-0.000 000 000 742 149 04 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 149 04(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
-0.000 000 000 742 149 04(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:

|-0.000 000 000 742 149 04| = 0.000 000 000 742 149 04


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: 0.000 000 000 742 149 04.

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) 0.000 000 000 742 149 04 × 2 = 0 + 0.000 000 001 484 298 08;
  • 2) 0.000 000 001 484 298 08 × 2 = 0 + 0.000 000 002 968 596 16;
  • 3) 0.000 000 002 968 596 16 × 2 = 0 + 0.000 000 005 937 192 32;
  • 4) 0.000 000 005 937 192 32 × 2 = 0 + 0.000 000 011 874 384 64;
  • 5) 0.000 000 011 874 384 64 × 2 = 0 + 0.000 000 023 748 769 28;
  • 6) 0.000 000 023 748 769 28 × 2 = 0 + 0.000 000 047 497 538 56;
  • 7) 0.000 000 047 497 538 56 × 2 = 0 + 0.000 000 094 995 077 12;
  • 8) 0.000 000 094 995 077 12 × 2 = 0 + 0.000 000 189 990 154 24;
  • 9) 0.000 000 189 990 154 24 × 2 = 0 + 0.000 000 379 980 308 48;
  • 10) 0.000 000 379 980 308 48 × 2 = 0 + 0.000 000 759 960 616 96;
  • 11) 0.000 000 759 960 616 96 × 2 = 0 + 0.000 001 519 921 233 92;
  • 12) 0.000 001 519 921 233 92 × 2 = 0 + 0.000 003 039 842 467 84;
  • 13) 0.000 003 039 842 467 84 × 2 = 0 + 0.000 006 079 684 935 68;
  • 14) 0.000 006 079 684 935 68 × 2 = 0 + 0.000 012 159 369 871 36;
  • 15) 0.000 012 159 369 871 36 × 2 = 0 + 0.000 024 318 739 742 72;
  • 16) 0.000 024 318 739 742 72 × 2 = 0 + 0.000 048 637 479 485 44;
  • 17) 0.000 048 637 479 485 44 × 2 = 0 + 0.000 097 274 958 970 88;
  • 18) 0.000 097 274 958 970 88 × 2 = 0 + 0.000 194 549 917 941 76;
  • 19) 0.000 194 549 917 941 76 × 2 = 0 + 0.000 389 099 835 883 52;
  • 20) 0.000 389 099 835 883 52 × 2 = 0 + 0.000 778 199 671 767 04;
  • 21) 0.000 778 199 671 767 04 × 2 = 0 + 0.001 556 399 343 534 08;
  • 22) 0.001 556 399 343 534 08 × 2 = 0 + 0.003 112 798 687 068 16;
  • 23) 0.003 112 798 687 068 16 × 2 = 0 + 0.006 225 597 374 136 32;
  • 24) 0.006 225 597 374 136 32 × 2 = 0 + 0.012 451 194 748 272 64;
  • 25) 0.012 451 194 748 272 64 × 2 = 0 + 0.024 902 389 496 545 28;
  • 26) 0.024 902 389 496 545 28 × 2 = 0 + 0.049 804 778 993 090 56;
  • 27) 0.049 804 778 993 090 56 × 2 = 0 + 0.099 609 557 986 181 12;
  • 28) 0.099 609 557 986 181 12 × 2 = 0 + 0.199 219 115 972 362 24;
  • 29) 0.199 219 115 972 362 24 × 2 = 0 + 0.398 438 231 944 724 48;
  • 30) 0.398 438 231 944 724 48 × 2 = 0 + 0.796 876 463 889 448 96;
  • 31) 0.796 876 463 889 448 96 × 2 = 1 + 0.593 752 927 778 897 92;
  • 32) 0.593 752 927 778 897 92 × 2 = 1 + 0.187 505 855 557 795 84;
  • 33) 0.187 505 855 557 795 84 × 2 = 0 + 0.375 011 711 115 591 68;
  • 34) 0.375 011 711 115 591 68 × 2 = 0 + 0.750 023 422 231 183 36;
  • 35) 0.750 023 422 231 183 36 × 2 = 1 + 0.500 046 844 462 366 72;
  • 36) 0.500 046 844 462 366 72 × 2 = 1 + 0.000 093 688 924 733 44;
  • 37) 0.000 093 688 924 733 44 × 2 = 0 + 0.000 187 377 849 466 88;
  • 38) 0.000 187 377 849 466 88 × 2 = 0 + 0.000 374 755 698 933 76;
  • 39) 0.000 374 755 698 933 76 × 2 = 0 + 0.000 749 511 397 867 52;
  • 40) 0.000 749 511 397 867 52 × 2 = 0 + 0.001 499 022 795 735 04;
  • 41) 0.001 499 022 795 735 04 × 2 = 0 + 0.002 998 045 591 470 08;
  • 42) 0.002 998 045 591 470 08 × 2 = 0 + 0.005 996 091 182 940 16;
  • 43) 0.005 996 091 182 940 16 × 2 = 0 + 0.011 992 182 365 880 32;
  • 44) 0.011 992 182 365 880 32 × 2 = 0 + 0.023 984 364 731 760 64;
  • 45) 0.023 984 364 731 760 64 × 2 = 0 + 0.047 968 729 463 521 28;
  • 46) 0.047 968 729 463 521 28 × 2 = 0 + 0.095 937 458 927 042 56;
  • 47) 0.095 937 458 927 042 56 × 2 = 0 + 0.191 874 917 854 085 12;
  • 48) 0.191 874 917 854 085 12 × 2 = 0 + 0.383 749 835 708 170 24;
  • 49) 0.383 749 835 708 170 24 × 2 = 0 + 0.767 499 671 416 340 48;
  • 50) 0.767 499 671 416 340 48 × 2 = 1 + 0.534 999 342 832 680 96;
  • 51) 0.534 999 342 832 680 96 × 2 = 1 + 0.069 998 685 665 361 92;
  • 52) 0.069 998 685 665 361 92 × 2 = 0 + 0.139 997 371 330 723 84;
  • 53) 0.139 997 371 330 723 84 × 2 = 0 + 0.279 994 742 661 447 68;
  • 54) 0.279 994 742 661 447 68 × 2 = 0 + 0.559 989 485 322 895 36;

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:


0.000 000 000 742 149 04(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 00(2)

6. Positive number before normalization:

0.000 000 000 742 149 04(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 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:


0.000 000 000 742 149 04(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0110 00(2) × 20 =

1.1001 1000 0000 0000 0011 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 1000 0000 0000 0011 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 1100 0000 0000 0001 1000 =

100 1100 0000 0000 0001 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 1100 0000 0000 0001 1000


Decimal number -0.000 000 000 742 149 04 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0001 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