-0.042 419 142 298 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 298 7(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
-0.042 419 142 298 7(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:

|-0.042 419 142 298 7| = 0.042 419 142 298 7


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: 0.042 419 142 298 7.

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) 0.042 419 142 298 7 × 2 = 0 + 0.084 838 284 597 4;
  • 2) 0.084 838 284 597 4 × 2 = 0 + 0.169 676 569 194 8;
  • 3) 0.169 676 569 194 8 × 2 = 0 + 0.339 353 138 389 6;
  • 4) 0.339 353 138 389 6 × 2 = 0 + 0.678 706 276 779 2;
  • 5) 0.678 706 276 779 2 × 2 = 1 + 0.357 412 553 558 4;
  • 6) 0.357 412 553 558 4 × 2 = 0 + 0.714 825 107 116 8;
  • 7) 0.714 825 107 116 8 × 2 = 1 + 0.429 650 214 233 6;
  • 8) 0.429 650 214 233 6 × 2 = 0 + 0.859 300 428 467 2;
  • 9) 0.859 300 428 467 2 × 2 = 1 + 0.718 600 856 934 4;
  • 10) 0.718 600 856 934 4 × 2 = 1 + 0.437 201 713 868 8;
  • 11) 0.437 201 713 868 8 × 2 = 0 + 0.874 403 427 737 6;
  • 12) 0.874 403 427 737 6 × 2 = 1 + 0.748 806 855 475 2;
  • 13) 0.748 806 855 475 2 × 2 = 1 + 0.497 613 710 950 4;
  • 14) 0.497 613 710 950 4 × 2 = 0 + 0.995 227 421 900 8;
  • 15) 0.995 227 421 900 8 × 2 = 1 + 0.990 454 843 801 6;
  • 16) 0.990 454 843 801 6 × 2 = 1 + 0.980 909 687 603 2;
  • 17) 0.980 909 687 603 2 × 2 = 1 + 0.961 819 375 206 4;
  • 18) 0.961 819 375 206 4 × 2 = 1 + 0.923 638 750 412 8;
  • 19) 0.923 638 750 412 8 × 2 = 1 + 0.847 277 500 825 6;
  • 20) 0.847 277 500 825 6 × 2 = 1 + 0.694 555 001 651 2;
  • 21) 0.694 555 001 651 2 × 2 = 1 + 0.389 110 003 302 4;
  • 22) 0.389 110 003 302 4 × 2 = 0 + 0.778 220 006 604 8;
  • 23) 0.778 220 006 604 8 × 2 = 1 + 0.556 440 013 209 6;
  • 24) 0.556 440 013 209 6 × 2 = 1 + 0.112 880 026 419 2;
  • 25) 0.112 880 026 419 2 × 2 = 0 + 0.225 760 052 838 4;
  • 26) 0.225 760 052 838 4 × 2 = 0 + 0.451 520 105 676 8;
  • 27) 0.451 520 105 676 8 × 2 = 0 + 0.903 040 211 353 6;
  • 28) 0.903 040 211 353 6 × 2 = 1 + 0.806 080 422 707 2;
  • 29) 0.806 080 422 707 2 × 2 = 1 + 0.612 160 845 414 4;
  • 30) 0.612 160 845 414 4 × 2 = 1 + 0.224 321 690 828 8;
  • 31) 0.224 321 690 828 8 × 2 = 0 + 0.448 643 381 657 6;
  • 32) 0.448 643 381 657 6 × 2 = 0 + 0.897 286 763 315 2;
  • 33) 0.897 286 763 315 2 × 2 = 1 + 0.794 573 526 630 4;
  • 34) 0.794 573 526 630 4 × 2 = 1 + 0.589 147 053 260 8;
  • 35) 0.589 147 053 260 8 × 2 = 1 + 0.178 294 106 521 6;
  • 36) 0.178 294 106 521 6 × 2 = 0 + 0.356 588 213 043 2;
  • 37) 0.356 588 213 043 2 × 2 = 0 + 0.713 176 426 086 4;
  • 38) 0.713 176 426 086 4 × 2 = 1 + 0.426 352 852 172 8;
  • 39) 0.426 352 852 172 8 × 2 = 0 + 0.852 705 704 345 6;
  • 40) 0.852 705 704 345 6 × 2 = 1 + 0.705 411 408 691 2;
  • 41) 0.705 411 408 691 2 × 2 = 1 + 0.410 822 817 382 4;
  • 42) 0.410 822 817 382 4 × 2 = 0 + 0.821 645 634 764 8;
  • 43) 0.821 645 634 764 8 × 2 = 1 + 0.643 291 269 529 6;
  • 44) 0.643 291 269 529 6 × 2 = 1 + 0.286 582 539 059 2;
  • 45) 0.286 582 539 059 2 × 2 = 0 + 0.573 165 078 118 4;
  • 46) 0.573 165 078 118 4 × 2 = 1 + 0.146 330 156 236 8;
  • 47) 0.146 330 156 236 8 × 2 = 0 + 0.292 660 312 473 6;
  • 48) 0.292 660 312 473 6 × 2 = 0 + 0.585 320 624 947 2;
  • 49) 0.585 320 624 947 2 × 2 = 1 + 0.170 641 249 894 4;
  • 50) 0.170 641 249 894 4 × 2 = 0 + 0.341 282 499 788 8;
  • 51) 0.341 282 499 788 8 × 2 = 0 + 0.682 564 999 577 6;
  • 52) 0.682 564 999 577 6 × 2 = 1 + 0.365 129 999 155 2;
  • 53) 0.365 129 999 155 2 × 2 = 0 + 0.730 259 998 310 4;
  • 54) 0.730 259 998 310 4 × 2 = 1 + 0.460 519 996 620 8;
  • 55) 0.460 519 996 620 8 × 2 = 0 + 0.921 039 993 241 6;
  • 56) 0.921 039 993 241 6 × 2 = 1 + 0.842 079 986 483 2;
  • 57) 0.842 079 986 483 2 × 2 = 1 + 0.684 159 972 966 4;

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:


0.042 419 142 298 7(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 0101 1011 0100 1001 0101 1(2)

6. Positive number before normalization:

0.042 419 142 298 7(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 0101 1011 0100 1001 0101 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:


0.042 419 142 298 7(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 0101 1011 0100 1001 0101 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1110 0101 1011 0100 1001 0101 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1100 1011 0110 1001 0010 1011(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):
1.0101 1011 0111 1111 0110 0011 1001 1100 1011 0110 1001 0010 1011


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. 0101 1011 0111 1111 0110 0011 1001 1100 1011 0110 1001 0010 1011 =

0101 1011 0111 1111 0110 0011 1001 1100 1011 0110 1001 0010 1011


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) =
0101 1011 0111 1111 0110 0011 1001 1100 1011 0110 1001 0010 1011


Decimal number -0.042 419 142 298 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1001 1100 1011 0110 1001 0010 1011

» Convert decimal -0.042 419 142 303 1 to 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