Decimal to 64 Bit IEEE 754 Binary: Convert Number -0.033 343 988 072 6 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.033 343 988 072 6(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (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 6| = 0.033 343 988 072 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: 0.033 343 988 072 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) 0.033 343 988 072 6 × 2 = 0 + 0.066 687 976 145 2;
  • 2) 0.066 687 976 145 2 × 2 = 0 + 0.133 375 952 290 4;
  • 3) 0.133 375 952 290 4 × 2 = 0 + 0.266 751 904 580 8;
  • 4) 0.266 751 904 580 8 × 2 = 0 + 0.533 503 809 161 6;
  • 5) 0.533 503 809 161 6 × 2 = 1 + 0.067 007 618 323 2;
  • 6) 0.067 007 618 323 2 × 2 = 0 + 0.134 015 236 646 4;
  • 7) 0.134 015 236 646 4 × 2 = 0 + 0.268 030 473 292 8;
  • 8) 0.268 030 473 292 8 × 2 = 0 + 0.536 060 946 585 6;
  • 9) 0.536 060 946 585 6 × 2 = 1 + 0.072 121 893 171 2;
  • 10) 0.072 121 893 171 2 × 2 = 0 + 0.144 243 786 342 4;
  • 11) 0.144 243 786 342 4 × 2 = 0 + 0.288 487 572 684 8;
  • 12) 0.288 487 572 684 8 × 2 = 0 + 0.576 975 145 369 6;
  • 13) 0.576 975 145 369 6 × 2 = 1 + 0.153 950 290 739 2;
  • 14) 0.153 950 290 739 2 × 2 = 0 + 0.307 900 581 478 4;
  • 15) 0.307 900 581 478 4 × 2 = 0 + 0.615 801 162 956 8;
  • 16) 0.615 801 162 956 8 × 2 = 1 + 0.231 602 325 913 6;
  • 17) 0.231 602 325 913 6 × 2 = 0 + 0.463 204 651 827 2;
  • 18) 0.463 204 651 827 2 × 2 = 0 + 0.926 409 303 654 4;
  • 19) 0.926 409 303 654 4 × 2 = 1 + 0.852 818 607 308 8;
  • 20) 0.852 818 607 308 8 × 2 = 1 + 0.705 637 214 617 6;
  • 21) 0.705 637 214 617 6 × 2 = 1 + 0.411 274 429 235 2;
  • 22) 0.411 274 429 235 2 × 2 = 0 + 0.822 548 858 470 4;
  • 23) 0.822 548 858 470 4 × 2 = 1 + 0.645 097 716 940 8;
  • 24) 0.645 097 716 940 8 × 2 = 1 + 0.290 195 433 881 6;
  • 25) 0.290 195 433 881 6 × 2 = 0 + 0.580 390 867 763 2;
  • 26) 0.580 390 867 763 2 × 2 = 1 + 0.160 781 735 526 4;
  • 27) 0.160 781 735 526 4 × 2 = 0 + 0.321 563 471 052 8;
  • 28) 0.321 563 471 052 8 × 2 = 0 + 0.643 126 942 105 6;
  • 29) 0.643 126 942 105 6 × 2 = 1 + 0.286 253 884 211 2;
  • 30) 0.286 253 884 211 2 × 2 = 0 + 0.572 507 768 422 4;
  • 31) 0.572 507 768 422 4 × 2 = 1 + 0.145 015 536 844 8;
  • 32) 0.145 015 536 844 8 × 2 = 0 + 0.290 031 073 689 6;
  • 33) 0.290 031 073 689 6 × 2 = 0 + 0.580 062 147 379 2;
  • 34) 0.580 062 147 379 2 × 2 = 1 + 0.160 124 294 758 4;
  • 35) 0.160 124 294 758 4 × 2 = 0 + 0.320 248 589 516 8;
  • 36) 0.320 248 589 516 8 × 2 = 0 + 0.640 497 179 033 6;
  • 37) 0.640 497 179 033 6 × 2 = 1 + 0.280 994 358 067 2;
  • 38) 0.280 994 358 067 2 × 2 = 0 + 0.561 988 716 134 4;
  • 39) 0.561 988 716 134 4 × 2 = 1 + 0.123 977 432 268 8;
  • 40) 0.123 977 432 268 8 × 2 = 0 + 0.247 954 864 537 6;
  • 41) 0.247 954 864 537 6 × 2 = 0 + 0.495 909 729 075 2;
  • 42) 0.495 909 729 075 2 × 2 = 0 + 0.991 819 458 150 4;
  • 43) 0.991 819 458 150 4 × 2 = 1 + 0.983 638 916 300 8;
  • 44) 0.983 638 916 300 8 × 2 = 1 + 0.967 277 832 601 6;
  • 45) 0.967 277 832 601 6 × 2 = 1 + 0.934 555 665 203 2;
  • 46) 0.934 555 665 203 2 × 2 = 1 + 0.869 111 330 406 4;
  • 47) 0.869 111 330 406 4 × 2 = 1 + 0.738 222 660 812 8;
  • 48) 0.738 222 660 812 8 × 2 = 1 + 0.476 445 321 625 6;
  • 49) 0.476 445 321 625 6 × 2 = 0 + 0.952 890 643 251 2;
  • 50) 0.952 890 643 251 2 × 2 = 1 + 0.905 781 286 502 4;
  • 51) 0.905 781 286 502 4 × 2 = 1 + 0.811 562 573 004 8;
  • 52) 0.811 562 573 004 8 × 2 = 1 + 0.623 125 146 009 6;
  • 53) 0.623 125 146 009 6 × 2 = 1 + 0.246 250 292 019 2;
  • 54) 0.246 250 292 019 2 × 2 = 0 + 0.492 500 584 038 4;
  • 55) 0.492 500 584 038 4 × 2 = 0 + 0.985 001 168 076 8;
  • 56) 0.985 001 168 076 8 × 2 = 1 + 0.970 002 336 153 6;
  • 57) 0.970 002 336 153 6 × 2 = 1 + 0.940 004 672 307 2;

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 6(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0011 1111 0111 1001 1(2)

6. Positive number before normalization:

0.033 343 988 072 6(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0011 1111 0111 1001 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 6(10) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0011 1111 0111 1001 1(2) =

0.0000 1000 1000 1001 0011 1011 0100 1010 0100 1010 0011 1111 0111 1001 1(2) × 20 =

1.0001 0001 0010 0111 0110 1001 0100 1001 0100 0111 1110 1111 0011(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 0100 0111 1110 1111 0011


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 0100 0111 1110 1111 0011 =

0001 0001 0010 0111 0110 1001 0100 1001 0100 0111 1110 1111 0011


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 0100 0111 1110 1111 0011


The base ten decimal number -0.033 343 988 072 6 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0001 0001 0010 0111 0110 1001 0100 1001 0100 0111 1110 1111 0011

» Decimal number in base ten -0.033 343 988 078 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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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