-0.000 282 006 106 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 106(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
-0.000 282 006 106(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:

|-0.000 282 006 106| = 0.000 282 006 106


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: 0.000 282 006 106.

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) 0.000 282 006 106 × 2 = 0 + 0.000 564 012 212;
  • 2) 0.000 564 012 212 × 2 = 0 + 0.001 128 024 424;
  • 3) 0.001 128 024 424 × 2 = 0 + 0.002 256 048 848;
  • 4) 0.002 256 048 848 × 2 = 0 + 0.004 512 097 696;
  • 5) 0.004 512 097 696 × 2 = 0 + 0.009 024 195 392;
  • 6) 0.009 024 195 392 × 2 = 0 + 0.018 048 390 784;
  • 7) 0.018 048 390 784 × 2 = 0 + 0.036 096 781 568;
  • 8) 0.036 096 781 568 × 2 = 0 + 0.072 193 563 136;
  • 9) 0.072 193 563 136 × 2 = 0 + 0.144 387 126 272;
  • 10) 0.144 387 126 272 × 2 = 0 + 0.288 774 252 544;
  • 11) 0.288 774 252 544 × 2 = 0 + 0.577 548 505 088;
  • 12) 0.577 548 505 088 × 2 = 1 + 0.155 097 010 176;
  • 13) 0.155 097 010 176 × 2 = 0 + 0.310 194 020 352;
  • 14) 0.310 194 020 352 × 2 = 0 + 0.620 388 040 704;
  • 15) 0.620 388 040 704 × 2 = 1 + 0.240 776 081 408;
  • 16) 0.240 776 081 408 × 2 = 0 + 0.481 552 162 816;
  • 17) 0.481 552 162 816 × 2 = 0 + 0.963 104 325 632;
  • 18) 0.963 104 325 632 × 2 = 1 + 0.926 208 651 264;
  • 19) 0.926 208 651 264 × 2 = 1 + 0.852 417 302 528;
  • 20) 0.852 417 302 528 × 2 = 1 + 0.704 834 605 056;
  • 21) 0.704 834 605 056 × 2 = 1 + 0.409 669 210 112;
  • 22) 0.409 669 210 112 × 2 = 0 + 0.819 338 420 224;
  • 23) 0.819 338 420 224 × 2 = 1 + 0.638 676 840 448;
  • 24) 0.638 676 840 448 × 2 = 1 + 0.277 353 680 896;
  • 25) 0.277 353 680 896 × 2 = 0 + 0.554 707 361 792;
  • 26) 0.554 707 361 792 × 2 = 1 + 0.109 414 723 584;
  • 27) 0.109 414 723 584 × 2 = 0 + 0.218 829 447 168;
  • 28) 0.218 829 447 168 × 2 = 0 + 0.437 658 894 336;
  • 29) 0.437 658 894 336 × 2 = 0 + 0.875 317 788 672;
  • 30) 0.875 317 788 672 × 2 = 1 + 0.750 635 577 344;
  • 31) 0.750 635 577 344 × 2 = 1 + 0.501 271 154 688;
  • 32) 0.501 271 154 688 × 2 = 1 + 0.002 542 309 376;
  • 33) 0.002 542 309 376 × 2 = 0 + 0.005 084 618 752;
  • 34) 0.005 084 618 752 × 2 = 0 + 0.010 169 237 504;
  • 35) 0.010 169 237 504 × 2 = 0 + 0.020 338 475 008;
  • 36) 0.020 338 475 008 × 2 = 0 + 0.040 676 950 016;
  • 37) 0.040 676 950 016 × 2 = 0 + 0.081 353 900 032;
  • 38) 0.081 353 900 032 × 2 = 0 + 0.162 707 800 064;
  • 39) 0.162 707 800 064 × 2 = 0 + 0.325 415 600 128;
  • 40) 0.325 415 600 128 × 2 = 0 + 0.650 831 200 256;
  • 41) 0.650 831 200 256 × 2 = 1 + 0.301 662 400 512;
  • 42) 0.301 662 400 512 × 2 = 0 + 0.603 324 801 024;
  • 43) 0.603 324 801 024 × 2 = 1 + 0.206 649 602 048;
  • 44) 0.206 649 602 048 × 2 = 0 + 0.413 299 204 096;
  • 45) 0.413 299 204 096 × 2 = 0 + 0.826 598 408 192;
  • 46) 0.826 598 408 192 × 2 = 1 + 0.653 196 816 384;
  • 47) 0.653 196 816 384 × 2 = 1 + 0.306 393 632 768;
  • 48) 0.306 393 632 768 × 2 = 0 + 0.612 787 265 536;
  • 49) 0.612 787 265 536 × 2 = 1 + 0.225 574 531 072;
  • 50) 0.225 574 531 072 × 2 = 0 + 0.451 149 062 144;
  • 51) 0.451 149 062 144 × 2 = 0 + 0.902 298 124 288;
  • 52) 0.902 298 124 288 × 2 = 1 + 0.804 596 248 576;
  • 53) 0.804 596 248 576 × 2 = 1 + 0.609 192 497 152;
  • 54) 0.609 192 497 152 × 2 = 1 + 0.218 384 994 304;
  • 55) 0.218 384 994 304 × 2 = 0 + 0.436 769 988 608;
  • 56) 0.436 769 988 608 × 2 = 0 + 0.873 539 977 216;
  • 57) 0.873 539 977 216 × 2 = 1 + 0.747 079 954 432;
  • 58) 0.747 079 954 432 × 2 = 1 + 0.494 159 908 864;
  • 59) 0.494 159 908 864 × 2 = 0 + 0.988 319 817 728;
  • 60) 0.988 319 817 728 × 2 = 1 + 0.976 639 635 456;
  • 61) 0.976 639 635 456 × 2 = 1 + 0.953 279 270 912;
  • 62) 0.953 279 270 912 × 2 = 1 + 0.906 558 541 824;
  • 63) 0.906 558 541 824 × 2 = 1 + 0.813 117 083 648;
  • 64) 0.813 117 083 648 × 2 = 1 + 0.626 234 167 296;

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:


0.000 282 006 106(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0000 0000 1010 0110 1001 1100 1101 1111(2)

6. Positive number before normalization:

0.000 282 006 106(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0000 0000 1010 0110 1001 1100 1101 1111(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:


0.000 282 006 106(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0000 0000 1010 0110 1001 1100 1101 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 0000 0000 1010 0110 1001 1100 1101 1111(2) × 20 =

1.0010 0111 1011 0100 0111 0000 0000 1010 0110 1001 1100 1101 1111(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 0111 0000 0000 1010 0110 1001 1100 1101 1111


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 0111 0000 0000 1010 0110 1001 1100 1101 1111 =

0010 0111 1011 0100 0111 0000 0000 1010 0110 1001 1100 1101 1111


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 0111 0000 0000 1010 0110 1001 1100 1101 1111


Decimal number -0.000 282 006 106 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 0000 0000 1010 0110 1001 1100 1101 1111

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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