-0.000 000 000 742 147 675 66 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 66(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 675 66(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 675 66| = 0.000 000 000 742 147 675 66


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: 0.000 000 000 742 147 675 66.

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) 0.000 000 000 742 147 675 66 × 2 = 0 + 0.000 000 001 484 295 351 32;
  • 2) 0.000 000 001 484 295 351 32 × 2 = 0 + 0.000 000 002 968 590 702 64;
  • 3) 0.000 000 002 968 590 702 64 × 2 = 0 + 0.000 000 005 937 181 405 28;
  • 4) 0.000 000 005 937 181 405 28 × 2 = 0 + 0.000 000 011 874 362 810 56;
  • 5) 0.000 000 011 874 362 810 56 × 2 = 0 + 0.000 000 023 748 725 621 12;
  • 6) 0.000 000 023 748 725 621 12 × 2 = 0 + 0.000 000 047 497 451 242 24;
  • 7) 0.000 000 047 497 451 242 24 × 2 = 0 + 0.000 000 094 994 902 484 48;
  • 8) 0.000 000 094 994 902 484 48 × 2 = 0 + 0.000 000 189 989 804 968 96;
  • 9) 0.000 000 189 989 804 968 96 × 2 = 0 + 0.000 000 379 979 609 937 92;
  • 10) 0.000 000 379 979 609 937 92 × 2 = 0 + 0.000 000 759 959 219 875 84;
  • 11) 0.000 000 759 959 219 875 84 × 2 = 0 + 0.000 001 519 918 439 751 68;
  • 12) 0.000 001 519 918 439 751 68 × 2 = 0 + 0.000 003 039 836 879 503 36;
  • 13) 0.000 003 039 836 879 503 36 × 2 = 0 + 0.000 006 079 673 759 006 72;
  • 14) 0.000 006 079 673 759 006 72 × 2 = 0 + 0.000 012 159 347 518 013 44;
  • 15) 0.000 012 159 347 518 013 44 × 2 = 0 + 0.000 024 318 695 036 026 88;
  • 16) 0.000 024 318 695 036 026 88 × 2 = 0 + 0.000 048 637 390 072 053 76;
  • 17) 0.000 048 637 390 072 053 76 × 2 = 0 + 0.000 097 274 780 144 107 52;
  • 18) 0.000 097 274 780 144 107 52 × 2 = 0 + 0.000 194 549 560 288 215 04;
  • 19) 0.000 194 549 560 288 215 04 × 2 = 0 + 0.000 389 099 120 576 430 08;
  • 20) 0.000 389 099 120 576 430 08 × 2 = 0 + 0.000 778 198 241 152 860 16;
  • 21) 0.000 778 198 241 152 860 16 × 2 = 0 + 0.001 556 396 482 305 720 32;
  • 22) 0.001 556 396 482 305 720 32 × 2 = 0 + 0.003 112 792 964 611 440 64;
  • 23) 0.003 112 792 964 611 440 64 × 2 = 0 + 0.006 225 585 929 222 881 28;
  • 24) 0.006 225 585 929 222 881 28 × 2 = 0 + 0.012 451 171 858 445 762 56;
  • 25) 0.012 451 171 858 445 762 56 × 2 = 0 + 0.024 902 343 716 891 525 12;
  • 26) 0.024 902 343 716 891 525 12 × 2 = 0 + 0.049 804 687 433 783 050 24;
  • 27) 0.049 804 687 433 783 050 24 × 2 = 0 + 0.099 609 374 867 566 100 48;
  • 28) 0.099 609 374 867 566 100 48 × 2 = 0 + 0.199 218 749 735 132 200 96;
  • 29) 0.199 218 749 735 132 200 96 × 2 = 0 + 0.398 437 499 470 264 401 92;
  • 30) 0.398 437 499 470 264 401 92 × 2 = 0 + 0.796 874 998 940 528 803 84;
  • 31) 0.796 874 998 940 528 803 84 × 2 = 1 + 0.593 749 997 881 057 607 68;
  • 32) 0.593 749 997 881 057 607 68 × 2 = 1 + 0.187 499 995 762 115 215 36;
  • 33) 0.187 499 995 762 115 215 36 × 2 = 0 + 0.374 999 991 524 230 430 72;
  • 34) 0.374 999 991 524 230 430 72 × 2 = 0 + 0.749 999 983 048 460 861 44;
  • 35) 0.749 999 983 048 460 861 44 × 2 = 1 + 0.499 999 966 096 921 722 88;
  • 36) 0.499 999 966 096 921 722 88 × 2 = 0 + 0.999 999 932 193 843 445 76;
  • 37) 0.999 999 932 193 843 445 76 × 2 = 1 + 0.999 999 864 387 686 891 52;
  • 38) 0.999 999 864 387 686 891 52 × 2 = 1 + 0.999 999 728 775 373 783 04;
  • 39) 0.999 999 728 775 373 783 04 × 2 = 1 + 0.999 999 457 550 747 566 08;
  • 40) 0.999 999 457 550 747 566 08 × 2 = 1 + 0.999 998 915 101 495 132 16;
  • 41) 0.999 998 915 101 495 132 16 × 2 = 1 + 0.999 997 830 202 990 264 32;
  • 42) 0.999 997 830 202 990 264 32 × 2 = 1 + 0.999 995 660 405 980 528 64;
  • 43) 0.999 995 660 405 980 528 64 × 2 = 1 + 0.999 991 320 811 961 057 28;
  • 44) 0.999 991 320 811 961 057 28 × 2 = 1 + 0.999 982 641 623 922 114 56;
  • 45) 0.999 982 641 623 922 114 56 × 2 = 1 + 0.999 965 283 247 844 229 12;
  • 46) 0.999 965 283 247 844 229 12 × 2 = 1 + 0.999 930 566 495 688 458 24;
  • 47) 0.999 930 566 495 688 458 24 × 2 = 1 + 0.999 861 132 991 376 916 48;
  • 48) 0.999 861 132 991 376 916 48 × 2 = 1 + 0.999 722 265 982 753 832 96;
  • 49) 0.999 722 265 982 753 832 96 × 2 = 1 + 0.999 444 531 965 507 665 92;
  • 50) 0.999 444 531 965 507 665 92 × 2 = 1 + 0.998 889 063 931 015 331 84;
  • 51) 0.998 889 063 931 015 331 84 × 2 = 1 + 0.997 778 127 862 030 663 68;
  • 52) 0.997 778 127 862 030 663 68 × 2 = 1 + 0.995 556 255 724 061 327 36;
  • 53) 0.995 556 255 724 061 327 36 × 2 = 1 + 0.991 112 511 448 122 654 72;
  • 54) 0.991 112 511 448 122 654 72 × 2 = 1 + 0.982 225 022 896 245 309 44;

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:


0.000 000 000 742 147 675 66(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 675 66(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 675 66(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 0000(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 675 66 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111