-0.000 282 006 369 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 369(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
-0.000 282 006 369(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:

|-0.000 282 006 369| = 0.000 282 006 369


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: 0.000 282 006 369.

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) 0.000 282 006 369 × 2 = 0 + 0.000 564 012 738;
  • 2) 0.000 564 012 738 × 2 = 0 + 0.001 128 025 476;
  • 3) 0.001 128 025 476 × 2 = 0 + 0.002 256 050 952;
  • 4) 0.002 256 050 952 × 2 = 0 + 0.004 512 101 904;
  • 5) 0.004 512 101 904 × 2 = 0 + 0.009 024 203 808;
  • 6) 0.009 024 203 808 × 2 = 0 + 0.018 048 407 616;
  • 7) 0.018 048 407 616 × 2 = 0 + 0.036 096 815 232;
  • 8) 0.036 096 815 232 × 2 = 0 + 0.072 193 630 464;
  • 9) 0.072 193 630 464 × 2 = 0 + 0.144 387 260 928;
  • 10) 0.144 387 260 928 × 2 = 0 + 0.288 774 521 856;
  • 11) 0.288 774 521 856 × 2 = 0 + 0.577 549 043 712;
  • 12) 0.577 549 043 712 × 2 = 1 + 0.155 098 087 424;
  • 13) 0.155 098 087 424 × 2 = 0 + 0.310 196 174 848;
  • 14) 0.310 196 174 848 × 2 = 0 + 0.620 392 349 696;
  • 15) 0.620 392 349 696 × 2 = 1 + 0.240 784 699 392;
  • 16) 0.240 784 699 392 × 2 = 0 + 0.481 569 398 784;
  • 17) 0.481 569 398 784 × 2 = 0 + 0.963 138 797 568;
  • 18) 0.963 138 797 568 × 2 = 1 + 0.926 277 595 136;
  • 19) 0.926 277 595 136 × 2 = 1 + 0.852 555 190 272;
  • 20) 0.852 555 190 272 × 2 = 1 + 0.705 110 380 544;
  • 21) 0.705 110 380 544 × 2 = 1 + 0.410 220 761 088;
  • 22) 0.410 220 761 088 × 2 = 0 + 0.820 441 522 176;
  • 23) 0.820 441 522 176 × 2 = 1 + 0.640 883 044 352;
  • 24) 0.640 883 044 352 × 2 = 1 + 0.281 766 088 704;
  • 25) 0.281 766 088 704 × 2 = 0 + 0.563 532 177 408;
  • 26) 0.563 532 177 408 × 2 = 1 + 0.127 064 354 816;
  • 27) 0.127 064 354 816 × 2 = 0 + 0.254 128 709 632;
  • 28) 0.254 128 709 632 × 2 = 0 + 0.508 257 419 264;
  • 29) 0.508 257 419 264 × 2 = 1 + 0.016 514 838 528;
  • 30) 0.016 514 838 528 × 2 = 0 + 0.033 029 677 056;
  • 31) 0.033 029 677 056 × 2 = 0 + 0.066 059 354 112;
  • 32) 0.066 059 354 112 × 2 = 0 + 0.132 118 708 224;
  • 33) 0.132 118 708 224 × 2 = 0 + 0.264 237 416 448;
  • 34) 0.264 237 416 448 × 2 = 0 + 0.528 474 832 896;
  • 35) 0.528 474 832 896 × 2 = 1 + 0.056 949 665 792;
  • 36) 0.056 949 665 792 × 2 = 0 + 0.113 899 331 584;
  • 37) 0.113 899 331 584 × 2 = 0 + 0.227 798 663 168;
  • 38) 0.227 798 663 168 × 2 = 0 + 0.455 597 326 336;
  • 39) 0.455 597 326 336 × 2 = 0 + 0.911 194 652 672;
  • 40) 0.911 194 652 672 × 2 = 1 + 0.822 389 305 344;
  • 41) 0.822 389 305 344 × 2 = 1 + 0.644 778 610 688;
  • 42) 0.644 778 610 688 × 2 = 1 + 0.289 557 221 376;
  • 43) 0.289 557 221 376 × 2 = 0 + 0.579 114 442 752;
  • 44) 0.579 114 442 752 × 2 = 1 + 0.158 228 885 504;
  • 45) 0.158 228 885 504 × 2 = 0 + 0.316 457 771 008;
  • 46) 0.316 457 771 008 × 2 = 0 + 0.632 915 542 016;
  • 47) 0.632 915 542 016 × 2 = 1 + 0.265 831 084 032;
  • 48) 0.265 831 084 032 × 2 = 0 + 0.531 662 168 064;
  • 49) 0.531 662 168 064 × 2 = 1 + 0.063 324 336 128;
  • 50) 0.063 324 336 128 × 2 = 0 + 0.126 648 672 256;
  • 51) 0.126 648 672 256 × 2 = 0 + 0.253 297 344 512;
  • 52) 0.253 297 344 512 × 2 = 0 + 0.506 594 689 024;
  • 53) 0.506 594 689 024 × 2 = 1 + 0.013 189 378 048;
  • 54) 0.013 189 378 048 × 2 = 0 + 0.026 378 756 096;
  • 55) 0.026 378 756 096 × 2 = 0 + 0.052 757 512 192;
  • 56) 0.052 757 512 192 × 2 = 0 + 0.105 515 024 384;
  • 57) 0.105 515 024 384 × 2 = 0 + 0.211 030 048 768;
  • 58) 0.211 030 048 768 × 2 = 0 + 0.422 060 097 536;
  • 59) 0.422 060 097 536 × 2 = 0 + 0.844 120 195 072;
  • 60) 0.844 120 195 072 × 2 = 1 + 0.688 240 390 144;
  • 61) 0.688 240 390 144 × 2 = 1 + 0.376 480 780 288;
  • 62) 0.376 480 780 288 × 2 = 0 + 0.752 961 560 576;
  • 63) 0.752 961 560 576 × 2 = 1 + 0.505 923 121 152;
  • 64) 0.505 923 121 152 × 2 = 1 + 0.011 846 242 304;

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:


0.000 282 006 369(10) =

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

6. Positive number before normalization:

0.000 282 006 369(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0010 0001 1101 0010 1000 1000 0001 1011(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:


0.000 282 006 369(10) =

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

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

1.0010 0111 1011 0100 1000 0010 0001 1101 0010 1000 1000 0001 1011(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 0100 1000 0010 0001 1101 0010 1000 1000 0001 1011


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 0100 1000 0010 0001 1101 0010 1000 1000 0001 1011 =

0010 0111 1011 0100 1000 0010 0001 1101 0010 1000 1000 0001 1011


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 0100 1000 0010 0001 1101 0010 1000 1000 0001 1011


Decimal number -0.000 282 006 369 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0010 0001 1101 0010 1000 1000 0001 1011

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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