-0.000 282 006 462 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 462(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
-0.000 282 006 462(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:

|-0.000 282 006 462| = 0.000 282 006 462


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: 0.000 282 006 462.

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) 0.000 282 006 462 × 2 = 0 + 0.000 564 012 924;
  • 2) 0.000 564 012 924 × 2 = 0 + 0.001 128 025 848;
  • 3) 0.001 128 025 848 × 2 = 0 + 0.002 256 051 696;
  • 4) 0.002 256 051 696 × 2 = 0 + 0.004 512 103 392;
  • 5) 0.004 512 103 392 × 2 = 0 + 0.009 024 206 784;
  • 6) 0.009 024 206 784 × 2 = 0 + 0.018 048 413 568;
  • 7) 0.018 048 413 568 × 2 = 0 + 0.036 096 827 136;
  • 8) 0.036 096 827 136 × 2 = 0 + 0.072 193 654 272;
  • 9) 0.072 193 654 272 × 2 = 0 + 0.144 387 308 544;
  • 10) 0.144 387 308 544 × 2 = 0 + 0.288 774 617 088;
  • 11) 0.288 774 617 088 × 2 = 0 + 0.577 549 234 176;
  • 12) 0.577 549 234 176 × 2 = 1 + 0.155 098 468 352;
  • 13) 0.155 098 468 352 × 2 = 0 + 0.310 196 936 704;
  • 14) 0.310 196 936 704 × 2 = 0 + 0.620 393 873 408;
  • 15) 0.620 393 873 408 × 2 = 1 + 0.240 787 746 816;
  • 16) 0.240 787 746 816 × 2 = 0 + 0.481 575 493 632;
  • 17) 0.481 575 493 632 × 2 = 0 + 0.963 150 987 264;
  • 18) 0.963 150 987 264 × 2 = 1 + 0.926 301 974 528;
  • 19) 0.926 301 974 528 × 2 = 1 + 0.852 603 949 056;
  • 20) 0.852 603 949 056 × 2 = 1 + 0.705 207 898 112;
  • 21) 0.705 207 898 112 × 2 = 1 + 0.410 415 796 224;
  • 22) 0.410 415 796 224 × 2 = 0 + 0.820 831 592 448;
  • 23) 0.820 831 592 448 × 2 = 1 + 0.641 663 184 896;
  • 24) 0.641 663 184 896 × 2 = 1 + 0.283 326 369 792;
  • 25) 0.283 326 369 792 × 2 = 0 + 0.566 652 739 584;
  • 26) 0.566 652 739 584 × 2 = 1 + 0.133 305 479 168;
  • 27) 0.133 305 479 168 × 2 = 0 + 0.266 610 958 336;
  • 28) 0.266 610 958 336 × 2 = 0 + 0.533 221 916 672;
  • 29) 0.533 221 916 672 × 2 = 1 + 0.066 443 833 344;
  • 30) 0.066 443 833 344 × 2 = 0 + 0.132 887 666 688;
  • 31) 0.132 887 666 688 × 2 = 0 + 0.265 775 333 376;
  • 32) 0.265 775 333 376 × 2 = 0 + 0.531 550 666 752;
  • 33) 0.531 550 666 752 × 2 = 1 + 0.063 101 333 504;
  • 34) 0.063 101 333 504 × 2 = 0 + 0.126 202 667 008;
  • 35) 0.126 202 667 008 × 2 = 0 + 0.252 405 334 016;
  • 36) 0.252 405 334 016 × 2 = 0 + 0.504 810 668 032;
  • 37) 0.504 810 668 032 × 2 = 1 + 0.009 621 336 064;
  • 38) 0.009 621 336 064 × 2 = 0 + 0.019 242 672 128;
  • 39) 0.019 242 672 128 × 2 = 0 + 0.038 485 344 256;
  • 40) 0.038 485 344 256 × 2 = 0 + 0.076 970 688 512;
  • 41) 0.076 970 688 512 × 2 = 0 + 0.153 941 377 024;
  • 42) 0.153 941 377 024 × 2 = 0 + 0.307 882 754 048;
  • 43) 0.307 882 754 048 × 2 = 0 + 0.615 765 508 096;
  • 44) 0.615 765 508 096 × 2 = 1 + 0.231 531 016 192;
  • 45) 0.231 531 016 192 × 2 = 0 + 0.463 062 032 384;
  • 46) 0.463 062 032 384 × 2 = 0 + 0.926 124 064 768;
  • 47) 0.926 124 064 768 × 2 = 1 + 0.852 248 129 536;
  • 48) 0.852 248 129 536 × 2 = 1 + 0.704 496 259 072;
  • 49) 0.704 496 259 072 × 2 = 1 + 0.408 992 518 144;
  • 50) 0.408 992 518 144 × 2 = 0 + 0.817 985 036 288;
  • 51) 0.817 985 036 288 × 2 = 1 + 0.635 970 072 576;
  • 52) 0.635 970 072 576 × 2 = 1 + 0.271 940 145 152;
  • 53) 0.271 940 145 152 × 2 = 0 + 0.543 880 290 304;
  • 54) 0.543 880 290 304 × 2 = 1 + 0.087 760 580 608;
  • 55) 0.087 760 580 608 × 2 = 0 + 0.175 521 161 216;
  • 56) 0.175 521 161 216 × 2 = 0 + 0.351 042 322 432;
  • 57) 0.351 042 322 432 × 2 = 0 + 0.702 084 644 864;
  • 58) 0.702 084 644 864 × 2 = 1 + 0.404 169 289 728;
  • 59) 0.404 169 289 728 × 2 = 0 + 0.808 338 579 456;
  • 60) 0.808 338 579 456 × 2 = 1 + 0.616 677 158 912;
  • 61) 0.616 677 158 912 × 2 = 1 + 0.233 354 317 824;
  • 62) 0.233 354 317 824 × 2 = 0 + 0.466 708 635 648;
  • 63) 0.466 708 635 648 × 2 = 0 + 0.933 417 271 296;
  • 64) 0.933 417 271 296 × 2 = 1 + 0.866 834 542 592;

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:


0.000 282 006 462(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1000 0001 0011 1011 0100 0101 1001(2)

6. Positive number before normalization:

0.000 282 006 462(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1000 0001 0011 1011 0100 0101 1001(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:


0.000 282 006 462(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1000 0001 0011 1011 0100 0101 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 1000 0001 0011 1011 0100 0101 1001(2) × 20 =

1.0010 0111 1011 0100 1000 1000 1000 0001 0011 1011 0100 0101 1001(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 1000 1000 0001 0011 1011 0100 0101 1001


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 1000 1000 0001 0011 1011 0100 0101 1001 =

0010 0111 1011 0100 1000 1000 1000 0001 0011 1011 0100 0101 1001


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 1000 1000 0001 0011 1011 0100 0101 1001


Decimal number -0.000 282 006 462 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 1000 1000 0001 0011 1011 0100 0101 1001

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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