-0.042 419 142 310 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 310 2(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
-0.042 419 142 310 2(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:

|-0.042 419 142 310 2| = 0.042 419 142 310 2


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: 0.042 419 142 310 2.

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) 0.042 419 142 310 2 × 2 = 0 + 0.084 838 284 620 4;
  • 2) 0.084 838 284 620 4 × 2 = 0 + 0.169 676 569 240 8;
  • 3) 0.169 676 569 240 8 × 2 = 0 + 0.339 353 138 481 6;
  • 4) 0.339 353 138 481 6 × 2 = 0 + 0.678 706 276 963 2;
  • 5) 0.678 706 276 963 2 × 2 = 1 + 0.357 412 553 926 4;
  • 6) 0.357 412 553 926 4 × 2 = 0 + 0.714 825 107 852 8;
  • 7) 0.714 825 107 852 8 × 2 = 1 + 0.429 650 215 705 6;
  • 8) 0.429 650 215 705 6 × 2 = 0 + 0.859 300 431 411 2;
  • 9) 0.859 300 431 411 2 × 2 = 1 + 0.718 600 862 822 4;
  • 10) 0.718 600 862 822 4 × 2 = 1 + 0.437 201 725 644 8;
  • 11) 0.437 201 725 644 8 × 2 = 0 + 0.874 403 451 289 6;
  • 12) 0.874 403 451 289 6 × 2 = 1 + 0.748 806 902 579 2;
  • 13) 0.748 806 902 579 2 × 2 = 1 + 0.497 613 805 158 4;
  • 14) 0.497 613 805 158 4 × 2 = 0 + 0.995 227 610 316 8;
  • 15) 0.995 227 610 316 8 × 2 = 1 + 0.990 455 220 633 6;
  • 16) 0.990 455 220 633 6 × 2 = 1 + 0.980 910 441 267 2;
  • 17) 0.980 910 441 267 2 × 2 = 1 + 0.961 820 882 534 4;
  • 18) 0.961 820 882 534 4 × 2 = 1 + 0.923 641 765 068 8;
  • 19) 0.923 641 765 068 8 × 2 = 1 + 0.847 283 530 137 6;
  • 20) 0.847 283 530 137 6 × 2 = 1 + 0.694 567 060 275 2;
  • 21) 0.694 567 060 275 2 × 2 = 1 + 0.389 134 120 550 4;
  • 22) 0.389 134 120 550 4 × 2 = 0 + 0.778 268 241 100 8;
  • 23) 0.778 268 241 100 8 × 2 = 1 + 0.556 536 482 201 6;
  • 24) 0.556 536 482 201 6 × 2 = 1 + 0.113 072 964 403 2;
  • 25) 0.113 072 964 403 2 × 2 = 0 + 0.226 145 928 806 4;
  • 26) 0.226 145 928 806 4 × 2 = 0 + 0.452 291 857 612 8;
  • 27) 0.452 291 857 612 8 × 2 = 0 + 0.904 583 715 225 6;
  • 28) 0.904 583 715 225 6 × 2 = 1 + 0.809 167 430 451 2;
  • 29) 0.809 167 430 451 2 × 2 = 1 + 0.618 334 860 902 4;
  • 30) 0.618 334 860 902 4 × 2 = 1 + 0.236 669 721 804 8;
  • 31) 0.236 669 721 804 8 × 2 = 0 + 0.473 339 443 609 6;
  • 32) 0.473 339 443 609 6 × 2 = 0 + 0.946 678 887 219 2;
  • 33) 0.946 678 887 219 2 × 2 = 1 + 0.893 357 774 438 4;
  • 34) 0.893 357 774 438 4 × 2 = 1 + 0.786 715 548 876 8;
  • 35) 0.786 715 548 876 8 × 2 = 1 + 0.573 431 097 753 6;
  • 36) 0.573 431 097 753 6 × 2 = 1 + 0.146 862 195 507 2;
  • 37) 0.146 862 195 507 2 × 2 = 0 + 0.293 724 391 014 4;
  • 38) 0.293 724 391 014 4 × 2 = 0 + 0.587 448 782 028 8;
  • 39) 0.587 448 782 028 8 × 2 = 1 + 0.174 897 564 057 6;
  • 40) 0.174 897 564 057 6 × 2 = 0 + 0.349 795 128 115 2;
  • 41) 0.349 795 128 115 2 × 2 = 0 + 0.699 590 256 230 4;
  • 42) 0.699 590 256 230 4 × 2 = 1 + 0.399 180 512 460 8;
  • 43) 0.399 180 512 460 8 × 2 = 0 + 0.798 361 024 921 6;
  • 44) 0.798 361 024 921 6 × 2 = 1 + 0.596 722 049 843 2;
  • 45) 0.596 722 049 843 2 × 2 = 1 + 0.193 444 099 686 4;
  • 46) 0.193 444 099 686 4 × 2 = 0 + 0.386 888 199 372 8;
  • 47) 0.386 888 199 372 8 × 2 = 0 + 0.773 776 398 745 6;
  • 48) 0.773 776 398 745 6 × 2 = 1 + 0.547 552 797 491 2;
  • 49) 0.547 552 797 491 2 × 2 = 1 + 0.095 105 594 982 4;
  • 50) 0.095 105 594 982 4 × 2 = 0 + 0.190 211 189 964 8;
  • 51) 0.190 211 189 964 8 × 2 = 0 + 0.380 422 379 929 6;
  • 52) 0.380 422 379 929 6 × 2 = 0 + 0.760 844 759 859 2;
  • 53) 0.760 844 759 859 2 × 2 = 1 + 0.521 689 519 718 4;
  • 54) 0.521 689 519 718 4 × 2 = 1 + 0.043 379 039 436 8;
  • 55) 0.043 379 039 436 8 × 2 = 0 + 0.086 758 078 873 6;
  • 56) 0.086 758 078 873 6 × 2 = 0 + 0.173 516 157 747 2;
  • 57) 0.173 516 157 747 2 × 2 = 0 + 0.347 032 315 494 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:


0.042 419 142 310 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0010 0101 1001 1000 1100 0(2)

6. Positive number before normalization:

0.042 419 142 310 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0010 0101 1001 1000 1100 0(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:


0.042 419 142 310 2(10) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0010 0101 1001 1000 1100 0(2) =

0.0000 1010 1101 1011 1111 1011 0001 1100 1111 0010 0101 1001 1000 1100 0(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1001 1110 0100 1011 0011 0001 1000(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 1110 0100 1011 0011 0001 1000


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 1110 0100 1011 0011 0001 1000 =

0101 1011 0111 1111 0110 0011 1001 1110 0100 1011 0011 0001 1000


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 1110 0100 1011 0011 0001 1000


Decimal number -0.042 419 142 310 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1001 1110 0100 1011 0011 0001 1000

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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