-0.000 282 005 801 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 801(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
-0.000 282 005 801(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:

|-0.000 282 005 801| = 0.000 282 005 801


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: 0.000 282 005 801.

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) 0.000 282 005 801 × 2 = 0 + 0.000 564 011 602;
  • 2) 0.000 564 011 602 × 2 = 0 + 0.001 128 023 204;
  • 3) 0.001 128 023 204 × 2 = 0 + 0.002 256 046 408;
  • 4) 0.002 256 046 408 × 2 = 0 + 0.004 512 092 816;
  • 5) 0.004 512 092 816 × 2 = 0 + 0.009 024 185 632;
  • 6) 0.009 024 185 632 × 2 = 0 + 0.018 048 371 264;
  • 7) 0.018 048 371 264 × 2 = 0 + 0.036 096 742 528;
  • 8) 0.036 096 742 528 × 2 = 0 + 0.072 193 485 056;
  • 9) 0.072 193 485 056 × 2 = 0 + 0.144 386 970 112;
  • 10) 0.144 386 970 112 × 2 = 0 + 0.288 773 940 224;
  • 11) 0.288 773 940 224 × 2 = 0 + 0.577 547 880 448;
  • 12) 0.577 547 880 448 × 2 = 1 + 0.155 095 760 896;
  • 13) 0.155 095 760 896 × 2 = 0 + 0.310 191 521 792;
  • 14) 0.310 191 521 792 × 2 = 0 + 0.620 383 043 584;
  • 15) 0.620 383 043 584 × 2 = 1 + 0.240 766 087 168;
  • 16) 0.240 766 087 168 × 2 = 0 + 0.481 532 174 336;
  • 17) 0.481 532 174 336 × 2 = 0 + 0.963 064 348 672;
  • 18) 0.963 064 348 672 × 2 = 1 + 0.926 128 697 344;
  • 19) 0.926 128 697 344 × 2 = 1 + 0.852 257 394 688;
  • 20) 0.852 257 394 688 × 2 = 1 + 0.704 514 789 376;
  • 21) 0.704 514 789 376 × 2 = 1 + 0.409 029 578 752;
  • 22) 0.409 029 578 752 × 2 = 0 + 0.818 059 157 504;
  • 23) 0.818 059 157 504 × 2 = 1 + 0.636 118 315 008;
  • 24) 0.636 118 315 008 × 2 = 1 + 0.272 236 630 016;
  • 25) 0.272 236 630 016 × 2 = 0 + 0.544 473 260 032;
  • 26) 0.544 473 260 032 × 2 = 1 + 0.088 946 520 064;
  • 27) 0.088 946 520 064 × 2 = 0 + 0.177 893 040 128;
  • 28) 0.177 893 040 128 × 2 = 0 + 0.355 786 080 256;
  • 29) 0.355 786 080 256 × 2 = 0 + 0.711 572 160 512;
  • 30) 0.711 572 160 512 × 2 = 1 + 0.423 144 321 024;
  • 31) 0.423 144 321 024 × 2 = 0 + 0.846 288 642 048;
  • 32) 0.846 288 642 048 × 2 = 1 + 0.692 577 284 096;
  • 33) 0.692 577 284 096 × 2 = 1 + 0.385 154 568 192;
  • 34) 0.385 154 568 192 × 2 = 0 + 0.770 309 136 384;
  • 35) 0.770 309 136 384 × 2 = 1 + 0.540 618 272 768;
  • 36) 0.540 618 272 768 × 2 = 1 + 0.081 236 545 536;
  • 37) 0.081 236 545 536 × 2 = 0 + 0.162 473 091 072;
  • 38) 0.162 473 091 072 × 2 = 0 + 0.324 946 182 144;
  • 39) 0.324 946 182 144 × 2 = 0 + 0.649 892 364 288;
  • 40) 0.649 892 364 288 × 2 = 1 + 0.299 784 728 576;
  • 41) 0.299 784 728 576 × 2 = 0 + 0.599 569 457 152;
  • 42) 0.599 569 457 152 × 2 = 1 + 0.199 138 914 304;
  • 43) 0.199 138 914 304 × 2 = 0 + 0.398 277 828 608;
  • 44) 0.398 277 828 608 × 2 = 0 + 0.796 555 657 216;
  • 45) 0.796 555 657 216 × 2 = 1 + 0.593 111 314 432;
  • 46) 0.593 111 314 432 × 2 = 1 + 0.186 222 628 864;
  • 47) 0.186 222 628 864 × 2 = 0 + 0.372 445 257 728;
  • 48) 0.372 445 257 728 × 2 = 0 + 0.744 890 515 456;
  • 49) 0.744 890 515 456 × 2 = 1 + 0.489 781 030 912;
  • 50) 0.489 781 030 912 × 2 = 0 + 0.979 562 061 824;
  • 51) 0.979 562 061 824 × 2 = 1 + 0.959 124 123 648;
  • 52) 0.959 124 123 648 × 2 = 1 + 0.918 248 247 296;
  • 53) 0.918 248 247 296 × 2 = 1 + 0.836 496 494 592;
  • 54) 0.836 496 494 592 × 2 = 1 + 0.672 992 989 184;
  • 55) 0.672 992 989 184 × 2 = 1 + 0.345 985 978 368;
  • 56) 0.345 985 978 368 × 2 = 0 + 0.691 971 956 736;
  • 57) 0.691 971 956 736 × 2 = 1 + 0.383 943 913 472;
  • 58) 0.383 943 913 472 × 2 = 0 + 0.767 887 826 944;
  • 59) 0.767 887 826 944 × 2 = 1 + 0.535 775 653 888;
  • 60) 0.535 775 653 888 × 2 = 1 + 0.071 551 307 776;
  • 61) 0.071 551 307 776 × 2 = 0 + 0.143 102 615 552;
  • 62) 0.143 102 615 552 × 2 = 0 + 0.286 205 231 104;
  • 63) 0.286 205 231 104 × 2 = 0 + 0.572 410 462 208;
  • 64) 0.572 410 462 208 × 2 = 1 + 0.144 820 924 416;

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:


0.000 282 005 801(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0001 0100 1100 1011 1110 1011 0001(2)

6. Positive number before normalization:

0.000 282 005 801(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0001 0100 1100 1011 1110 1011 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:


0.000 282 005 801(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0001 0100 1100 1011 1110 1011 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1011 0001 0100 1100 1011 1110 1011 0001(2) × 20 =

1.0010 0111 1011 0100 0101 1011 0001 0100 1100 1011 1110 1011 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 1011 0001 0100 1100 1011 1110 1011 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 1011 0001 0100 1100 1011 1110 1011 0001 =

0010 0111 1011 0100 0101 1011 0001 0100 1100 1011 1110 1011 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 1011 0001 0100 1100 1011 1110 1011 0001


Decimal number -0.000 282 005 801 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1011 0001 0100 1100 1011 1110 1011 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