-0.000 000 000 742 147 444 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 444(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
-0.000 000 000 742 147 444(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:

|-0.000 000 000 742 147 444| = 0.000 000 000 742 147 444


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: 0.000 000 000 742 147 444.

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) 0.000 000 000 742 147 444 × 2 = 0 + 0.000 000 001 484 294 888;
  • 2) 0.000 000 001 484 294 888 × 2 = 0 + 0.000 000 002 968 589 776;
  • 3) 0.000 000 002 968 589 776 × 2 = 0 + 0.000 000 005 937 179 552;
  • 4) 0.000 000 005 937 179 552 × 2 = 0 + 0.000 000 011 874 359 104;
  • 5) 0.000 000 011 874 359 104 × 2 = 0 + 0.000 000 023 748 718 208;
  • 6) 0.000 000 023 748 718 208 × 2 = 0 + 0.000 000 047 497 436 416;
  • 7) 0.000 000 047 497 436 416 × 2 = 0 + 0.000 000 094 994 872 832;
  • 8) 0.000 000 094 994 872 832 × 2 = 0 + 0.000 000 189 989 745 664;
  • 9) 0.000 000 189 989 745 664 × 2 = 0 + 0.000 000 379 979 491 328;
  • 10) 0.000 000 379 979 491 328 × 2 = 0 + 0.000 000 759 958 982 656;
  • 11) 0.000 000 759 958 982 656 × 2 = 0 + 0.000 001 519 917 965 312;
  • 12) 0.000 001 519 917 965 312 × 2 = 0 + 0.000 003 039 835 930 624;
  • 13) 0.000 003 039 835 930 624 × 2 = 0 + 0.000 006 079 671 861 248;
  • 14) 0.000 006 079 671 861 248 × 2 = 0 + 0.000 012 159 343 722 496;
  • 15) 0.000 012 159 343 722 496 × 2 = 0 + 0.000 024 318 687 444 992;
  • 16) 0.000 024 318 687 444 992 × 2 = 0 + 0.000 048 637 374 889 984;
  • 17) 0.000 048 637 374 889 984 × 2 = 0 + 0.000 097 274 749 779 968;
  • 18) 0.000 097 274 749 779 968 × 2 = 0 + 0.000 194 549 499 559 936;
  • 19) 0.000 194 549 499 559 936 × 2 = 0 + 0.000 389 098 999 119 872;
  • 20) 0.000 389 098 999 119 872 × 2 = 0 + 0.000 778 197 998 239 744;
  • 21) 0.000 778 197 998 239 744 × 2 = 0 + 0.001 556 395 996 479 488;
  • 22) 0.001 556 395 996 479 488 × 2 = 0 + 0.003 112 791 992 958 976;
  • 23) 0.003 112 791 992 958 976 × 2 = 0 + 0.006 225 583 985 917 952;
  • 24) 0.006 225 583 985 917 952 × 2 = 0 + 0.012 451 167 971 835 904;
  • 25) 0.012 451 167 971 835 904 × 2 = 0 + 0.024 902 335 943 671 808;
  • 26) 0.024 902 335 943 671 808 × 2 = 0 + 0.049 804 671 887 343 616;
  • 27) 0.049 804 671 887 343 616 × 2 = 0 + 0.099 609 343 774 687 232;
  • 28) 0.099 609 343 774 687 232 × 2 = 0 + 0.199 218 687 549 374 464;
  • 29) 0.199 218 687 549 374 464 × 2 = 0 + 0.398 437 375 098 748 928;
  • 30) 0.398 437 375 098 748 928 × 2 = 0 + 0.796 874 750 197 497 856;
  • 31) 0.796 874 750 197 497 856 × 2 = 1 + 0.593 749 500 394 995 712;
  • 32) 0.593 749 500 394 995 712 × 2 = 1 + 0.187 499 000 789 991 424;
  • 33) 0.187 499 000 789 991 424 × 2 = 0 + 0.374 998 001 579 982 848;
  • 34) 0.374 998 001 579 982 848 × 2 = 0 + 0.749 996 003 159 965 696;
  • 35) 0.749 996 003 159 965 696 × 2 = 1 + 0.499 992 006 319 931 392;
  • 36) 0.499 992 006 319 931 392 × 2 = 0 + 0.999 984 012 639 862 784;
  • 37) 0.999 984 012 639 862 784 × 2 = 1 + 0.999 968 025 279 725 568;
  • 38) 0.999 968 025 279 725 568 × 2 = 1 + 0.999 936 050 559 451 136;
  • 39) 0.999 936 050 559 451 136 × 2 = 1 + 0.999 872 101 118 902 272;
  • 40) 0.999 872 101 118 902 272 × 2 = 1 + 0.999 744 202 237 804 544;
  • 41) 0.999 744 202 237 804 544 × 2 = 1 + 0.999 488 404 475 609 088;
  • 42) 0.999 488 404 475 609 088 × 2 = 1 + 0.998 976 808 951 218 176;
  • 43) 0.998 976 808 951 218 176 × 2 = 1 + 0.997 953 617 902 436 352;
  • 44) 0.997 953 617 902 436 352 × 2 = 1 + 0.995 907 235 804 872 704;
  • 45) 0.995 907 235 804 872 704 × 2 = 1 + 0.991 814 471 609 745 408;
  • 46) 0.991 814 471 609 745 408 × 2 = 1 + 0.983 628 943 219 490 816;
  • 47) 0.983 628 943 219 490 816 × 2 = 1 + 0.967 257 886 438 981 632;
  • 48) 0.967 257 886 438 981 632 × 2 = 1 + 0.934 515 772 877 963 264;
  • 49) 0.934 515 772 877 963 264 × 2 = 1 + 0.869 031 545 755 926 528;
  • 50) 0.869 031 545 755 926 528 × 2 = 1 + 0.738 063 091 511 853 056;
  • 51) 0.738 063 091 511 853 056 × 2 = 1 + 0.476 126 183 023 706 112;
  • 52) 0.476 126 183 023 706 112 × 2 = 0 + 0.952 252 366 047 412 224;
  • 53) 0.952 252 366 047 412 224 × 2 = 1 + 0.904 504 732 094 824 448;
  • 54) 0.904 504 732 094 824 448 × 2 = 1 + 0.809 009 464 189 648 896;

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:


0.000 000 000 742 147 444(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 444(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 444(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 011


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 1011 =

100 1011 1111 1111 1111 1011


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 1011


Decimal number -0.000 000 000 742 147 444 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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