-0.000 000 000 742 146 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 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
-0.000 000 000 742 146 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:

|-0.000 000 000 742 146 3| = 0.000 000 000 742 146 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: 0.000 000 000 742 146 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) 0.000 000 000 742 146 3 × 2 = 0 + 0.000 000 001 484 292 6;
  • 2) 0.000 000 001 484 292 6 × 2 = 0 + 0.000 000 002 968 585 2;
  • 3) 0.000 000 002 968 585 2 × 2 = 0 + 0.000 000 005 937 170 4;
  • 4) 0.000 000 005 937 170 4 × 2 = 0 + 0.000 000 011 874 340 8;
  • 5) 0.000 000 011 874 340 8 × 2 = 0 + 0.000 000 023 748 681 6;
  • 6) 0.000 000 023 748 681 6 × 2 = 0 + 0.000 000 047 497 363 2;
  • 7) 0.000 000 047 497 363 2 × 2 = 0 + 0.000 000 094 994 726 4;
  • 8) 0.000 000 094 994 726 4 × 2 = 0 + 0.000 000 189 989 452 8;
  • 9) 0.000 000 189 989 452 8 × 2 = 0 + 0.000 000 379 978 905 6;
  • 10) 0.000 000 379 978 905 6 × 2 = 0 + 0.000 000 759 957 811 2;
  • 11) 0.000 000 759 957 811 2 × 2 = 0 + 0.000 001 519 915 622 4;
  • 12) 0.000 001 519 915 622 4 × 2 = 0 + 0.000 003 039 831 244 8;
  • 13) 0.000 003 039 831 244 8 × 2 = 0 + 0.000 006 079 662 489 6;
  • 14) 0.000 006 079 662 489 6 × 2 = 0 + 0.000 012 159 324 979 2;
  • 15) 0.000 012 159 324 979 2 × 2 = 0 + 0.000 024 318 649 958 4;
  • 16) 0.000 024 318 649 958 4 × 2 = 0 + 0.000 048 637 299 916 8;
  • 17) 0.000 048 637 299 916 8 × 2 = 0 + 0.000 097 274 599 833 6;
  • 18) 0.000 097 274 599 833 6 × 2 = 0 + 0.000 194 549 199 667 2;
  • 19) 0.000 194 549 199 667 2 × 2 = 0 + 0.000 389 098 399 334 4;
  • 20) 0.000 389 098 399 334 4 × 2 = 0 + 0.000 778 196 798 668 8;
  • 21) 0.000 778 196 798 668 8 × 2 = 0 + 0.001 556 393 597 337 6;
  • 22) 0.001 556 393 597 337 6 × 2 = 0 + 0.003 112 787 194 675 2;
  • 23) 0.003 112 787 194 675 2 × 2 = 0 + 0.006 225 574 389 350 4;
  • 24) 0.006 225 574 389 350 4 × 2 = 0 + 0.012 451 148 778 700 8;
  • 25) 0.012 451 148 778 700 8 × 2 = 0 + 0.024 902 297 557 401 6;
  • 26) 0.024 902 297 557 401 6 × 2 = 0 + 0.049 804 595 114 803 2;
  • 27) 0.049 804 595 114 803 2 × 2 = 0 + 0.099 609 190 229 606 4;
  • 28) 0.099 609 190 229 606 4 × 2 = 0 + 0.199 218 380 459 212 8;
  • 29) 0.199 218 380 459 212 8 × 2 = 0 + 0.398 436 760 918 425 6;
  • 30) 0.398 436 760 918 425 6 × 2 = 0 + 0.796 873 521 836 851 2;
  • 31) 0.796 873 521 836 851 2 × 2 = 1 + 0.593 747 043 673 702 4;
  • 32) 0.593 747 043 673 702 4 × 2 = 1 + 0.187 494 087 347 404 8;
  • 33) 0.187 494 087 347 404 8 × 2 = 0 + 0.374 988 174 694 809 6;
  • 34) 0.374 988 174 694 809 6 × 2 = 0 + 0.749 976 349 389 619 2;
  • 35) 0.749 976 349 389 619 2 × 2 = 1 + 0.499 952 698 779 238 4;
  • 36) 0.499 952 698 779 238 4 × 2 = 0 + 0.999 905 397 558 476 8;
  • 37) 0.999 905 397 558 476 8 × 2 = 1 + 0.999 810 795 116 953 6;
  • 38) 0.999 810 795 116 953 6 × 2 = 1 + 0.999 621 590 233 907 2;
  • 39) 0.999 621 590 233 907 2 × 2 = 1 + 0.999 243 180 467 814 4;
  • 40) 0.999 243 180 467 814 4 × 2 = 1 + 0.998 486 360 935 628 8;
  • 41) 0.998 486 360 935 628 8 × 2 = 1 + 0.996 972 721 871 257 6;
  • 42) 0.996 972 721 871 257 6 × 2 = 1 + 0.993 945 443 742 515 2;
  • 43) 0.993 945 443 742 515 2 × 2 = 1 + 0.987 890 887 485 030 4;
  • 44) 0.987 890 887 485 030 4 × 2 = 1 + 0.975 781 774 970 060 8;
  • 45) 0.975 781 774 970 060 8 × 2 = 1 + 0.951 563 549 940 121 6;
  • 46) 0.951 563 549 940 121 6 × 2 = 1 + 0.903 127 099 880 243 2;
  • 47) 0.903 127 099 880 243 2 × 2 = 1 + 0.806 254 199 760 486 4;
  • 48) 0.806 254 199 760 486 4 × 2 = 1 + 0.612 508 399 520 972 8;
  • 49) 0.612 508 399 520 972 8 × 2 = 1 + 0.225 016 799 041 945 6;
  • 50) 0.225 016 799 041 945 6 × 2 = 0 + 0.450 033 598 083 891 2;
  • 51) 0.450 033 598 083 891 2 × 2 = 0 + 0.900 067 196 167 782 4;
  • 52) 0.900 067 196 167 782 4 × 2 = 1 + 0.800 134 392 335 564 8;
  • 53) 0.800 134 392 335 564 8 × 2 = 1 + 0.600 268 784 671 129 6;
  • 54) 0.600 268 784 671 129 6 × 2 = 1 + 0.200 537 569 342 259 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:


0.000 000 000 742 146 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1001 11(2)

6. Positive number before normalization:

0.000 000 000 742 146 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1001 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:


0.000 000 000 742 146 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1001 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1001 11(2) × 20 =

1.1001 0111 1111 1111 1100 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 1100 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 1110 0111 =

100 1011 1111 1111 1110 0111


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 1110 0111


Decimal number -0.000 000 000 742 146 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 0111

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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