-0.000 000 000 742 147 673 08 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 08(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
-0.000 000 000 742 147 673 08(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:

|-0.000 000 000 742 147 673 08| = 0.000 000 000 742 147 673 08


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: 0.000 000 000 742 147 673 08.

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) 0.000 000 000 742 147 673 08 × 2 = 0 + 0.000 000 001 484 295 346 16;
  • 2) 0.000 000 001 484 295 346 16 × 2 = 0 + 0.000 000 002 968 590 692 32;
  • 3) 0.000 000 002 968 590 692 32 × 2 = 0 + 0.000 000 005 937 181 384 64;
  • 4) 0.000 000 005 937 181 384 64 × 2 = 0 + 0.000 000 011 874 362 769 28;
  • 5) 0.000 000 011 874 362 769 28 × 2 = 0 + 0.000 000 023 748 725 538 56;
  • 6) 0.000 000 023 748 725 538 56 × 2 = 0 + 0.000 000 047 497 451 077 12;
  • 7) 0.000 000 047 497 451 077 12 × 2 = 0 + 0.000 000 094 994 902 154 24;
  • 8) 0.000 000 094 994 902 154 24 × 2 = 0 + 0.000 000 189 989 804 308 48;
  • 9) 0.000 000 189 989 804 308 48 × 2 = 0 + 0.000 000 379 979 608 616 96;
  • 10) 0.000 000 379 979 608 616 96 × 2 = 0 + 0.000 000 759 959 217 233 92;
  • 11) 0.000 000 759 959 217 233 92 × 2 = 0 + 0.000 001 519 918 434 467 84;
  • 12) 0.000 001 519 918 434 467 84 × 2 = 0 + 0.000 003 039 836 868 935 68;
  • 13) 0.000 003 039 836 868 935 68 × 2 = 0 + 0.000 006 079 673 737 871 36;
  • 14) 0.000 006 079 673 737 871 36 × 2 = 0 + 0.000 012 159 347 475 742 72;
  • 15) 0.000 012 159 347 475 742 72 × 2 = 0 + 0.000 024 318 694 951 485 44;
  • 16) 0.000 024 318 694 951 485 44 × 2 = 0 + 0.000 048 637 389 902 970 88;
  • 17) 0.000 048 637 389 902 970 88 × 2 = 0 + 0.000 097 274 779 805 941 76;
  • 18) 0.000 097 274 779 805 941 76 × 2 = 0 + 0.000 194 549 559 611 883 52;
  • 19) 0.000 194 549 559 611 883 52 × 2 = 0 + 0.000 389 099 119 223 767 04;
  • 20) 0.000 389 099 119 223 767 04 × 2 = 0 + 0.000 778 198 238 447 534 08;
  • 21) 0.000 778 198 238 447 534 08 × 2 = 0 + 0.001 556 396 476 895 068 16;
  • 22) 0.001 556 396 476 895 068 16 × 2 = 0 + 0.003 112 792 953 790 136 32;
  • 23) 0.003 112 792 953 790 136 32 × 2 = 0 + 0.006 225 585 907 580 272 64;
  • 24) 0.006 225 585 907 580 272 64 × 2 = 0 + 0.012 451 171 815 160 545 28;
  • 25) 0.012 451 171 815 160 545 28 × 2 = 0 + 0.024 902 343 630 321 090 56;
  • 26) 0.024 902 343 630 321 090 56 × 2 = 0 + 0.049 804 687 260 642 181 12;
  • 27) 0.049 804 687 260 642 181 12 × 2 = 0 + 0.099 609 374 521 284 362 24;
  • 28) 0.099 609 374 521 284 362 24 × 2 = 0 + 0.199 218 749 042 568 724 48;
  • 29) 0.199 218 749 042 568 724 48 × 2 = 0 + 0.398 437 498 085 137 448 96;
  • 30) 0.398 437 498 085 137 448 96 × 2 = 0 + 0.796 874 996 170 274 897 92;
  • 31) 0.796 874 996 170 274 897 92 × 2 = 1 + 0.593 749 992 340 549 795 84;
  • 32) 0.593 749 992 340 549 795 84 × 2 = 1 + 0.187 499 984 681 099 591 68;
  • 33) 0.187 499 984 681 099 591 68 × 2 = 0 + 0.374 999 969 362 199 183 36;
  • 34) 0.374 999 969 362 199 183 36 × 2 = 0 + 0.749 999 938 724 398 366 72;
  • 35) 0.749 999 938 724 398 366 72 × 2 = 1 + 0.499 999 877 448 796 733 44;
  • 36) 0.499 999 877 448 796 733 44 × 2 = 0 + 0.999 999 754 897 593 466 88;
  • 37) 0.999 999 754 897 593 466 88 × 2 = 1 + 0.999 999 509 795 186 933 76;
  • 38) 0.999 999 509 795 186 933 76 × 2 = 1 + 0.999 999 019 590 373 867 52;
  • 39) 0.999 999 019 590 373 867 52 × 2 = 1 + 0.999 998 039 180 747 735 04;
  • 40) 0.999 998 039 180 747 735 04 × 2 = 1 + 0.999 996 078 361 495 470 08;
  • 41) 0.999 996 078 361 495 470 08 × 2 = 1 + 0.999 992 156 722 990 940 16;
  • 42) 0.999 992 156 722 990 940 16 × 2 = 1 + 0.999 984 313 445 981 880 32;
  • 43) 0.999 984 313 445 981 880 32 × 2 = 1 + 0.999 968 626 891 963 760 64;
  • 44) 0.999 968 626 891 963 760 64 × 2 = 1 + 0.999 937 253 783 927 521 28;
  • 45) 0.999 937 253 783 927 521 28 × 2 = 1 + 0.999 874 507 567 855 042 56;
  • 46) 0.999 874 507 567 855 042 56 × 2 = 1 + 0.999 749 015 135 710 085 12;
  • 47) 0.999 749 015 135 710 085 12 × 2 = 1 + 0.999 498 030 271 420 170 24;
  • 48) 0.999 498 030 271 420 170 24 × 2 = 1 + 0.998 996 060 542 840 340 48;
  • 49) 0.998 996 060 542 840 340 48 × 2 = 1 + 0.997 992 121 085 680 680 96;
  • 50) 0.997 992 121 085 680 680 96 × 2 = 1 + 0.995 984 242 171 361 361 92;
  • 51) 0.995 984 242 171 361 361 92 × 2 = 1 + 0.991 968 484 342 722 723 84;
  • 52) 0.991 968 484 342 722 723 84 × 2 = 1 + 0.983 936 968 685 445 447 68;
  • 53) 0.983 936 968 685 445 447 68 × 2 = 1 + 0.967 873 937 370 890 895 36;
  • 54) 0.967 873 937 370 890 895 36 × 2 = 1 + 0.935 747 874 741 781 790 72;

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:


0.000 000 000 742 147 673 08(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 08(10) =

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


0.000 000 000 742 147 673 08(10) =

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

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

1.1001 0111 1111 1111 1111 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 1111 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 1111 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 673 08 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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