-0.000 000 000 742 147 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 3(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 3(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 3| = 0.000 000 000 742 147 3


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 3.

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 3 × 2 = 0 + 0.000 000 001 484 294 6;
  • 2) 0.000 000 001 484 294 6 × 2 = 0 + 0.000 000 002 968 589 2;
  • 3) 0.000 000 002 968 589 2 × 2 = 0 + 0.000 000 005 937 178 4;
  • 4) 0.000 000 005 937 178 4 × 2 = 0 + 0.000 000 011 874 356 8;
  • 5) 0.000 000 011 874 356 8 × 2 = 0 + 0.000 000 023 748 713 6;
  • 6) 0.000 000 023 748 713 6 × 2 = 0 + 0.000 000 047 497 427 2;
  • 7) 0.000 000 047 497 427 2 × 2 = 0 + 0.000 000 094 994 854 4;
  • 8) 0.000 000 094 994 854 4 × 2 = 0 + 0.000 000 189 989 708 8;
  • 9) 0.000 000 189 989 708 8 × 2 = 0 + 0.000 000 379 979 417 6;
  • 10) 0.000 000 379 979 417 6 × 2 = 0 + 0.000 000 759 958 835 2;
  • 11) 0.000 000 759 958 835 2 × 2 = 0 + 0.000 001 519 917 670 4;
  • 12) 0.000 001 519 917 670 4 × 2 = 0 + 0.000 003 039 835 340 8;
  • 13) 0.000 003 039 835 340 8 × 2 = 0 + 0.000 006 079 670 681 6;
  • 14) 0.000 006 079 670 681 6 × 2 = 0 + 0.000 012 159 341 363 2;
  • 15) 0.000 012 159 341 363 2 × 2 = 0 + 0.000 024 318 682 726 4;
  • 16) 0.000 024 318 682 726 4 × 2 = 0 + 0.000 048 637 365 452 8;
  • 17) 0.000 048 637 365 452 8 × 2 = 0 + 0.000 097 274 730 905 6;
  • 18) 0.000 097 274 730 905 6 × 2 = 0 + 0.000 194 549 461 811 2;
  • 19) 0.000 194 549 461 811 2 × 2 = 0 + 0.000 389 098 923 622 4;
  • 20) 0.000 389 098 923 622 4 × 2 = 0 + 0.000 778 197 847 244 8;
  • 21) 0.000 778 197 847 244 8 × 2 = 0 + 0.001 556 395 694 489 6;
  • 22) 0.001 556 395 694 489 6 × 2 = 0 + 0.003 112 791 388 979 2;
  • 23) 0.003 112 791 388 979 2 × 2 = 0 + 0.006 225 582 777 958 4;
  • 24) 0.006 225 582 777 958 4 × 2 = 0 + 0.012 451 165 555 916 8;
  • 25) 0.012 451 165 555 916 8 × 2 = 0 + 0.024 902 331 111 833 6;
  • 26) 0.024 902 331 111 833 6 × 2 = 0 + 0.049 804 662 223 667 2;
  • 27) 0.049 804 662 223 667 2 × 2 = 0 + 0.099 609 324 447 334 4;
  • 28) 0.099 609 324 447 334 4 × 2 = 0 + 0.199 218 648 894 668 8;
  • 29) 0.199 218 648 894 668 8 × 2 = 0 + 0.398 437 297 789 337 6;
  • 30) 0.398 437 297 789 337 6 × 2 = 0 + 0.796 874 595 578 675 2;
  • 31) 0.796 874 595 578 675 2 × 2 = 1 + 0.593 749 191 157 350 4;
  • 32) 0.593 749 191 157 350 4 × 2 = 1 + 0.187 498 382 314 700 8;
  • 33) 0.187 498 382 314 700 8 × 2 = 0 + 0.374 996 764 629 401 6;
  • 34) 0.374 996 764 629 401 6 × 2 = 0 + 0.749 993 529 258 803 2;
  • 35) 0.749 993 529 258 803 2 × 2 = 1 + 0.499 987 058 517 606 4;
  • 36) 0.499 987 058 517 606 4 × 2 = 0 + 0.999 974 117 035 212 8;
  • 37) 0.999 974 117 035 212 8 × 2 = 1 + 0.999 948 234 070 425 6;
  • 38) 0.999 948 234 070 425 6 × 2 = 1 + 0.999 896 468 140 851 2;
  • 39) 0.999 896 468 140 851 2 × 2 = 1 + 0.999 792 936 281 702 4;
  • 40) 0.999 792 936 281 702 4 × 2 = 1 + 0.999 585 872 563 404 8;
  • 41) 0.999 585 872 563 404 8 × 2 = 1 + 0.999 171 745 126 809 6;
  • 42) 0.999 171 745 126 809 6 × 2 = 1 + 0.998 343 490 253 619 2;
  • 43) 0.998 343 490 253 619 2 × 2 = 1 + 0.996 686 980 507 238 4;
  • 44) 0.996 686 980 507 238 4 × 2 = 1 + 0.993 373 961 014 476 8;
  • 45) 0.993 373 961 014 476 8 × 2 = 1 + 0.986 747 922 028 953 6;
  • 46) 0.986 747 922 028 953 6 × 2 = 1 + 0.973 495 844 057 907 2;
  • 47) 0.973 495 844 057 907 2 × 2 = 1 + 0.946 991 688 115 814 4;
  • 48) 0.946 991 688 115 814 4 × 2 = 1 + 0.893 983 376 231 628 8;
  • 49) 0.893 983 376 231 628 8 × 2 = 1 + 0.787 966 752 463 257 6;
  • 50) 0.787 966 752 463 257 6 × 2 = 1 + 0.575 933 504 926 515 2;
  • 51) 0.575 933 504 926 515 2 × 2 = 1 + 0.151 867 009 853 030 4;
  • 52) 0.151 867 009 853 030 4 × 2 = 0 + 0.303 734 019 706 060 8;
  • 53) 0.303 734 019 706 060 8 × 2 = 0 + 0.607 468 039 412 121 6;
  • 54) 0.607 468 039 412 121 6 × 2 = 1 + 0.214 936 078 824 243 2;

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 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(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 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 01(2) × 20 =

1.1001 0111 1111 1111 1111 001(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 001


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 1001 =

100 1011 1111 1111 1111 1001


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 1001


Decimal number -0.000 000 000 742 147 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1001

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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