-0.000 282 005 920 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 920 6(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
-0.000 282 005 920 6(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:

|-0.000 282 005 920 6| = 0.000 282 005 920 6


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: 0.000 282 005 920 6.

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) 0.000 282 005 920 6 × 2 = 0 + 0.000 564 011 841 2;
  • 2) 0.000 564 011 841 2 × 2 = 0 + 0.001 128 023 682 4;
  • 3) 0.001 128 023 682 4 × 2 = 0 + 0.002 256 047 364 8;
  • 4) 0.002 256 047 364 8 × 2 = 0 + 0.004 512 094 729 6;
  • 5) 0.004 512 094 729 6 × 2 = 0 + 0.009 024 189 459 2;
  • 6) 0.009 024 189 459 2 × 2 = 0 + 0.018 048 378 918 4;
  • 7) 0.018 048 378 918 4 × 2 = 0 + 0.036 096 757 836 8;
  • 8) 0.036 096 757 836 8 × 2 = 0 + 0.072 193 515 673 6;
  • 9) 0.072 193 515 673 6 × 2 = 0 + 0.144 387 031 347 2;
  • 10) 0.144 387 031 347 2 × 2 = 0 + 0.288 774 062 694 4;
  • 11) 0.288 774 062 694 4 × 2 = 0 + 0.577 548 125 388 8;
  • 12) 0.577 548 125 388 8 × 2 = 1 + 0.155 096 250 777 6;
  • 13) 0.155 096 250 777 6 × 2 = 0 + 0.310 192 501 555 2;
  • 14) 0.310 192 501 555 2 × 2 = 0 + 0.620 385 003 110 4;
  • 15) 0.620 385 003 110 4 × 2 = 1 + 0.240 770 006 220 8;
  • 16) 0.240 770 006 220 8 × 2 = 0 + 0.481 540 012 441 6;
  • 17) 0.481 540 012 441 6 × 2 = 0 + 0.963 080 024 883 2;
  • 18) 0.963 080 024 883 2 × 2 = 1 + 0.926 160 049 766 4;
  • 19) 0.926 160 049 766 4 × 2 = 1 + 0.852 320 099 532 8;
  • 20) 0.852 320 099 532 8 × 2 = 1 + 0.704 640 199 065 6;
  • 21) 0.704 640 199 065 6 × 2 = 1 + 0.409 280 398 131 2;
  • 22) 0.409 280 398 131 2 × 2 = 0 + 0.818 560 796 262 4;
  • 23) 0.818 560 796 262 4 × 2 = 1 + 0.637 121 592 524 8;
  • 24) 0.637 121 592 524 8 × 2 = 1 + 0.274 243 185 049 6;
  • 25) 0.274 243 185 049 6 × 2 = 0 + 0.548 486 370 099 2;
  • 26) 0.548 486 370 099 2 × 2 = 1 + 0.096 972 740 198 4;
  • 27) 0.096 972 740 198 4 × 2 = 0 + 0.193 945 480 396 8;
  • 28) 0.193 945 480 396 8 × 2 = 0 + 0.387 890 960 793 6;
  • 29) 0.387 890 960 793 6 × 2 = 0 + 0.775 781 921 587 2;
  • 30) 0.775 781 921 587 2 × 2 = 1 + 0.551 563 843 174 4;
  • 31) 0.551 563 843 174 4 × 2 = 1 + 0.103 127 686 348 8;
  • 32) 0.103 127 686 348 8 × 2 = 0 + 0.206 255 372 697 6;
  • 33) 0.206 255 372 697 6 × 2 = 0 + 0.412 510 745 395 2;
  • 34) 0.412 510 745 395 2 × 2 = 0 + 0.825 021 490 790 4;
  • 35) 0.825 021 490 790 4 × 2 = 1 + 0.650 042 981 580 8;
  • 36) 0.650 042 981 580 8 × 2 = 1 + 0.300 085 963 161 6;
  • 37) 0.300 085 963 161 6 × 2 = 0 + 0.600 171 926 323 2;
  • 38) 0.600 171 926 323 2 × 2 = 1 + 0.200 343 852 646 4;
  • 39) 0.200 343 852 646 4 × 2 = 0 + 0.400 687 705 292 8;
  • 40) 0.400 687 705 292 8 × 2 = 0 + 0.801 375 410 585 6;
  • 41) 0.801 375 410 585 6 × 2 = 1 + 0.602 750 821 171 2;
  • 42) 0.602 750 821 171 2 × 2 = 1 + 0.205 501 642 342 4;
  • 43) 0.205 501 642 342 4 × 2 = 0 + 0.411 003 284 684 8;
  • 44) 0.411 003 284 684 8 × 2 = 0 + 0.822 006 569 369 6;
  • 45) 0.822 006 569 369 6 × 2 = 1 + 0.644 013 138 739 2;
  • 46) 0.644 013 138 739 2 × 2 = 1 + 0.288 026 277 478 4;
  • 47) 0.288 026 277 478 4 × 2 = 0 + 0.576 052 554 956 8;
  • 48) 0.576 052 554 956 8 × 2 = 1 + 0.152 105 109 913 6;
  • 49) 0.152 105 109 913 6 × 2 = 0 + 0.304 210 219 827 2;
  • 50) 0.304 210 219 827 2 × 2 = 0 + 0.608 420 439 654 4;
  • 51) 0.608 420 439 654 4 × 2 = 1 + 0.216 840 879 308 8;
  • 52) 0.216 840 879 308 8 × 2 = 0 + 0.433 681 758 617 6;
  • 53) 0.433 681 758 617 6 × 2 = 0 + 0.867 363 517 235 2;
  • 54) 0.867 363 517 235 2 × 2 = 1 + 0.734 727 034 470 4;
  • 55) 0.734 727 034 470 4 × 2 = 1 + 0.469 454 068 940 8;
  • 56) 0.469 454 068 940 8 × 2 = 0 + 0.938 908 137 881 6;
  • 57) 0.938 908 137 881 6 × 2 = 1 + 0.877 816 275 763 2;
  • 58) 0.877 816 275 763 2 × 2 = 1 + 0.755 632 551 526 4;
  • 59) 0.755 632 551 526 4 × 2 = 1 + 0.511 265 103 052 8;
  • 60) 0.511 265 103 052 8 × 2 = 1 + 0.022 530 206 105 6;
  • 61) 0.022 530 206 105 6 × 2 = 0 + 0.045 060 412 211 2;
  • 62) 0.045 060 412 211 2 × 2 = 0 + 0.090 120 824 422 4;
  • 63) 0.090 120 824 422 4 × 2 = 0 + 0.180 241 648 844 8;
  • 64) 0.180 241 648 844 8 × 2 = 0 + 0.360 483 297 689 6;

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:


0.000 282 005 920 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0100 1100 1101 0010 0110 1111 0000(2)

6. Positive number before normalization:

0.000 282 005 920 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0100 1100 1101 0010 0110 1111 0000(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:


0.000 282 005 920 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0100 1100 1101 0010 0110 1111 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0100 1100 1101 0010 0110 1111 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0100 1100 1101 0010 0110 1111 0000(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 0110 0011 0100 1100 1101 0010 0110 1111 0000


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 0110 0011 0100 1100 1101 0010 0110 1111 0000 =

0010 0111 1011 0100 0110 0011 0100 1100 1101 0010 0110 1111 0000


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 0110 0011 0100 1100 1101 0010 0110 1111 0000


Decimal number -0.000 282 005 920 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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