-0.042 419 142 327 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.042 419 142 327 4(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
-0.042 419 142 327 4(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:

|-0.042 419 142 327 4| = 0.042 419 142 327 4


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: 0.042 419 142 327 4.

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) 0.042 419 142 327 4 × 2 = 0 + 0.084 838 284 654 8;
  • 2) 0.084 838 284 654 8 × 2 = 0 + 0.169 676 569 309 6;
  • 3) 0.169 676 569 309 6 × 2 = 0 + 0.339 353 138 619 2;
  • 4) 0.339 353 138 619 2 × 2 = 0 + 0.678 706 277 238 4;
  • 5) 0.678 706 277 238 4 × 2 = 1 + 0.357 412 554 476 8;
  • 6) 0.357 412 554 476 8 × 2 = 0 + 0.714 825 108 953 6;
  • 7) 0.714 825 108 953 6 × 2 = 1 + 0.429 650 217 907 2;
  • 8) 0.429 650 217 907 2 × 2 = 0 + 0.859 300 435 814 4;
  • 9) 0.859 300 435 814 4 × 2 = 1 + 0.718 600 871 628 8;
  • 10) 0.718 600 871 628 8 × 2 = 1 + 0.437 201 743 257 6;
  • 11) 0.437 201 743 257 6 × 2 = 0 + 0.874 403 486 515 2;
  • 12) 0.874 403 486 515 2 × 2 = 1 + 0.748 806 973 030 4;
  • 13) 0.748 806 973 030 4 × 2 = 1 + 0.497 613 946 060 8;
  • 14) 0.497 613 946 060 8 × 2 = 0 + 0.995 227 892 121 6;
  • 15) 0.995 227 892 121 6 × 2 = 1 + 0.990 455 784 243 2;
  • 16) 0.990 455 784 243 2 × 2 = 1 + 0.980 911 568 486 4;
  • 17) 0.980 911 568 486 4 × 2 = 1 + 0.961 823 136 972 8;
  • 18) 0.961 823 136 972 8 × 2 = 1 + 0.923 646 273 945 6;
  • 19) 0.923 646 273 945 6 × 2 = 1 + 0.847 292 547 891 2;
  • 20) 0.847 292 547 891 2 × 2 = 1 + 0.694 585 095 782 4;
  • 21) 0.694 585 095 782 4 × 2 = 1 + 0.389 170 191 564 8;
  • 22) 0.389 170 191 564 8 × 2 = 0 + 0.778 340 383 129 6;
  • 23) 0.778 340 383 129 6 × 2 = 1 + 0.556 680 766 259 2;
  • 24) 0.556 680 766 259 2 × 2 = 1 + 0.113 361 532 518 4;
  • 25) 0.113 361 532 518 4 × 2 = 0 + 0.226 723 065 036 8;
  • 26) 0.226 723 065 036 8 × 2 = 0 + 0.453 446 130 073 6;
  • 27) 0.453 446 130 073 6 × 2 = 0 + 0.906 892 260 147 2;
  • 28) 0.906 892 260 147 2 × 2 = 1 + 0.813 784 520 294 4;
  • 29) 0.813 784 520 294 4 × 2 = 1 + 0.627 569 040 588 8;
  • 30) 0.627 569 040 588 8 × 2 = 1 + 0.255 138 081 177 6;
  • 31) 0.255 138 081 177 6 × 2 = 0 + 0.510 276 162 355 2;
  • 32) 0.510 276 162 355 2 × 2 = 1 + 0.020 552 324 710 4;
  • 33) 0.020 552 324 710 4 × 2 = 0 + 0.041 104 649 420 8;
  • 34) 0.041 104 649 420 8 × 2 = 0 + 0.082 209 298 841 6;
  • 35) 0.082 209 298 841 6 × 2 = 0 + 0.164 418 597 683 2;
  • 36) 0.164 418 597 683 2 × 2 = 0 + 0.328 837 195 366 4;
  • 37) 0.328 837 195 366 4 × 2 = 0 + 0.657 674 390 732 8;
  • 38) 0.657 674 390 732 8 × 2 = 1 + 0.315 348 781 465 6;
  • 39) 0.315 348 781 465 6 × 2 = 0 + 0.630 697 562 931 2;
  • 40) 0.630 697 562 931 2 × 2 = 1 + 0.261 395 125 862 4;
  • 41) 0.261 395 125 862 4 × 2 = 0 + 0.522 790 251 724 8;
  • 42) 0.522 790 251 724 8 × 2 = 1 + 0.045 580 503 449 6;
  • 43) 0.045 580 503 449 6 × 2 = 0 + 0.091 161 006 899 2;
  • 44) 0.091 161 006 899 2 × 2 = 0 + 0.182 322 013 798 4;
  • 45) 0.182 322 013 798 4 × 2 = 0 + 0.364 644 027 596 8;
  • 46) 0.364 644 027 596 8 × 2 = 0 + 0.729 288 055 193 6;
  • 47) 0.729 288 055 193 6 × 2 = 1 + 0.458 576 110 387 2;
  • 48) 0.458 576 110 387 2 × 2 = 0 + 0.917 152 220 774 4;
  • 49) 0.917 152 220 774 4 × 2 = 1 + 0.834 304 441 548 8;
  • 50) 0.834 304 441 548 8 × 2 = 1 + 0.668 608 883 097 6;
  • 51) 0.668 608 883 097 6 × 2 = 1 + 0.337 217 766 195 2;
  • 52) 0.337 217 766 195 2 × 2 = 0 + 0.674 435 532 390 4;
  • 53) 0.674 435 532 390 4 × 2 = 1 + 0.348 871 064 780 8;
  • 54) 0.348 871 064 780 8 × 2 = 0 + 0.697 742 129 561 6;
  • 55) 0.697 742 129 561 6 × 2 = 1 + 0.395 484 259 123 2;
  • 56) 0.395 484 259 123 2 × 2 = 0 + 0.790 968 518 246 4;
  • 57) 0.790 968 518 246 4 × 2 = 1 + 0.581 937 036 492 8;

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:


0.042 419 142 327 4(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0101 0100 0010 1110 1010 1(2)

6. Positive number before normalization:

0.042 419 142 327 4(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0101 0100 0010 1110 1010 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:


0.042 419 142 327 4(10) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0101 0100 0010 1110 1010 1(2) =

0.0000 1010 1101 1011 1111 1011 0001 1101 0000 0101 0100 0010 1110 1010 1(2) × 20 =

1.0101 1011 0111 1111 0110 0011 1010 0000 1010 1000 0101 1101 0101(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 1010 0000 1010 1000 0101 1101 0101


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 1010 0000 1010 1000 0101 1101 0101 =

0101 1011 0111 1111 0110 0011 1010 0000 1010 1000 0101 1101 0101


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 1010 0000 1010 1000 0101 1101 0101


Decimal number -0.042 419 142 327 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 0101 1011 0111 1111 0110 0011 1010 0000 1010 1000 0101 1101 0101

» Convert decimal -0.042 419 142 331 9 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