-0.033 343 988 072 21 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.033 343 988 072 21(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
-0.033 343 988 072 21(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:

|-0.033 343 988 072 21| = 0.033 343 988 072 21


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: 0.033 343 988 072 21.

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) 0.033 343 988 072 21 × 2 = 0 + 0.066 687 976 144 42;
  • 2) 0.066 687 976 144 42 × 2 = 0 + 0.133 375 952 288 84;
  • 3) 0.133 375 952 288 84 × 2 = 0 + 0.266 751 904 577 68;
  • 4) 0.266 751 904 577 68 × 2 = 0 + 0.533 503 809 155 36;
  • 5) 0.533 503 809 155 36 × 2 = 1 + 0.067 007 618 310 72;
  • 6) 0.067 007 618 310 72 × 2 = 0 + 0.134 015 236 621 44;
  • 7) 0.134 015 236 621 44 × 2 = 0 + 0.268 030 473 242 88;
  • 8) 0.268 030 473 242 88 × 2 = 0 + 0.536 060 946 485 76;
  • 9) 0.536 060 946 485 76 × 2 = 1 + 0.072 121 892 971 52;
  • 10) 0.072 121 892 971 52 × 2 = 0 + 0.144 243 785 943 04;
  • 11) 0.144 243 785 943 04 × 2 = 0 + 0.288 487 571 886 08;
  • 12) 0.288 487 571 886 08 × 2 = 0 + 0.576 975 143 772 16;
  • 13) 0.576 975 143 772 16 × 2 = 1 + 0.153 950 287 544 32;
  • 14) 0.153 950 287 544 32 × 2 = 0 + 0.307 900 575 088 64;
  • 15) 0.307 900 575 088 64 × 2 = 0 + 0.615 801 150 177 28;
  • 16) 0.615 801 150 177 28 × 2 = 1 + 0.231 602 300 354 56;
  • 17) 0.231 602 300 354 56 × 2 = 0 + 0.463 204 600 709 12;
  • 18) 0.463 204 600 709 12 × 2 = 0 + 0.926 409 201 418 24;
  • 19) 0.926 409 201 418 24 × 2 = 1 + 0.852 818 402 836 48;
  • 20) 0.852 818 402 836 48 × 2 = 1 + 0.705 636 805 672 96;
  • 21) 0.705 636 805 672 96 × 2 = 1 + 0.411 273 611 345 92;
  • 22) 0.411 273 611 345 92 × 2 = 0 + 0.822 547 222 691 84;
  • 23) 0.822 547 222 691 84 × 2 = 1 + 0.645 094 445 383 68;
  • 24) 0.645 094 445 383 68 × 2 = 1 + 0.290 188 890 767 36;
  • 25) 0.290 188 890 767 36 × 2 = 0 + 0.580 377 781 534 72;
  • 26) 0.580 377 781 534 72 × 2 = 1 + 0.160 755 563 069 44;
  • 27) 0.160 755 563 069 44 × 2 = 0 + 0.321 511 126 138 88;
  • 28) 0.321 511 126 138 88 × 2 = 0 + 0.643 022 252 277 76;
  • 29) 0.643 022 252 277 76 × 2 = 1 + 0.286 044 504 555 52;
  • 30) 0.286 044 504 555 52 × 2 = 0 + 0.572 089 009 111 04;
  • 31) 0.572 089 009 111 04 × 2 = 1 + 0.144 178 018 222 08;
  • 32) 0.144 178 018 222 08 × 2 = 0 + 0.288 356 036 444 16;
  • 33) 0.288 356 036 444 16 × 2 = 0 + 0.576 712 072 888 32;
  • 34) 0.576 712 072 888 32 × 2 = 1 + 0.153 424 145 776 64;
  • 35) 0.153 424 145 776 64 × 2 = 0 + 0.306 848 291 553 28;
  • 36) 0.306 848 291 553 28 × 2 = 0 + 0.613 696 583 106 56;
  • 37) 0.613 696 583 106 56 × 2 = 1 + 0.227 393 166 213 12;
  • 38) 0.227 393 166 213 12 × 2 = 0 + 0.454 786 332 426 24;
  • 39) 0.454 786 332 426 24 × 2 = 0 + 0.909 572 664 852 48;
  • 40) 0.909 572 664 852 48 × 2 = 1 + 0.819 145 329 704 96;
  • 41) 0.819 145 329 704 96 × 2 = 1 + 0.638 290 659 409 92;
  • 42) 0.638 290 659 409 92 × 2 = 1 + 0.276 581 318 819 84;
  • 43) 0.276 581 318 819 84 × 2 = 0 + 0.553 162 637 639 68;
  • 44) 0.553 162 637 639 68 × 2 = 1 + 0.106 325 275 279 36;
  • 45) 0.106 325 275 279 36 × 2 = 0 + 0.212 650 550 558 72;
  • 46) 0.212 650 550 558 72 × 2 = 0 + 0.425 301 101 117 44;
  • 47) 0.425 301 101 117 44 × 2 = 0 + 0.850 602 202 234 88;
  • 48) 0.850 602 202 234 88 × 2 = 1 + 0.701 204 404 469 76;
  • 49) 0.701 204 404 469 76 × 2 = 1 + 0.402 408 808 939 52;
  • 50) 0.402 408 808 939 52 × 2 = 0 + 0.804 817 617 879 04;
  • 51) 0.804 817 617 879 04 × 2 = 1 + 0.609 635 235 758 08;
  • 52) 0.609 635 235 758 08 × 2 = 1 + 0.219 270 471 516 16;
  • 53) 0.219 270 471 516 16 × 2 = 0 + 0.438 540 943 032 32;
  • 54) 0.438 540 943 032 32 × 2 = 0 + 0.877 081 886 064 64;
  • 55) 0.877 081 886 064 64 × 2 = 1 + 0.754 163 772 129 28;
  • 56) 0.754 163 772 129 28 × 2 = 1 + 0.508 327 544 258 56;
  • 57) 0.508 327 544 258 56 × 2 = 1 + 0.016 655 088 517 12;

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:


0.033 343 988 072 21(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 1101 0001 1011 0011 1(2)

6. Positive number before normalization:

0.033 343 988 072 21(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 1101 0001 1011 0011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.033 343 988 072 21(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 1101 0001 1011 0011 1(2) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1001 1101 0001 1011 0011 1(2) × 20 =

1.0001 0001 0010 0111 0110 1001 0100 1001 0011 1010 0011 0110 0111(2) × 2-5


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): -5

Mantissa (not normalized):
1.0001 0001 0010 0111 0110 1001 0100 1001 0011 1010 0011 0110 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 0001 0001 0010 0111 0110 1001 0100 1001 0011 1010 0011 0110 0111 =

0001 0001 0010 0111 0110 1001 0100 1001 0011 1010 0011 0110 0111


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 1010

Mantissa (52 bits) =
0001 0001 0010 0111 0110 1001 0100 1001 0011 1010 0011 0110 0111


Decimal number -0.033 343 988 072 21 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0001 0001 0010 0111 0110 1001 0100 1001 0011 1010 0011 0110 0111

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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