-0.000 000 000 742 147 676 12 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 12(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
-0.000 000 000 742 147 676 12(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:

|-0.000 000 000 742 147 676 12| = 0.000 000 000 742 147 676 12


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: 0.000 000 000 742 147 676 12.

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) 0.000 000 000 742 147 676 12 × 2 = 0 + 0.000 000 001 484 295 352 24;
  • 2) 0.000 000 001 484 295 352 24 × 2 = 0 + 0.000 000 002 968 590 704 48;
  • 3) 0.000 000 002 968 590 704 48 × 2 = 0 + 0.000 000 005 937 181 408 96;
  • 4) 0.000 000 005 937 181 408 96 × 2 = 0 + 0.000 000 011 874 362 817 92;
  • 5) 0.000 000 011 874 362 817 92 × 2 = 0 + 0.000 000 023 748 725 635 84;
  • 6) 0.000 000 023 748 725 635 84 × 2 = 0 + 0.000 000 047 497 451 271 68;
  • 7) 0.000 000 047 497 451 271 68 × 2 = 0 + 0.000 000 094 994 902 543 36;
  • 8) 0.000 000 094 994 902 543 36 × 2 = 0 + 0.000 000 189 989 805 086 72;
  • 9) 0.000 000 189 989 805 086 72 × 2 = 0 + 0.000 000 379 979 610 173 44;
  • 10) 0.000 000 379 979 610 173 44 × 2 = 0 + 0.000 000 759 959 220 346 88;
  • 11) 0.000 000 759 959 220 346 88 × 2 = 0 + 0.000 001 519 918 440 693 76;
  • 12) 0.000 001 519 918 440 693 76 × 2 = 0 + 0.000 003 039 836 881 387 52;
  • 13) 0.000 003 039 836 881 387 52 × 2 = 0 + 0.000 006 079 673 762 775 04;
  • 14) 0.000 006 079 673 762 775 04 × 2 = 0 + 0.000 012 159 347 525 550 08;
  • 15) 0.000 012 159 347 525 550 08 × 2 = 0 + 0.000 024 318 695 051 100 16;
  • 16) 0.000 024 318 695 051 100 16 × 2 = 0 + 0.000 048 637 390 102 200 32;
  • 17) 0.000 048 637 390 102 200 32 × 2 = 0 + 0.000 097 274 780 204 400 64;
  • 18) 0.000 097 274 780 204 400 64 × 2 = 0 + 0.000 194 549 560 408 801 28;
  • 19) 0.000 194 549 560 408 801 28 × 2 = 0 + 0.000 389 099 120 817 602 56;
  • 20) 0.000 389 099 120 817 602 56 × 2 = 0 + 0.000 778 198 241 635 205 12;
  • 21) 0.000 778 198 241 635 205 12 × 2 = 0 + 0.001 556 396 483 270 410 24;
  • 22) 0.001 556 396 483 270 410 24 × 2 = 0 + 0.003 112 792 966 540 820 48;
  • 23) 0.003 112 792 966 540 820 48 × 2 = 0 + 0.006 225 585 933 081 640 96;
  • 24) 0.006 225 585 933 081 640 96 × 2 = 0 + 0.012 451 171 866 163 281 92;
  • 25) 0.012 451 171 866 163 281 92 × 2 = 0 + 0.024 902 343 732 326 563 84;
  • 26) 0.024 902 343 732 326 563 84 × 2 = 0 + 0.049 804 687 464 653 127 68;
  • 27) 0.049 804 687 464 653 127 68 × 2 = 0 + 0.099 609 374 929 306 255 36;
  • 28) 0.099 609 374 929 306 255 36 × 2 = 0 + 0.199 218 749 858 612 510 72;
  • 29) 0.199 218 749 858 612 510 72 × 2 = 0 + 0.398 437 499 717 225 021 44;
  • 30) 0.398 437 499 717 225 021 44 × 2 = 0 + 0.796 874 999 434 450 042 88;
  • 31) 0.796 874 999 434 450 042 88 × 2 = 1 + 0.593 749 998 868 900 085 76;
  • 32) 0.593 749 998 868 900 085 76 × 2 = 1 + 0.187 499 997 737 800 171 52;
  • 33) 0.187 499 997 737 800 171 52 × 2 = 0 + 0.374 999 995 475 600 343 04;
  • 34) 0.374 999 995 475 600 343 04 × 2 = 0 + 0.749 999 990 951 200 686 08;
  • 35) 0.749 999 990 951 200 686 08 × 2 = 1 + 0.499 999 981 902 401 372 16;
  • 36) 0.499 999 981 902 401 372 16 × 2 = 0 + 0.999 999 963 804 802 744 32;
  • 37) 0.999 999 963 804 802 744 32 × 2 = 1 + 0.999 999 927 609 605 488 64;
  • 38) 0.999 999 927 609 605 488 64 × 2 = 1 + 0.999 999 855 219 210 977 28;
  • 39) 0.999 999 855 219 210 977 28 × 2 = 1 + 0.999 999 710 438 421 954 56;
  • 40) 0.999 999 710 438 421 954 56 × 2 = 1 + 0.999 999 420 876 843 909 12;
  • 41) 0.999 999 420 876 843 909 12 × 2 = 1 + 0.999 998 841 753 687 818 24;
  • 42) 0.999 998 841 753 687 818 24 × 2 = 1 + 0.999 997 683 507 375 636 48;
  • 43) 0.999 997 683 507 375 636 48 × 2 = 1 + 0.999 995 367 014 751 272 96;
  • 44) 0.999 995 367 014 751 272 96 × 2 = 1 + 0.999 990 734 029 502 545 92;
  • 45) 0.999 990 734 029 502 545 92 × 2 = 1 + 0.999 981 468 059 005 091 84;
  • 46) 0.999 981 468 059 005 091 84 × 2 = 1 + 0.999 962 936 118 010 183 68;
  • 47) 0.999 962 936 118 010 183 68 × 2 = 1 + 0.999 925 872 236 020 367 36;
  • 48) 0.999 925 872 236 020 367 36 × 2 = 1 + 0.999 851 744 472 040 734 72;
  • 49) 0.999 851 744 472 040 734 72 × 2 = 1 + 0.999 703 488 944 081 469 44;
  • 50) 0.999 703 488 944 081 469 44 × 2 = 1 + 0.999 406 977 888 162 938 88;
  • 51) 0.999 406 977 888 162 938 88 × 2 = 1 + 0.998 813 955 776 325 877 76;
  • 52) 0.998 813 955 776 325 877 76 × 2 = 1 + 0.997 627 911 552 651 755 52;
  • 53) 0.997 627 911 552 651 755 52 × 2 = 1 + 0.995 255 823 105 303 511 04;
  • 54) 0.995 255 823 105 303 511 04 × 2 = 1 + 0.990 511 646 210 607 022 08;

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:


0.000 000 000 742 147 676 12(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 12(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:


0.000 000 000 742 147 676 12(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 -0.000 000 000 742 147 676 12 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