-0.000 000 000 742 147 24 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 24(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 24(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 24| = 0.000 000 000 742 147 24


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 24.

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 24 × 2 = 0 + 0.000 000 001 484 294 48;
  • 2) 0.000 000 001 484 294 48 × 2 = 0 + 0.000 000 002 968 588 96;
  • 3) 0.000 000 002 968 588 96 × 2 = 0 + 0.000 000 005 937 177 92;
  • 4) 0.000 000 005 937 177 92 × 2 = 0 + 0.000 000 011 874 355 84;
  • 5) 0.000 000 011 874 355 84 × 2 = 0 + 0.000 000 023 748 711 68;
  • 6) 0.000 000 023 748 711 68 × 2 = 0 + 0.000 000 047 497 423 36;
  • 7) 0.000 000 047 497 423 36 × 2 = 0 + 0.000 000 094 994 846 72;
  • 8) 0.000 000 094 994 846 72 × 2 = 0 + 0.000 000 189 989 693 44;
  • 9) 0.000 000 189 989 693 44 × 2 = 0 + 0.000 000 379 979 386 88;
  • 10) 0.000 000 379 979 386 88 × 2 = 0 + 0.000 000 759 958 773 76;
  • 11) 0.000 000 759 958 773 76 × 2 = 0 + 0.000 001 519 917 547 52;
  • 12) 0.000 001 519 917 547 52 × 2 = 0 + 0.000 003 039 835 095 04;
  • 13) 0.000 003 039 835 095 04 × 2 = 0 + 0.000 006 079 670 190 08;
  • 14) 0.000 006 079 670 190 08 × 2 = 0 + 0.000 012 159 340 380 16;
  • 15) 0.000 012 159 340 380 16 × 2 = 0 + 0.000 024 318 680 760 32;
  • 16) 0.000 024 318 680 760 32 × 2 = 0 + 0.000 048 637 361 520 64;
  • 17) 0.000 048 637 361 520 64 × 2 = 0 + 0.000 097 274 723 041 28;
  • 18) 0.000 097 274 723 041 28 × 2 = 0 + 0.000 194 549 446 082 56;
  • 19) 0.000 194 549 446 082 56 × 2 = 0 + 0.000 389 098 892 165 12;
  • 20) 0.000 389 098 892 165 12 × 2 = 0 + 0.000 778 197 784 330 24;
  • 21) 0.000 778 197 784 330 24 × 2 = 0 + 0.001 556 395 568 660 48;
  • 22) 0.001 556 395 568 660 48 × 2 = 0 + 0.003 112 791 137 320 96;
  • 23) 0.003 112 791 137 320 96 × 2 = 0 + 0.006 225 582 274 641 92;
  • 24) 0.006 225 582 274 641 92 × 2 = 0 + 0.012 451 164 549 283 84;
  • 25) 0.012 451 164 549 283 84 × 2 = 0 + 0.024 902 329 098 567 68;
  • 26) 0.024 902 329 098 567 68 × 2 = 0 + 0.049 804 658 197 135 36;
  • 27) 0.049 804 658 197 135 36 × 2 = 0 + 0.099 609 316 394 270 72;
  • 28) 0.099 609 316 394 270 72 × 2 = 0 + 0.199 218 632 788 541 44;
  • 29) 0.199 218 632 788 541 44 × 2 = 0 + 0.398 437 265 577 082 88;
  • 30) 0.398 437 265 577 082 88 × 2 = 0 + 0.796 874 531 154 165 76;
  • 31) 0.796 874 531 154 165 76 × 2 = 1 + 0.593 749 062 308 331 52;
  • 32) 0.593 749 062 308 331 52 × 2 = 1 + 0.187 498 124 616 663 04;
  • 33) 0.187 498 124 616 663 04 × 2 = 0 + 0.374 996 249 233 326 08;
  • 34) 0.374 996 249 233 326 08 × 2 = 0 + 0.749 992 498 466 652 16;
  • 35) 0.749 992 498 466 652 16 × 2 = 1 + 0.499 984 996 933 304 32;
  • 36) 0.499 984 996 933 304 32 × 2 = 0 + 0.999 969 993 866 608 64;
  • 37) 0.999 969 993 866 608 64 × 2 = 1 + 0.999 939 987 733 217 28;
  • 38) 0.999 939 987 733 217 28 × 2 = 1 + 0.999 879 975 466 434 56;
  • 39) 0.999 879 975 466 434 56 × 2 = 1 + 0.999 759 950 932 869 12;
  • 40) 0.999 759 950 932 869 12 × 2 = 1 + 0.999 519 901 865 738 24;
  • 41) 0.999 519 901 865 738 24 × 2 = 1 + 0.999 039 803 731 476 48;
  • 42) 0.999 039 803 731 476 48 × 2 = 1 + 0.998 079 607 462 952 96;
  • 43) 0.998 079 607 462 952 96 × 2 = 1 + 0.996 159 214 925 905 92;
  • 44) 0.996 159 214 925 905 92 × 2 = 1 + 0.992 318 429 851 811 84;
  • 45) 0.992 318 429 851 811 84 × 2 = 1 + 0.984 636 859 703 623 68;
  • 46) 0.984 636 859 703 623 68 × 2 = 1 + 0.969 273 719 407 247 36;
  • 47) 0.969 273 719 407 247 36 × 2 = 1 + 0.938 547 438 814 494 72;
  • 48) 0.938 547 438 814 494 72 × 2 = 1 + 0.877 094 877 628 989 44;
  • 49) 0.877 094 877 628 989 44 × 2 = 1 + 0.754 189 755 257 978 88;
  • 50) 0.754 189 755 257 978 88 × 2 = 1 + 0.508 379 510 515 957 76;
  • 51) 0.508 379 510 515 957 76 × 2 = 1 + 0.016 759 021 031 915 52;
  • 52) 0.016 759 021 031 915 52 × 2 = 0 + 0.033 518 042 063 831 04;
  • 53) 0.033 518 042 063 831 04 × 2 = 0 + 0.067 036 084 127 662 08;
  • 54) 0.067 036 084 127 662 08 × 2 = 0 + 0.134 072 168 255 324 16;

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 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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 24(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 00(2) × 20 =

1.1001 0111 1111 1111 1111 000(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 000


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 1000 =

100 1011 1111 1111 1111 1000


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 1000


Decimal number -0.000 000 000 742 147 24 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1000

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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