-0.000 000 000 742 147 92 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 92(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 92(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 92| = 0.000 000 000 742 147 92


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 92.

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 92 × 2 = 0 + 0.000 000 001 484 295 84;
  • 2) 0.000 000 001 484 295 84 × 2 = 0 + 0.000 000 002 968 591 68;
  • 3) 0.000 000 002 968 591 68 × 2 = 0 + 0.000 000 005 937 183 36;
  • 4) 0.000 000 005 937 183 36 × 2 = 0 + 0.000 000 011 874 366 72;
  • 5) 0.000 000 011 874 366 72 × 2 = 0 + 0.000 000 023 748 733 44;
  • 6) 0.000 000 023 748 733 44 × 2 = 0 + 0.000 000 047 497 466 88;
  • 7) 0.000 000 047 497 466 88 × 2 = 0 + 0.000 000 094 994 933 76;
  • 8) 0.000 000 094 994 933 76 × 2 = 0 + 0.000 000 189 989 867 52;
  • 9) 0.000 000 189 989 867 52 × 2 = 0 + 0.000 000 379 979 735 04;
  • 10) 0.000 000 379 979 735 04 × 2 = 0 + 0.000 000 759 959 470 08;
  • 11) 0.000 000 759 959 470 08 × 2 = 0 + 0.000 001 519 918 940 16;
  • 12) 0.000 001 519 918 940 16 × 2 = 0 + 0.000 003 039 837 880 32;
  • 13) 0.000 003 039 837 880 32 × 2 = 0 + 0.000 006 079 675 760 64;
  • 14) 0.000 006 079 675 760 64 × 2 = 0 + 0.000 012 159 351 521 28;
  • 15) 0.000 012 159 351 521 28 × 2 = 0 + 0.000 024 318 703 042 56;
  • 16) 0.000 024 318 703 042 56 × 2 = 0 + 0.000 048 637 406 085 12;
  • 17) 0.000 048 637 406 085 12 × 2 = 0 + 0.000 097 274 812 170 24;
  • 18) 0.000 097 274 812 170 24 × 2 = 0 + 0.000 194 549 624 340 48;
  • 19) 0.000 194 549 624 340 48 × 2 = 0 + 0.000 389 099 248 680 96;
  • 20) 0.000 389 099 248 680 96 × 2 = 0 + 0.000 778 198 497 361 92;
  • 21) 0.000 778 198 497 361 92 × 2 = 0 + 0.001 556 396 994 723 84;
  • 22) 0.001 556 396 994 723 84 × 2 = 0 + 0.003 112 793 989 447 68;
  • 23) 0.003 112 793 989 447 68 × 2 = 0 + 0.006 225 587 978 895 36;
  • 24) 0.006 225 587 978 895 36 × 2 = 0 + 0.012 451 175 957 790 72;
  • 25) 0.012 451 175 957 790 72 × 2 = 0 + 0.024 902 351 915 581 44;
  • 26) 0.024 902 351 915 581 44 × 2 = 0 + 0.049 804 703 831 162 88;
  • 27) 0.049 804 703 831 162 88 × 2 = 0 + 0.099 609 407 662 325 76;
  • 28) 0.099 609 407 662 325 76 × 2 = 0 + 0.199 218 815 324 651 52;
  • 29) 0.199 218 815 324 651 52 × 2 = 0 + 0.398 437 630 649 303 04;
  • 30) 0.398 437 630 649 303 04 × 2 = 0 + 0.796 875 261 298 606 08;
  • 31) 0.796 875 261 298 606 08 × 2 = 1 + 0.593 750 522 597 212 16;
  • 32) 0.593 750 522 597 212 16 × 2 = 1 + 0.187 501 045 194 424 32;
  • 33) 0.187 501 045 194 424 32 × 2 = 0 + 0.375 002 090 388 848 64;
  • 34) 0.375 002 090 388 848 64 × 2 = 0 + 0.750 004 180 777 697 28;
  • 35) 0.750 004 180 777 697 28 × 2 = 1 + 0.500 008 361 555 394 56;
  • 36) 0.500 008 361 555 394 56 × 2 = 1 + 0.000 016 723 110 789 12;
  • 37) 0.000 016 723 110 789 12 × 2 = 0 + 0.000 033 446 221 578 24;
  • 38) 0.000 033 446 221 578 24 × 2 = 0 + 0.000 066 892 443 156 48;
  • 39) 0.000 066 892 443 156 48 × 2 = 0 + 0.000 133 784 886 312 96;
  • 40) 0.000 133 784 886 312 96 × 2 = 0 + 0.000 267 569 772 625 92;
  • 41) 0.000 267 569 772 625 92 × 2 = 0 + 0.000 535 139 545 251 84;
  • 42) 0.000 535 139 545 251 84 × 2 = 0 + 0.001 070 279 090 503 68;
  • 43) 0.001 070 279 090 503 68 × 2 = 0 + 0.002 140 558 181 007 36;
  • 44) 0.002 140 558 181 007 36 × 2 = 0 + 0.004 281 116 362 014 72;
  • 45) 0.004 281 116 362 014 72 × 2 = 0 + 0.008 562 232 724 029 44;
  • 46) 0.008 562 232 724 029 44 × 2 = 0 + 0.017 124 465 448 058 88;
  • 47) 0.017 124 465 448 058 88 × 2 = 0 + 0.034 248 930 896 117 76;
  • 48) 0.034 248 930 896 117 76 × 2 = 0 + 0.068 497 861 792 235 52;
  • 49) 0.068 497 861 792 235 52 × 2 = 0 + 0.136 995 723 584 471 04;
  • 50) 0.136 995 723 584 471 04 × 2 = 0 + 0.273 991 447 168 942 08;
  • 51) 0.273 991 447 168 942 08 × 2 = 0 + 0.547 982 894 337 884 16;
  • 52) 0.547 982 894 337 884 16 × 2 = 1 + 0.095 965 788 675 768 32;
  • 53) 0.095 965 788 675 768 32 × 2 = 0 + 0.191 931 577 351 536 64;
  • 54) 0.191 931 577 351 536 64 × 2 = 0 + 0.383 863 154 703 073 28;

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 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(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 92(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 00(2) × 20 =

1.1001 1000 0000 0000 0000 100(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 1000 0000 0000 0000 100


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 1100 0000 0000 0000 0100 =

100 1100 0000 0000 0000 0100


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 1100 0000 0000 0000 0100


Decimal number -0.000 000 000 742 147 92 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0100

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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