-0.000 000 000 742 147 675 27 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 27(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
-0.000 000 000 742 147 675 27(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:

|-0.000 000 000 742 147 675 27| = 0.000 000 000 742 147 675 27


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: 0.000 000 000 742 147 675 27.

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) 0.000 000 000 742 147 675 27 × 2 = 0 + 0.000 000 001 484 295 350 54;
  • 2) 0.000 000 001 484 295 350 54 × 2 = 0 + 0.000 000 002 968 590 701 08;
  • 3) 0.000 000 002 968 590 701 08 × 2 = 0 + 0.000 000 005 937 181 402 16;
  • 4) 0.000 000 005 937 181 402 16 × 2 = 0 + 0.000 000 011 874 362 804 32;
  • 5) 0.000 000 011 874 362 804 32 × 2 = 0 + 0.000 000 023 748 725 608 64;
  • 6) 0.000 000 023 748 725 608 64 × 2 = 0 + 0.000 000 047 497 451 217 28;
  • 7) 0.000 000 047 497 451 217 28 × 2 = 0 + 0.000 000 094 994 902 434 56;
  • 8) 0.000 000 094 994 902 434 56 × 2 = 0 + 0.000 000 189 989 804 869 12;
  • 9) 0.000 000 189 989 804 869 12 × 2 = 0 + 0.000 000 379 979 609 738 24;
  • 10) 0.000 000 379 979 609 738 24 × 2 = 0 + 0.000 000 759 959 219 476 48;
  • 11) 0.000 000 759 959 219 476 48 × 2 = 0 + 0.000 001 519 918 438 952 96;
  • 12) 0.000 001 519 918 438 952 96 × 2 = 0 + 0.000 003 039 836 877 905 92;
  • 13) 0.000 003 039 836 877 905 92 × 2 = 0 + 0.000 006 079 673 755 811 84;
  • 14) 0.000 006 079 673 755 811 84 × 2 = 0 + 0.000 012 159 347 511 623 68;
  • 15) 0.000 012 159 347 511 623 68 × 2 = 0 + 0.000 024 318 695 023 247 36;
  • 16) 0.000 024 318 695 023 247 36 × 2 = 0 + 0.000 048 637 390 046 494 72;
  • 17) 0.000 048 637 390 046 494 72 × 2 = 0 + 0.000 097 274 780 092 989 44;
  • 18) 0.000 097 274 780 092 989 44 × 2 = 0 + 0.000 194 549 560 185 978 88;
  • 19) 0.000 194 549 560 185 978 88 × 2 = 0 + 0.000 389 099 120 371 957 76;
  • 20) 0.000 389 099 120 371 957 76 × 2 = 0 + 0.000 778 198 240 743 915 52;
  • 21) 0.000 778 198 240 743 915 52 × 2 = 0 + 0.001 556 396 481 487 831 04;
  • 22) 0.001 556 396 481 487 831 04 × 2 = 0 + 0.003 112 792 962 975 662 08;
  • 23) 0.003 112 792 962 975 662 08 × 2 = 0 + 0.006 225 585 925 951 324 16;
  • 24) 0.006 225 585 925 951 324 16 × 2 = 0 + 0.012 451 171 851 902 648 32;
  • 25) 0.012 451 171 851 902 648 32 × 2 = 0 + 0.024 902 343 703 805 296 64;
  • 26) 0.024 902 343 703 805 296 64 × 2 = 0 + 0.049 804 687 407 610 593 28;
  • 27) 0.049 804 687 407 610 593 28 × 2 = 0 + 0.099 609 374 815 221 186 56;
  • 28) 0.099 609 374 815 221 186 56 × 2 = 0 + 0.199 218 749 630 442 373 12;
  • 29) 0.199 218 749 630 442 373 12 × 2 = 0 + 0.398 437 499 260 884 746 24;
  • 30) 0.398 437 499 260 884 746 24 × 2 = 0 + 0.796 874 998 521 769 492 48;
  • 31) 0.796 874 998 521 769 492 48 × 2 = 1 + 0.593 749 997 043 538 984 96;
  • 32) 0.593 749 997 043 538 984 96 × 2 = 1 + 0.187 499 994 087 077 969 92;
  • 33) 0.187 499 994 087 077 969 92 × 2 = 0 + 0.374 999 988 174 155 939 84;
  • 34) 0.374 999 988 174 155 939 84 × 2 = 0 + 0.749 999 976 348 311 879 68;
  • 35) 0.749 999 976 348 311 879 68 × 2 = 1 + 0.499 999 952 696 623 759 36;
  • 36) 0.499 999 952 696 623 759 36 × 2 = 0 + 0.999 999 905 393 247 518 72;
  • 37) 0.999 999 905 393 247 518 72 × 2 = 1 + 0.999 999 810 786 495 037 44;
  • 38) 0.999 999 810 786 495 037 44 × 2 = 1 + 0.999 999 621 572 990 074 88;
  • 39) 0.999 999 621 572 990 074 88 × 2 = 1 + 0.999 999 243 145 980 149 76;
  • 40) 0.999 999 243 145 980 149 76 × 2 = 1 + 0.999 998 486 291 960 299 52;
  • 41) 0.999 998 486 291 960 299 52 × 2 = 1 + 0.999 996 972 583 920 599 04;
  • 42) 0.999 996 972 583 920 599 04 × 2 = 1 + 0.999 993 945 167 841 198 08;
  • 43) 0.999 993 945 167 841 198 08 × 2 = 1 + 0.999 987 890 335 682 396 16;
  • 44) 0.999 987 890 335 682 396 16 × 2 = 1 + 0.999 975 780 671 364 792 32;
  • 45) 0.999 975 780 671 364 792 32 × 2 = 1 + 0.999 951 561 342 729 584 64;
  • 46) 0.999 951 561 342 729 584 64 × 2 = 1 + 0.999 903 122 685 459 169 28;
  • 47) 0.999 903 122 685 459 169 28 × 2 = 1 + 0.999 806 245 370 918 338 56;
  • 48) 0.999 806 245 370 918 338 56 × 2 = 1 + 0.999 612 490 741 836 677 12;
  • 49) 0.999 612 490 741 836 677 12 × 2 = 1 + 0.999 224 981 483 673 354 24;
  • 50) 0.999 224 981 483 673 354 24 × 2 = 1 + 0.998 449 962 967 346 708 48;
  • 51) 0.998 449 962 967 346 708 48 × 2 = 1 + 0.996 899 925 934 693 416 96;
  • 52) 0.996 899 925 934 693 416 96 × 2 = 1 + 0.993 799 851 869 386 833 92;
  • 53) 0.993 799 851 869 386 833 92 × 2 = 1 + 0.987 599 703 738 773 667 84;
  • 54) 0.987 599 703 738 773 667 84 × 2 = 1 + 0.975 199 407 477 547 335 68;

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:


0.000 000 000 742 147 675 27(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 27(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:


0.000 000 000 742 147 675 27(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 -0.000 000 000 742 147 675 27 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