-0.000 000 000 742 146 74 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 74(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 74(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 74| = 0.000 000 000 742 146 74


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 74.

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 74 × 2 = 0 + 0.000 000 001 484 293 48;
  • 2) 0.000 000 001 484 293 48 × 2 = 0 + 0.000 000 002 968 586 96;
  • 3) 0.000 000 002 968 586 96 × 2 = 0 + 0.000 000 005 937 173 92;
  • 4) 0.000 000 005 937 173 92 × 2 = 0 + 0.000 000 011 874 347 84;
  • 5) 0.000 000 011 874 347 84 × 2 = 0 + 0.000 000 023 748 695 68;
  • 6) 0.000 000 023 748 695 68 × 2 = 0 + 0.000 000 047 497 391 36;
  • 7) 0.000 000 047 497 391 36 × 2 = 0 + 0.000 000 094 994 782 72;
  • 8) 0.000 000 094 994 782 72 × 2 = 0 + 0.000 000 189 989 565 44;
  • 9) 0.000 000 189 989 565 44 × 2 = 0 + 0.000 000 379 979 130 88;
  • 10) 0.000 000 379 979 130 88 × 2 = 0 + 0.000 000 759 958 261 76;
  • 11) 0.000 000 759 958 261 76 × 2 = 0 + 0.000 001 519 916 523 52;
  • 12) 0.000 001 519 916 523 52 × 2 = 0 + 0.000 003 039 833 047 04;
  • 13) 0.000 003 039 833 047 04 × 2 = 0 + 0.000 006 079 666 094 08;
  • 14) 0.000 006 079 666 094 08 × 2 = 0 + 0.000 012 159 332 188 16;
  • 15) 0.000 012 159 332 188 16 × 2 = 0 + 0.000 024 318 664 376 32;
  • 16) 0.000 024 318 664 376 32 × 2 = 0 + 0.000 048 637 328 752 64;
  • 17) 0.000 048 637 328 752 64 × 2 = 0 + 0.000 097 274 657 505 28;
  • 18) 0.000 097 274 657 505 28 × 2 = 0 + 0.000 194 549 315 010 56;
  • 19) 0.000 194 549 315 010 56 × 2 = 0 + 0.000 389 098 630 021 12;
  • 20) 0.000 389 098 630 021 12 × 2 = 0 + 0.000 778 197 260 042 24;
  • 21) 0.000 778 197 260 042 24 × 2 = 0 + 0.001 556 394 520 084 48;
  • 22) 0.001 556 394 520 084 48 × 2 = 0 + 0.003 112 789 040 168 96;
  • 23) 0.003 112 789 040 168 96 × 2 = 0 + 0.006 225 578 080 337 92;
  • 24) 0.006 225 578 080 337 92 × 2 = 0 + 0.012 451 156 160 675 84;
  • 25) 0.012 451 156 160 675 84 × 2 = 0 + 0.024 902 312 321 351 68;
  • 26) 0.024 902 312 321 351 68 × 2 = 0 + 0.049 804 624 642 703 36;
  • 27) 0.049 804 624 642 703 36 × 2 = 0 + 0.099 609 249 285 406 72;
  • 28) 0.099 609 249 285 406 72 × 2 = 0 + 0.199 218 498 570 813 44;
  • 29) 0.199 218 498 570 813 44 × 2 = 0 + 0.398 436 997 141 626 88;
  • 30) 0.398 436 997 141 626 88 × 2 = 0 + 0.796 873 994 283 253 76;
  • 31) 0.796 873 994 283 253 76 × 2 = 1 + 0.593 747 988 566 507 52;
  • 32) 0.593 747 988 566 507 52 × 2 = 1 + 0.187 495 977 133 015 04;
  • 33) 0.187 495 977 133 015 04 × 2 = 0 + 0.374 991 954 266 030 08;
  • 34) 0.374 991 954 266 030 08 × 2 = 0 + 0.749 983 908 532 060 16;
  • 35) 0.749 983 908 532 060 16 × 2 = 1 + 0.499 967 817 064 120 32;
  • 36) 0.499 967 817 064 120 32 × 2 = 0 + 0.999 935 634 128 240 64;
  • 37) 0.999 935 634 128 240 64 × 2 = 1 + 0.999 871 268 256 481 28;
  • 38) 0.999 871 268 256 481 28 × 2 = 1 + 0.999 742 536 512 962 56;
  • 39) 0.999 742 536 512 962 56 × 2 = 1 + 0.999 485 073 025 925 12;
  • 40) 0.999 485 073 025 925 12 × 2 = 1 + 0.998 970 146 051 850 24;
  • 41) 0.998 970 146 051 850 24 × 2 = 1 + 0.997 940 292 103 700 48;
  • 42) 0.997 940 292 103 700 48 × 2 = 1 + 0.995 880 584 207 400 96;
  • 43) 0.995 880 584 207 400 96 × 2 = 1 + 0.991 761 168 414 801 92;
  • 44) 0.991 761 168 414 801 92 × 2 = 1 + 0.983 522 336 829 603 84;
  • 45) 0.983 522 336 829 603 84 × 2 = 1 + 0.967 044 673 659 207 68;
  • 46) 0.967 044 673 659 207 68 × 2 = 1 + 0.934 089 347 318 415 36;
  • 47) 0.934 089 347 318 415 36 × 2 = 1 + 0.868 178 694 636 830 72;
  • 48) 0.868 178 694 636 830 72 × 2 = 1 + 0.736 357 389 273 661 44;
  • 49) 0.736 357 389 273 661 44 × 2 = 1 + 0.472 714 778 547 322 88;
  • 50) 0.472 714 778 547 322 88 × 2 = 0 + 0.945 429 557 094 645 76;
  • 51) 0.945 429 557 094 645 76 × 2 = 1 + 0.890 859 114 189 291 52;
  • 52) 0.890 859 114 189 291 52 × 2 = 1 + 0.781 718 228 378 583 04;
  • 53) 0.781 718 228 378 583 04 × 2 = 1 + 0.563 436 456 757 166 08;
  • 54) 0.563 436 456 757 166 08 × 2 = 1 + 0.126 872 913 514 332 16;

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 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 11(2)

6. Positive number before normalization:

0.000 000 000 742 146 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 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 74(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 11(2) × 20 =

1.1001 0111 1111 1111 1101 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 1101 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 1111 =

100 1011 1111 1111 1110 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 1110 1111


Decimal number -0.000 000 000 742 146 74 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 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