-0.000 282 409 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 409(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
-0.000 282 409(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:

|-0.000 282 409| = 0.000 282 409


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: 0.000 282 409.

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) 0.000 282 409 × 2 = 0 + 0.000 564 818;
  • 2) 0.000 564 818 × 2 = 0 + 0.001 129 636;
  • 3) 0.001 129 636 × 2 = 0 + 0.002 259 272;
  • 4) 0.002 259 272 × 2 = 0 + 0.004 518 544;
  • 5) 0.004 518 544 × 2 = 0 + 0.009 037 088;
  • 6) 0.009 037 088 × 2 = 0 + 0.018 074 176;
  • 7) 0.018 074 176 × 2 = 0 + 0.036 148 352;
  • 8) 0.036 148 352 × 2 = 0 + 0.072 296 704;
  • 9) 0.072 296 704 × 2 = 0 + 0.144 593 408;
  • 10) 0.144 593 408 × 2 = 0 + 0.289 186 816;
  • 11) 0.289 186 816 × 2 = 0 + 0.578 373 632;
  • 12) 0.578 373 632 × 2 = 1 + 0.156 747 264;
  • 13) 0.156 747 264 × 2 = 0 + 0.313 494 528;
  • 14) 0.313 494 528 × 2 = 0 + 0.626 989 056;
  • 15) 0.626 989 056 × 2 = 1 + 0.253 978 112;
  • 16) 0.253 978 112 × 2 = 0 + 0.507 956 224;
  • 17) 0.507 956 224 × 2 = 1 + 0.015 912 448;
  • 18) 0.015 912 448 × 2 = 0 + 0.031 824 896;
  • 19) 0.031 824 896 × 2 = 0 + 0.063 649 792;
  • 20) 0.063 649 792 × 2 = 0 + 0.127 299 584;
  • 21) 0.127 299 584 × 2 = 0 + 0.254 599 168;
  • 22) 0.254 599 168 × 2 = 0 + 0.509 198 336;
  • 23) 0.509 198 336 × 2 = 1 + 0.018 396 672;
  • 24) 0.018 396 672 × 2 = 0 + 0.036 793 344;
  • 25) 0.036 793 344 × 2 = 0 + 0.073 586 688;
  • 26) 0.073 586 688 × 2 = 0 + 0.147 173 376;
  • 27) 0.147 173 376 × 2 = 0 + 0.294 346 752;
  • 28) 0.294 346 752 × 2 = 0 + 0.588 693 504;
  • 29) 0.588 693 504 × 2 = 1 + 0.177 387 008;
  • 30) 0.177 387 008 × 2 = 0 + 0.354 774 016;
  • 31) 0.354 774 016 × 2 = 0 + 0.709 548 032;
  • 32) 0.709 548 032 × 2 = 1 + 0.419 096 064;
  • 33) 0.419 096 064 × 2 = 0 + 0.838 192 128;
  • 34) 0.838 192 128 × 2 = 1 + 0.676 384 256;
  • 35) 0.676 384 256 × 2 = 1 + 0.352 768 512;
  • 36) 0.352 768 512 × 2 = 0 + 0.705 537 024;
  • 37) 0.705 537 024 × 2 = 1 + 0.411 074 048;
  • 38) 0.411 074 048 × 2 = 0 + 0.822 148 096;
  • 39) 0.822 148 096 × 2 = 1 + 0.644 296 192;
  • 40) 0.644 296 192 × 2 = 1 + 0.288 592 384;
  • 41) 0.288 592 384 × 2 = 0 + 0.577 184 768;
  • 42) 0.577 184 768 × 2 = 1 + 0.154 369 536;
  • 43) 0.154 369 536 × 2 = 0 + 0.308 739 072;
  • 44) 0.308 739 072 × 2 = 0 + 0.617 478 144;
  • 45) 0.617 478 144 × 2 = 1 + 0.234 956 288;
  • 46) 0.234 956 288 × 2 = 0 + 0.469 912 576;
  • 47) 0.469 912 576 × 2 = 0 + 0.939 825 152;
  • 48) 0.939 825 152 × 2 = 1 + 0.879 650 304;
  • 49) 0.879 650 304 × 2 = 1 + 0.759 300 608;
  • 50) 0.759 300 608 × 2 = 1 + 0.518 601 216;
  • 51) 0.518 601 216 × 2 = 1 + 0.037 202 432;
  • 52) 0.037 202 432 × 2 = 0 + 0.074 404 864;
  • 53) 0.074 404 864 × 2 = 0 + 0.148 809 728;
  • 54) 0.148 809 728 × 2 = 0 + 0.297 619 456;
  • 55) 0.297 619 456 × 2 = 0 + 0.595 238 912;
  • 56) 0.595 238 912 × 2 = 1 + 0.190 477 824;
  • 57) 0.190 477 824 × 2 = 0 + 0.380 955 648;
  • 58) 0.380 955 648 × 2 = 0 + 0.761 911 296;
  • 59) 0.761 911 296 × 2 = 1 + 0.523 822 592;
  • 60) 0.523 822 592 × 2 = 1 + 0.047 645 184;
  • 61) 0.047 645 184 × 2 = 0 + 0.095 290 368;
  • 62) 0.095 290 368 × 2 = 0 + 0.190 580 736;
  • 63) 0.190 580 736 × 2 = 0 + 0.381 161 472;
  • 64) 0.381 161 472 × 2 = 0 + 0.762 322 944;

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:


0.000 282 409(10) =

0.0000 0000 0001 0010 1000 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000(2)

6. Positive number before normalization:

0.000 282 409(10) =

0.0000 0000 0001 0010 1000 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000(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:


0.000 282 409(10) =

0.0000 0000 0001 0010 1000 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000(2) =

0.0000 0000 0001 0010 1000 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000(2) × 20 =

1.0010 1000 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000(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 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000


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 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000 =

0010 1000 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000


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 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000


Decimal number -0.000 282 409 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0010 0000 1001 0110 1011 0100 1001 1110 0001 0011 0000

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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