-0.000 000 000 742 147 16 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 16(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 16(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 16| = 0.000 000 000 742 147 16


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 16.

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 16 × 2 = 0 + 0.000 000 001 484 294 32;
  • 2) 0.000 000 001 484 294 32 × 2 = 0 + 0.000 000 002 968 588 64;
  • 3) 0.000 000 002 968 588 64 × 2 = 0 + 0.000 000 005 937 177 28;
  • 4) 0.000 000 005 937 177 28 × 2 = 0 + 0.000 000 011 874 354 56;
  • 5) 0.000 000 011 874 354 56 × 2 = 0 + 0.000 000 023 748 709 12;
  • 6) 0.000 000 023 748 709 12 × 2 = 0 + 0.000 000 047 497 418 24;
  • 7) 0.000 000 047 497 418 24 × 2 = 0 + 0.000 000 094 994 836 48;
  • 8) 0.000 000 094 994 836 48 × 2 = 0 + 0.000 000 189 989 672 96;
  • 9) 0.000 000 189 989 672 96 × 2 = 0 + 0.000 000 379 979 345 92;
  • 10) 0.000 000 379 979 345 92 × 2 = 0 + 0.000 000 759 958 691 84;
  • 11) 0.000 000 759 958 691 84 × 2 = 0 + 0.000 001 519 917 383 68;
  • 12) 0.000 001 519 917 383 68 × 2 = 0 + 0.000 003 039 834 767 36;
  • 13) 0.000 003 039 834 767 36 × 2 = 0 + 0.000 006 079 669 534 72;
  • 14) 0.000 006 079 669 534 72 × 2 = 0 + 0.000 012 159 339 069 44;
  • 15) 0.000 012 159 339 069 44 × 2 = 0 + 0.000 024 318 678 138 88;
  • 16) 0.000 024 318 678 138 88 × 2 = 0 + 0.000 048 637 356 277 76;
  • 17) 0.000 048 637 356 277 76 × 2 = 0 + 0.000 097 274 712 555 52;
  • 18) 0.000 097 274 712 555 52 × 2 = 0 + 0.000 194 549 425 111 04;
  • 19) 0.000 194 549 425 111 04 × 2 = 0 + 0.000 389 098 850 222 08;
  • 20) 0.000 389 098 850 222 08 × 2 = 0 + 0.000 778 197 700 444 16;
  • 21) 0.000 778 197 700 444 16 × 2 = 0 + 0.001 556 395 400 888 32;
  • 22) 0.001 556 395 400 888 32 × 2 = 0 + 0.003 112 790 801 776 64;
  • 23) 0.003 112 790 801 776 64 × 2 = 0 + 0.006 225 581 603 553 28;
  • 24) 0.006 225 581 603 553 28 × 2 = 0 + 0.012 451 163 207 106 56;
  • 25) 0.012 451 163 207 106 56 × 2 = 0 + 0.024 902 326 414 213 12;
  • 26) 0.024 902 326 414 213 12 × 2 = 0 + 0.049 804 652 828 426 24;
  • 27) 0.049 804 652 828 426 24 × 2 = 0 + 0.099 609 305 656 852 48;
  • 28) 0.099 609 305 656 852 48 × 2 = 0 + 0.199 218 611 313 704 96;
  • 29) 0.199 218 611 313 704 96 × 2 = 0 + 0.398 437 222 627 409 92;
  • 30) 0.398 437 222 627 409 92 × 2 = 0 + 0.796 874 445 254 819 84;
  • 31) 0.796 874 445 254 819 84 × 2 = 1 + 0.593 748 890 509 639 68;
  • 32) 0.593 748 890 509 639 68 × 2 = 1 + 0.187 497 781 019 279 36;
  • 33) 0.187 497 781 019 279 36 × 2 = 0 + 0.374 995 562 038 558 72;
  • 34) 0.374 995 562 038 558 72 × 2 = 0 + 0.749 991 124 077 117 44;
  • 35) 0.749 991 124 077 117 44 × 2 = 1 + 0.499 982 248 154 234 88;
  • 36) 0.499 982 248 154 234 88 × 2 = 0 + 0.999 964 496 308 469 76;
  • 37) 0.999 964 496 308 469 76 × 2 = 1 + 0.999 928 992 616 939 52;
  • 38) 0.999 928 992 616 939 52 × 2 = 1 + 0.999 857 985 233 879 04;
  • 39) 0.999 857 985 233 879 04 × 2 = 1 + 0.999 715 970 467 758 08;
  • 40) 0.999 715 970 467 758 08 × 2 = 1 + 0.999 431 940 935 516 16;
  • 41) 0.999 431 940 935 516 16 × 2 = 1 + 0.998 863 881 871 032 32;
  • 42) 0.998 863 881 871 032 32 × 2 = 1 + 0.997 727 763 742 064 64;
  • 43) 0.997 727 763 742 064 64 × 2 = 1 + 0.995 455 527 484 129 28;
  • 44) 0.995 455 527 484 129 28 × 2 = 1 + 0.990 911 054 968 258 56;
  • 45) 0.990 911 054 968 258 56 × 2 = 1 + 0.981 822 109 936 517 12;
  • 46) 0.981 822 109 936 517 12 × 2 = 1 + 0.963 644 219 873 034 24;
  • 47) 0.963 644 219 873 034 24 × 2 = 1 + 0.927 288 439 746 068 48;
  • 48) 0.927 288 439 746 068 48 × 2 = 1 + 0.854 576 879 492 136 96;
  • 49) 0.854 576 879 492 136 96 × 2 = 1 + 0.709 153 758 984 273 92;
  • 50) 0.709 153 758 984 273 92 × 2 = 1 + 0.418 307 517 968 547 84;
  • 51) 0.418 307 517 968 547 84 × 2 = 0 + 0.836 615 035 937 095 68;
  • 52) 0.836 615 035 937 095 68 × 2 = 1 + 0.673 230 071 874 191 36;
  • 53) 0.673 230 071 874 191 36 × 2 = 1 + 0.346 460 143 748 382 72;
  • 54) 0.346 460 143 748 382 72 × 2 = 0 + 0.692 920 287 496 765 44;

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 16(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 16(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 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:


0.000 000 000 742 147 16(10) =

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

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

1.1001 0111 1111 1111 1110 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 1111 1110 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 1111 0110 =

100 1011 1111 1111 1111 0110


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 0110


Decimal number -0.000 000 000 742 147 16 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 0110

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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