-0.000 000 000 742 121 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 121 1(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
-0.000 000 000 742 121 1(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:

|-0.000 000 000 742 121 1| = 0.000 000 000 742 121 1


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: 0.000 000 000 742 121 1.

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) 0.000 000 000 742 121 1 × 2 = 0 + 0.000 000 001 484 242 2;
  • 2) 0.000 000 001 484 242 2 × 2 = 0 + 0.000 000 002 968 484 4;
  • 3) 0.000 000 002 968 484 4 × 2 = 0 + 0.000 000 005 936 968 8;
  • 4) 0.000 000 005 936 968 8 × 2 = 0 + 0.000 000 011 873 937 6;
  • 5) 0.000 000 011 873 937 6 × 2 = 0 + 0.000 000 023 747 875 2;
  • 6) 0.000 000 023 747 875 2 × 2 = 0 + 0.000 000 047 495 750 4;
  • 7) 0.000 000 047 495 750 4 × 2 = 0 + 0.000 000 094 991 500 8;
  • 8) 0.000 000 094 991 500 8 × 2 = 0 + 0.000 000 189 983 001 6;
  • 9) 0.000 000 189 983 001 6 × 2 = 0 + 0.000 000 379 966 003 2;
  • 10) 0.000 000 379 966 003 2 × 2 = 0 + 0.000 000 759 932 006 4;
  • 11) 0.000 000 759 932 006 4 × 2 = 0 + 0.000 001 519 864 012 8;
  • 12) 0.000 001 519 864 012 8 × 2 = 0 + 0.000 003 039 728 025 6;
  • 13) 0.000 003 039 728 025 6 × 2 = 0 + 0.000 006 079 456 051 2;
  • 14) 0.000 006 079 456 051 2 × 2 = 0 + 0.000 012 158 912 102 4;
  • 15) 0.000 012 158 912 102 4 × 2 = 0 + 0.000 024 317 824 204 8;
  • 16) 0.000 024 317 824 204 8 × 2 = 0 + 0.000 048 635 648 409 6;
  • 17) 0.000 048 635 648 409 6 × 2 = 0 + 0.000 097 271 296 819 2;
  • 18) 0.000 097 271 296 819 2 × 2 = 0 + 0.000 194 542 593 638 4;
  • 19) 0.000 194 542 593 638 4 × 2 = 0 + 0.000 389 085 187 276 8;
  • 20) 0.000 389 085 187 276 8 × 2 = 0 + 0.000 778 170 374 553 6;
  • 21) 0.000 778 170 374 553 6 × 2 = 0 + 0.001 556 340 749 107 2;
  • 22) 0.001 556 340 749 107 2 × 2 = 0 + 0.003 112 681 498 214 4;
  • 23) 0.003 112 681 498 214 4 × 2 = 0 + 0.006 225 362 996 428 8;
  • 24) 0.006 225 362 996 428 8 × 2 = 0 + 0.012 450 725 992 857 6;
  • 25) 0.012 450 725 992 857 6 × 2 = 0 + 0.024 901 451 985 715 2;
  • 26) 0.024 901 451 985 715 2 × 2 = 0 + 0.049 802 903 971 430 4;
  • 27) 0.049 802 903 971 430 4 × 2 = 0 + 0.099 605 807 942 860 8;
  • 28) 0.099 605 807 942 860 8 × 2 = 0 + 0.199 211 615 885 721 6;
  • 29) 0.199 211 615 885 721 6 × 2 = 0 + 0.398 423 231 771 443 2;
  • 30) 0.398 423 231 771 443 2 × 2 = 0 + 0.796 846 463 542 886 4;
  • 31) 0.796 846 463 542 886 4 × 2 = 1 + 0.593 692 927 085 772 8;
  • 32) 0.593 692 927 085 772 8 × 2 = 1 + 0.187 385 854 171 545 6;
  • 33) 0.187 385 854 171 545 6 × 2 = 0 + 0.374 771 708 343 091 2;
  • 34) 0.374 771 708 343 091 2 × 2 = 0 + 0.749 543 416 686 182 4;
  • 35) 0.749 543 416 686 182 4 × 2 = 1 + 0.499 086 833 372 364 8;
  • 36) 0.499 086 833 372 364 8 × 2 = 0 + 0.998 173 666 744 729 6;
  • 37) 0.998 173 666 744 729 6 × 2 = 1 + 0.996 347 333 489 459 2;
  • 38) 0.996 347 333 489 459 2 × 2 = 1 + 0.992 694 666 978 918 4;
  • 39) 0.992 694 666 978 918 4 × 2 = 1 + 0.985 389 333 957 836 8;
  • 40) 0.985 389 333 957 836 8 × 2 = 1 + 0.970 778 667 915 673 6;
  • 41) 0.970 778 667 915 673 6 × 2 = 1 + 0.941 557 335 831 347 2;
  • 42) 0.941 557 335 831 347 2 × 2 = 1 + 0.883 114 671 662 694 4;
  • 43) 0.883 114 671 662 694 4 × 2 = 1 + 0.766 229 343 325 388 8;
  • 44) 0.766 229 343 325 388 8 × 2 = 1 + 0.532 458 686 650 777 6;
  • 45) 0.532 458 686 650 777 6 × 2 = 1 + 0.064 917 373 301 555 2;
  • 46) 0.064 917 373 301 555 2 × 2 = 0 + 0.129 834 746 603 110 4;
  • 47) 0.129 834 746 603 110 4 × 2 = 0 + 0.259 669 493 206 220 8;
  • 48) 0.259 669 493 206 220 8 × 2 = 0 + 0.519 338 986 412 441 6;
  • 49) 0.519 338 986 412 441 6 × 2 = 1 + 0.038 677 972 824 883 2;
  • 50) 0.038 677 972 824 883 2 × 2 = 0 + 0.077 355 945 649 766 4;
  • 51) 0.077 355 945 649 766 4 × 2 = 0 + 0.154 711 891 299 532 8;
  • 52) 0.154 711 891 299 532 8 × 2 = 0 + 0.309 423 782 599 065 6;
  • 53) 0.309 423 782 599 065 6 × 2 = 0 + 0.618 847 565 198 131 2;
  • 54) 0.618 847 565 198 131 2 × 2 = 1 + 0.237 695 130 396 262 4;

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:


0.000 000 000 742 121 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 1000 01(2)

6. Positive number before normalization:

0.000 000 000 742 121 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 1000 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:


0.000 000 000 742 121 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 1000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1000 1000 01(2) × 20 =

1.1001 0111 1111 1100 0100 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 1100 0100 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 1110 0010 0001 =

100 1011 1111 1110 0010 0001


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 1110 0010 0001


Decimal number -0.000 000 000 742 121 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 0010 0001

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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