-0.000 282 006 459 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 459(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
-0.000 282 006 459(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:

|-0.000 282 006 459| = 0.000 282 006 459


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: 0.000 282 006 459.

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) 0.000 282 006 459 × 2 = 0 + 0.000 564 012 918;
  • 2) 0.000 564 012 918 × 2 = 0 + 0.001 128 025 836;
  • 3) 0.001 128 025 836 × 2 = 0 + 0.002 256 051 672;
  • 4) 0.002 256 051 672 × 2 = 0 + 0.004 512 103 344;
  • 5) 0.004 512 103 344 × 2 = 0 + 0.009 024 206 688;
  • 6) 0.009 024 206 688 × 2 = 0 + 0.018 048 413 376;
  • 7) 0.018 048 413 376 × 2 = 0 + 0.036 096 826 752;
  • 8) 0.036 096 826 752 × 2 = 0 + 0.072 193 653 504;
  • 9) 0.072 193 653 504 × 2 = 0 + 0.144 387 307 008;
  • 10) 0.144 387 307 008 × 2 = 0 + 0.288 774 614 016;
  • 11) 0.288 774 614 016 × 2 = 0 + 0.577 549 228 032;
  • 12) 0.577 549 228 032 × 2 = 1 + 0.155 098 456 064;
  • 13) 0.155 098 456 064 × 2 = 0 + 0.310 196 912 128;
  • 14) 0.310 196 912 128 × 2 = 0 + 0.620 393 824 256;
  • 15) 0.620 393 824 256 × 2 = 1 + 0.240 787 648 512;
  • 16) 0.240 787 648 512 × 2 = 0 + 0.481 575 297 024;
  • 17) 0.481 575 297 024 × 2 = 0 + 0.963 150 594 048;
  • 18) 0.963 150 594 048 × 2 = 1 + 0.926 301 188 096;
  • 19) 0.926 301 188 096 × 2 = 1 + 0.852 602 376 192;
  • 20) 0.852 602 376 192 × 2 = 1 + 0.705 204 752 384;
  • 21) 0.705 204 752 384 × 2 = 1 + 0.410 409 504 768;
  • 22) 0.410 409 504 768 × 2 = 0 + 0.820 819 009 536;
  • 23) 0.820 819 009 536 × 2 = 1 + 0.641 638 019 072;
  • 24) 0.641 638 019 072 × 2 = 1 + 0.283 276 038 144;
  • 25) 0.283 276 038 144 × 2 = 0 + 0.566 552 076 288;
  • 26) 0.566 552 076 288 × 2 = 1 + 0.133 104 152 576;
  • 27) 0.133 104 152 576 × 2 = 0 + 0.266 208 305 152;
  • 28) 0.266 208 305 152 × 2 = 0 + 0.532 416 610 304;
  • 29) 0.532 416 610 304 × 2 = 1 + 0.064 833 220 608;
  • 30) 0.064 833 220 608 × 2 = 0 + 0.129 666 441 216;
  • 31) 0.129 666 441 216 × 2 = 0 + 0.259 332 882 432;
  • 32) 0.259 332 882 432 × 2 = 0 + 0.518 665 764 864;
  • 33) 0.518 665 764 864 × 2 = 1 + 0.037 331 529 728;
  • 34) 0.037 331 529 728 × 2 = 0 + 0.074 663 059 456;
  • 35) 0.074 663 059 456 × 2 = 0 + 0.149 326 118 912;
  • 36) 0.149 326 118 912 × 2 = 0 + 0.298 652 237 824;
  • 37) 0.298 652 237 824 × 2 = 0 + 0.597 304 475 648;
  • 38) 0.597 304 475 648 × 2 = 1 + 0.194 608 951 296;
  • 39) 0.194 608 951 296 × 2 = 0 + 0.389 217 902 592;
  • 40) 0.389 217 902 592 × 2 = 0 + 0.778 435 805 184;
  • 41) 0.778 435 805 184 × 2 = 1 + 0.556 871 610 368;
  • 42) 0.556 871 610 368 × 2 = 1 + 0.113 743 220 736;
  • 43) 0.113 743 220 736 × 2 = 0 + 0.227 486 441 472;
  • 44) 0.227 486 441 472 × 2 = 0 + 0.454 972 882 944;
  • 45) 0.454 972 882 944 × 2 = 0 + 0.909 945 765 888;
  • 46) 0.909 945 765 888 × 2 = 1 + 0.819 891 531 776;
  • 47) 0.819 891 531 776 × 2 = 1 + 0.639 783 063 552;
  • 48) 0.639 783 063 552 × 2 = 1 + 0.279 566 127 104;
  • 49) 0.279 566 127 104 × 2 = 0 + 0.559 132 254 208;
  • 50) 0.559 132 254 208 × 2 = 1 + 0.118 264 508 416;
  • 51) 0.118 264 508 416 × 2 = 0 + 0.236 529 016 832;
  • 52) 0.236 529 016 832 × 2 = 0 + 0.473 058 033 664;
  • 53) 0.473 058 033 664 × 2 = 0 + 0.946 116 067 328;
  • 54) 0.946 116 067 328 × 2 = 1 + 0.892 232 134 656;
  • 55) 0.892 232 134 656 × 2 = 1 + 0.784 464 269 312;
  • 56) 0.784 464 269 312 × 2 = 1 + 0.568 928 538 624;
  • 57) 0.568 928 538 624 × 2 = 1 + 0.137 857 077 248;
  • 58) 0.137 857 077 248 × 2 = 0 + 0.275 714 154 496;
  • 59) 0.275 714 154 496 × 2 = 0 + 0.551 428 308 992;
  • 60) 0.551 428 308 992 × 2 = 1 + 0.102 856 617 984;
  • 61) 0.102 856 617 984 × 2 = 0 + 0.205 713 235 968;
  • 62) 0.205 713 235 968 × 2 = 0 + 0.411 426 471 936;
  • 63) 0.411 426 471 936 × 2 = 0 + 0.822 852 943 872;
  • 64) 0.822 852 943 872 × 2 = 1 + 0.645 705 887 744;

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:


0.000 282 006 459(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0100 1100 0111 0100 0111 1001 0001(2)

6. Positive number before normalization:

0.000 282 006 459(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0100 1100 0111 0100 0111 1001 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:


0.000 282 006 459(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0100 1100 0111 0100 0111 1001 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 1000 0100 1100 0111 0100 0111 1001 0001(2) × 20 =

1.0010 0111 1011 0100 1000 1000 0100 1100 0111 0100 0111 1001 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 1000 1000 0100 1100 0111 0100 0111 1001 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 1000 1000 0100 1100 0111 0100 0111 1001 0001 =

0010 0111 1011 0100 1000 1000 0100 1100 0111 0100 0111 1001 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 1000 1000 0100 1100 0111 0100 0111 1001 0001


Decimal number -0.000 282 006 459 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 1000 0100 1100 0111 0100 0111 1001 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