208 × 2 = 0 + 0.288 828 416;
  • 11) 0.288 828 416 × 2 = 0 + 0.577 656 832;
  • 12) 0.577 656 832 × 2 = 1 + 0.155 313 664;
  • 13) 0.155 313 664 × 2 = 0 + 0.310 627 328;
  • 14) 0.310 627 328 × 2 = 0 + 0.621 254 656;
  • 15) 0.621 254 656 × 2 = 1 + 0.242 509 312;
  • 16) 0.242 509 312 × 2 = 0 + 0.485 018 624;
  • 17) 0.485 018 624 × 2 = 0 + 0.970 037 248;
  • 18) 0.970 037 248 × 2 = 1 + 0.940 074 496;
  • 19) 0.940 074 496 × 2 = 1 + 0.880 148 992;
  • 20) 0.880 148 992 × 2 = 1 + 0.760 297 984;
  • 21) 0.760 297 984 × 2 = 1 + 0.520 595 968;
  • 22) 0.520 595 968 × 2 = 1 + 0.041 191 936;
  • 23) 0.041 191 936 × 2 = 0 + 0.082 383 872;
  • 24) 0.082 383 872 × 2 = 0 + 0.164 767 744;
  • 25) 0.164 767 744 × 2 = 0 + 0.329 535 488;
  • 26) 0.329 535 488 × 2 = 0 + 0.659 070 976;
  • 27) 0.659 070 976 × 2 = 1 + 0.318 141 952;
  • 28) 0.318 141 952 × 2 = 0 + 0.636 283 904;
  • 29) 0.636 283 904 × 2 = 1 + 0.272 567 808;
  • 30) 0.272 567 808 × 2 = 0 + 0.545 135 616;
  • 31) 0.545 135 616 × 2 = 1 + 0.090 271 232;
  • 32) 0.090 271 232 × 2 = 0 + 0.180 542 464;
  • 33) 0.180 542 464 × 2 = 0 + 0.361 084 928;
  • 34) 0.361 084 928 × 2 = 0 + 0.722 169 856;
  • 35) 0.722 169 856 × 2 = 1 + 0.444 339 712;
  • 36) 0.444 339 712 × 2 = 0 + 0.888 679 424;
  • 37) 0.888 679 424 × 2 = 1 + 0.777 358 848;
  • 38) 0.777 358 848 × 2 = 1 + 0.554 717 696;
  • 39) 0.554 717 696 × 2 = 1 + 0.109 435 392;
  • 40) 0.109 435 392 × 2 = 0 + 0.218 870 784;
  • 41) 0.218 870 784 × 2 = 0 + 0.437 741 568;
  • 42) 0.437 741 568 × 2 = 0 + 0.875 483 136;
  • 43) 0.875 483 136 × 2 = 1 + 0.750 966 272;
  • 44) 0.750 966 272 × 2 = 1 + 0.501 932 544;
  • 45) 0.501 932 544 × 2 = 1 + 0.003 865 088;
  • 46) 0.003 865 088 × 2 = 0 + 0.007 730 176;
  • 47) 0.007 730 176 × 2 = 0 + 0.015 460 352;
  • 48) 0.015 460 352 × 2 = 0 + 0.030 920 704;
  • 49) 0.030 920 704 × 2 = 0 + 0.061 841 408;
  • 50) 0.061 841 408 × 2 = 0 + 0.123 682 816;
  • 51) 0.123 682 816 × 2 = 0 + 0.247 365 632;
  • 52) 0.247 365 632 × 2 = 0 + 0.494 731 264;
  • 53) 0.494 731 264 × 2 = 0 + 0.989 462 528;
  • 54) 0.989 462 528 × 2 = 1 + 0.978 925 056;
  • 55) 0.978 925 056 × 2 = 1 + 0.957 850 112;
  • 56) 0.957 850 112 × 2 = 1 + 0.915 700 224;
  • 57) 0.915 700 224 × 2 = 1 + 0.831 400 448;
  • 58) 0.831 400 448 × 2 = 1 + 0.662 800 896;
  • 59) 0.662 800 896 × 2 = 1 + 0.325 601 792;
  • 60) 0.325 601 792 × 2 = 0 + 0.651 203 584;
  • 61) 0.651 203 584 × 2 = 1 + 0.302 407 168;
  • 62) 0.302 407 168 × 2 = 0 + 0.604 814 336;
  • 63) 0.604 814 336 × 2 = 1 + 0.209 628 672;
  • 64) 0.209 628 672 × 2 = 0 + 0.419 257 344;

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:


0.000 282 059(10) =

0.0000 0000 0001 0010 0111 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010(2)

6. Positive number before normalization:

0.000 282 059(10) =

0.0000 0000 0001 0010 0111 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010(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:


0.000 282 059(10) =

0.0000 0000 0001 0010 0111 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010(2) =

0.0000 0000 0001 0010 0111 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010(2) × 20 =

1.0010 0111 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010(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 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010


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 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010 =

0010 0111 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010


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 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010


Decimal number -0.000 282 059 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 0010 1010 0010 1110 0011 1000 0000 0111 1110 1010

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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