-0.000 000 000 742 138 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 138 7(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
-0.000 000 000 742 138 7(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:

|-0.000 000 000 742 138 7| = 0.000 000 000 742 138 7


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: 0.000 000 000 742 138 7.

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) 0.000 000 000 742 138 7 × 2 = 0 + 0.000 000 001 484 277 4;
  • 2) 0.000 000 001 484 277 4 × 2 = 0 + 0.000 000 002 968 554 8;
  • 3) 0.000 000 002 968 554 8 × 2 = 0 + 0.000 000 005 937 109 6;
  • 4) 0.000 000 005 937 109 6 × 2 = 0 + 0.000 000 011 874 219 2;
  • 5) 0.000 000 011 874 219 2 × 2 = 0 + 0.000 000 023 748 438 4;
  • 6) 0.000 000 023 748 438 4 × 2 = 0 + 0.000 000 047 496 876 8;
  • 7) 0.000 000 047 496 876 8 × 2 = 0 + 0.000 000 094 993 753 6;
  • 8) 0.000 000 094 993 753 6 × 2 = 0 + 0.000 000 189 987 507 2;
  • 9) 0.000 000 189 987 507 2 × 2 = 0 + 0.000 000 379 975 014 4;
  • 10) 0.000 000 379 975 014 4 × 2 = 0 + 0.000 000 759 950 028 8;
  • 11) 0.000 000 759 950 028 8 × 2 = 0 + 0.000 001 519 900 057 6;
  • 12) 0.000 001 519 900 057 6 × 2 = 0 + 0.000 003 039 800 115 2;
  • 13) 0.000 003 039 800 115 2 × 2 = 0 + 0.000 006 079 600 230 4;
  • 14) 0.000 006 079 600 230 4 × 2 = 0 + 0.000 012 159 200 460 8;
  • 15) 0.000 012 159 200 460 8 × 2 = 0 + 0.000 024 318 400 921 6;
  • 16) 0.000 024 318 400 921 6 × 2 = 0 + 0.000 048 636 801 843 2;
  • 17) 0.000 048 636 801 843 2 × 2 = 0 + 0.000 097 273 603 686 4;
  • 18) 0.000 097 273 603 686 4 × 2 = 0 + 0.000 194 547 207 372 8;
  • 19) 0.000 194 547 207 372 8 × 2 = 0 + 0.000 389 094 414 745 6;
  • 20) 0.000 389 094 414 745 6 × 2 = 0 + 0.000 778 188 829 491 2;
  • 21) 0.000 778 188 829 491 2 × 2 = 0 + 0.001 556 377 658 982 4;
  • 22) 0.001 556 377 658 982 4 × 2 = 0 + 0.003 112 755 317 964 8;
  • 23) 0.003 112 755 317 964 8 × 2 = 0 + 0.006 225 510 635 929 6;
  • 24) 0.006 225 510 635 929 6 × 2 = 0 + 0.012 451 021 271 859 2;
  • 25) 0.012 451 021 271 859 2 × 2 = 0 + 0.024 902 042 543 718 4;
  • 26) 0.024 902 042 543 718 4 × 2 = 0 + 0.049 804 085 087 436 8;
  • 27) 0.049 804 085 087 436 8 × 2 = 0 + 0.099 608 170 174 873 6;
  • 28) 0.099 608 170 174 873 6 × 2 = 0 + 0.199 216 340 349 747 2;
  • 29) 0.199 216 340 349 747 2 × 2 = 0 + 0.398 432 680 699 494 4;
  • 30) 0.398 432 680 699 494 4 × 2 = 0 + 0.796 865 361 398 988 8;
  • 31) 0.796 865 361 398 988 8 × 2 = 1 + 0.593 730 722 797 977 6;
  • 32) 0.593 730 722 797 977 6 × 2 = 1 + 0.187 461 445 595 955 2;
  • 33) 0.187 461 445 595 955 2 × 2 = 0 + 0.374 922 891 191 910 4;
  • 34) 0.374 922 891 191 910 4 × 2 = 0 + 0.749 845 782 383 820 8;
  • 35) 0.749 845 782 383 820 8 × 2 = 1 + 0.499 691 564 767 641 6;
  • 36) 0.499 691 564 767 641 6 × 2 = 0 + 0.999 383 129 535 283 2;
  • 37) 0.999 383 129 535 283 2 × 2 = 1 + 0.998 766 259 070 566 4;
  • 38) 0.998 766 259 070 566 4 × 2 = 1 + 0.997 532 518 141 132 8;
  • 39) 0.997 532 518 141 132 8 × 2 = 1 + 0.995 065 036 282 265 6;
  • 40) 0.995 065 036 282 265 6 × 2 = 1 + 0.990 130 072 564 531 2;
  • 41) 0.990 130 072 564 531 2 × 2 = 1 + 0.980 260 145 129 062 4;
  • 42) 0.980 260 145 129 062 4 × 2 = 1 + 0.960 520 290 258 124 8;
  • 43) 0.960 520 290 258 124 8 × 2 = 1 + 0.921 040 580 516 249 6;
  • 44) 0.921 040 580 516 249 6 × 2 = 1 + 0.842 081 161 032 499 2;
  • 45) 0.842 081 161 032 499 2 × 2 = 1 + 0.684 162 322 064 998 4;
  • 46) 0.684 162 322 064 998 4 × 2 = 1 + 0.368 324 644 129 996 8;
  • 47) 0.368 324 644 129 996 8 × 2 = 0 + 0.736 649 288 259 993 6;
  • 48) 0.736 649 288 259 993 6 × 2 = 1 + 0.473 298 576 519 987 2;
  • 49) 0.473 298 576 519 987 2 × 2 = 0 + 0.946 597 153 039 974 4;
  • 50) 0.946 597 153 039 974 4 × 2 = 1 + 0.893 194 306 079 948 8;
  • 51) 0.893 194 306 079 948 8 × 2 = 1 + 0.786 388 612 159 897 6;
  • 52) 0.786 388 612 159 897 6 × 2 = 1 + 0.572 777 224 319 795 2;
  • 53) 0.572 777 224 319 795 2 × 2 = 1 + 0.145 554 448 639 590 4;
  • 54) 0.145 554 448 639 590 4 × 2 = 0 + 0.291 108 897 279 180 8;

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:


0.000 000 000 742 138 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0111 10(2)

6. Positive number before normalization:

0.000 000 000 742 138 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0111 10(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:


0.000 000 000 742 138 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0111 10(2) × 20 =

1.1001 0111 1111 1110 1011 110(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 1110 1011 110


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 0101 1110 =

100 1011 1111 1111 0101 1110


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 0101 1110


Decimal number -0.000 000 000 742 138 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 0101 1110

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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