-0.000 282 005 702 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 702(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
-0.000 282 005 702(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:

|-0.000 282 005 702| = 0.000 282 005 702


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: 0.000 282 005 702.

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) 0.000 282 005 702 × 2 = 0 + 0.000 564 011 404;
  • 2) 0.000 564 011 404 × 2 = 0 + 0.001 128 022 808;
  • 3) 0.001 128 022 808 × 2 = 0 + 0.002 256 045 616;
  • 4) 0.002 256 045 616 × 2 = 0 + 0.004 512 091 232;
  • 5) 0.004 512 091 232 × 2 = 0 + 0.009 024 182 464;
  • 6) 0.009 024 182 464 × 2 = 0 + 0.018 048 364 928;
  • 7) 0.018 048 364 928 × 2 = 0 + 0.036 096 729 856;
  • 8) 0.036 096 729 856 × 2 = 0 + 0.072 193 459 712;
  • 9) 0.072 193 459 712 × 2 = 0 + 0.144 386 919 424;
  • 10) 0.144 386 919 424 × 2 = 0 + 0.288 773 838 848;
  • 11) 0.288 773 838 848 × 2 = 0 + 0.577 547 677 696;
  • 12) 0.577 547 677 696 × 2 = 1 + 0.155 095 355 392;
  • 13) 0.155 095 355 392 × 2 = 0 + 0.310 190 710 784;
  • 14) 0.310 190 710 784 × 2 = 0 + 0.620 381 421 568;
  • 15) 0.620 381 421 568 × 2 = 1 + 0.240 762 843 136;
  • 16) 0.240 762 843 136 × 2 = 0 + 0.481 525 686 272;
  • 17) 0.481 525 686 272 × 2 = 0 + 0.963 051 372 544;
  • 18) 0.963 051 372 544 × 2 = 1 + 0.926 102 745 088;
  • 19) 0.926 102 745 088 × 2 = 1 + 0.852 205 490 176;
  • 20) 0.852 205 490 176 × 2 = 1 + 0.704 410 980 352;
  • 21) 0.704 410 980 352 × 2 = 1 + 0.408 821 960 704;
  • 22) 0.408 821 960 704 × 2 = 0 + 0.817 643 921 408;
  • 23) 0.817 643 921 408 × 2 = 1 + 0.635 287 842 816;
  • 24) 0.635 287 842 816 × 2 = 1 + 0.270 575 685 632;
  • 25) 0.270 575 685 632 × 2 = 0 + 0.541 151 371 264;
  • 26) 0.541 151 371 264 × 2 = 1 + 0.082 302 742 528;
  • 27) 0.082 302 742 528 × 2 = 0 + 0.164 605 485 056;
  • 28) 0.164 605 485 056 × 2 = 0 + 0.329 210 970 112;
  • 29) 0.329 210 970 112 × 2 = 0 + 0.658 421 940 224;
  • 30) 0.658 421 940 224 × 2 = 1 + 0.316 843 880 448;
  • 31) 0.316 843 880 448 × 2 = 0 + 0.633 687 760 896;
  • 32) 0.633 687 760 896 × 2 = 1 + 0.267 375 521 792;
  • 33) 0.267 375 521 792 × 2 = 0 + 0.534 751 043 584;
  • 34) 0.534 751 043 584 × 2 = 1 + 0.069 502 087 168;
  • 35) 0.069 502 087 168 × 2 = 0 + 0.139 004 174 336;
  • 36) 0.139 004 174 336 × 2 = 0 + 0.278 008 348 672;
  • 37) 0.278 008 348 672 × 2 = 0 + 0.556 016 697 344;
  • 38) 0.556 016 697 344 × 2 = 1 + 0.112 033 394 688;
  • 39) 0.112 033 394 688 × 2 = 0 + 0.224 066 789 376;
  • 40) 0.224 066 789 376 × 2 = 0 + 0.448 133 578 752;
  • 41) 0.448 133 578 752 × 2 = 0 + 0.896 267 157 504;
  • 42) 0.896 267 157 504 × 2 = 1 + 0.792 534 315 008;
  • 43) 0.792 534 315 008 × 2 = 1 + 0.585 068 630 016;
  • 44) 0.585 068 630 016 × 2 = 1 + 0.170 137 260 032;
  • 45) 0.170 137 260 032 × 2 = 0 + 0.340 274 520 064;
  • 46) 0.340 274 520 064 × 2 = 0 + 0.680 549 040 128;
  • 47) 0.680 549 040 128 × 2 = 1 + 0.361 098 080 256;
  • 48) 0.361 098 080 256 × 2 = 0 + 0.722 196 160 512;
  • 49) 0.722 196 160 512 × 2 = 1 + 0.444 392 321 024;
  • 50) 0.444 392 321 024 × 2 = 0 + 0.888 784 642 048;
  • 51) 0.888 784 642 048 × 2 = 1 + 0.777 569 284 096;
  • 52) 0.777 569 284 096 × 2 = 1 + 0.555 138 568 192;
  • 53) 0.555 138 568 192 × 2 = 1 + 0.110 277 136 384;
  • 54) 0.110 277 136 384 × 2 = 0 + 0.220 554 272 768;
  • 55) 0.220 554 272 768 × 2 = 0 + 0.441 108 545 536;
  • 56) 0.441 108 545 536 × 2 = 0 + 0.882 217 091 072;
  • 57) 0.882 217 091 072 × 2 = 1 + 0.764 434 182 144;
  • 58) 0.764 434 182 144 × 2 = 1 + 0.528 868 364 288;
  • 59) 0.528 868 364 288 × 2 = 1 + 0.057 736 728 576;
  • 60) 0.057 736 728 576 × 2 = 0 + 0.115 473 457 152;
  • 61) 0.115 473 457 152 × 2 = 0 + 0.230 946 914 304;
  • 62) 0.230 946 914 304 × 2 = 0 + 0.461 893 828 608;
  • 63) 0.461 893 828 608 × 2 = 0 + 0.923 787 657 216;
  • 64) 0.923 787 657 216 × 2 = 1 + 0.847 575 314 432;

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:


0.000 282 005 702(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0100 0111 0010 1011 1000 1110 0001(2)

6. Positive number before normalization:

0.000 282 005 702(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0100 0111 0010 1011 1000 1110 0001(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:


0.000 282 005 702(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0100 0111 0010 1011 1000 1110 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0100 0100 0111 0010 1011 1000 1110 0001(2) × 20 =

1.0010 0111 1011 0100 0101 0100 0100 0111 0010 1011 1000 1110 0001(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 0101 0100 0100 0111 0010 1011 1000 1110 0001


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 0101 0100 0100 0111 0010 1011 1000 1110 0001 =

0010 0111 1011 0100 0101 0100 0100 0111 0010 1011 1000 1110 0001


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 0101 0100 0100 0111 0010 1011 1000 1110 0001


Decimal number -0.000 282 005 702 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0100 0100 0111 0010 1011 1000 1110 0001

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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