-0.000 000 000 742 147 07 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 07(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 07(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 07| = 0.000 000 000 742 147 07


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 07.

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 07 × 2 = 0 + 0.000 000 001 484 294 14;
  • 2) 0.000 000 001 484 294 14 × 2 = 0 + 0.000 000 002 968 588 28;
  • 3) 0.000 000 002 968 588 28 × 2 = 0 + 0.000 000 005 937 176 56;
  • 4) 0.000 000 005 937 176 56 × 2 = 0 + 0.000 000 011 874 353 12;
  • 5) 0.000 000 011 874 353 12 × 2 = 0 + 0.000 000 023 748 706 24;
  • 6) 0.000 000 023 748 706 24 × 2 = 0 + 0.000 000 047 497 412 48;
  • 7) 0.000 000 047 497 412 48 × 2 = 0 + 0.000 000 094 994 824 96;
  • 8) 0.000 000 094 994 824 96 × 2 = 0 + 0.000 000 189 989 649 92;
  • 9) 0.000 000 189 989 649 92 × 2 = 0 + 0.000 000 379 979 299 84;
  • 10) 0.000 000 379 979 299 84 × 2 = 0 + 0.000 000 759 958 599 68;
  • 11) 0.000 000 759 958 599 68 × 2 = 0 + 0.000 001 519 917 199 36;
  • 12) 0.000 001 519 917 199 36 × 2 = 0 + 0.000 003 039 834 398 72;
  • 13) 0.000 003 039 834 398 72 × 2 = 0 + 0.000 006 079 668 797 44;
  • 14) 0.000 006 079 668 797 44 × 2 = 0 + 0.000 012 159 337 594 88;
  • 15) 0.000 012 159 337 594 88 × 2 = 0 + 0.000 024 318 675 189 76;
  • 16) 0.000 024 318 675 189 76 × 2 = 0 + 0.000 048 637 350 379 52;
  • 17) 0.000 048 637 350 379 52 × 2 = 0 + 0.000 097 274 700 759 04;
  • 18) 0.000 097 274 700 759 04 × 2 = 0 + 0.000 194 549 401 518 08;
  • 19) 0.000 194 549 401 518 08 × 2 = 0 + 0.000 389 098 803 036 16;
  • 20) 0.000 389 098 803 036 16 × 2 = 0 + 0.000 778 197 606 072 32;
  • 21) 0.000 778 197 606 072 32 × 2 = 0 + 0.001 556 395 212 144 64;
  • 22) 0.001 556 395 212 144 64 × 2 = 0 + 0.003 112 790 424 289 28;
  • 23) 0.003 112 790 424 289 28 × 2 = 0 + 0.006 225 580 848 578 56;
  • 24) 0.006 225 580 848 578 56 × 2 = 0 + 0.012 451 161 697 157 12;
  • 25) 0.012 451 161 697 157 12 × 2 = 0 + 0.024 902 323 394 314 24;
  • 26) 0.024 902 323 394 314 24 × 2 = 0 + 0.049 804 646 788 628 48;
  • 27) 0.049 804 646 788 628 48 × 2 = 0 + 0.099 609 293 577 256 96;
  • 28) 0.099 609 293 577 256 96 × 2 = 0 + 0.199 218 587 154 513 92;
  • 29) 0.199 218 587 154 513 92 × 2 = 0 + 0.398 437 174 309 027 84;
  • 30) 0.398 437 174 309 027 84 × 2 = 0 + 0.796 874 348 618 055 68;
  • 31) 0.796 874 348 618 055 68 × 2 = 1 + 0.593 748 697 236 111 36;
  • 32) 0.593 748 697 236 111 36 × 2 = 1 + 0.187 497 394 472 222 72;
  • 33) 0.187 497 394 472 222 72 × 2 = 0 + 0.374 994 788 944 445 44;
  • 34) 0.374 994 788 944 445 44 × 2 = 0 + 0.749 989 577 888 890 88;
  • 35) 0.749 989 577 888 890 88 × 2 = 1 + 0.499 979 155 777 781 76;
  • 36) 0.499 979 155 777 781 76 × 2 = 0 + 0.999 958 311 555 563 52;
  • 37) 0.999 958 311 555 563 52 × 2 = 1 + 0.999 916 623 111 127 04;
  • 38) 0.999 916 623 111 127 04 × 2 = 1 + 0.999 833 246 222 254 08;
  • 39) 0.999 833 246 222 254 08 × 2 = 1 + 0.999 666 492 444 508 16;
  • 40) 0.999 666 492 444 508 16 × 2 = 1 + 0.999 332 984 889 016 32;
  • 41) 0.999 332 984 889 016 32 × 2 = 1 + 0.998 665 969 778 032 64;
  • 42) 0.998 665 969 778 032 64 × 2 = 1 + 0.997 331 939 556 065 28;
  • 43) 0.997 331 939 556 065 28 × 2 = 1 + 0.994 663 879 112 130 56;
  • 44) 0.994 663 879 112 130 56 × 2 = 1 + 0.989 327 758 224 261 12;
  • 45) 0.989 327 758 224 261 12 × 2 = 1 + 0.978 655 516 448 522 24;
  • 46) 0.978 655 516 448 522 24 × 2 = 1 + 0.957 311 032 897 044 48;
  • 47) 0.957 311 032 897 044 48 × 2 = 1 + 0.914 622 065 794 088 96;
  • 48) 0.914 622 065 794 088 96 × 2 = 1 + 0.829 244 131 588 177 92;
  • 49) 0.829 244 131 588 177 92 × 2 = 1 + 0.658 488 263 176 355 84;
  • 50) 0.658 488 263 176 355 84 × 2 = 1 + 0.316 976 526 352 711 68;
  • 51) 0.316 976 526 352 711 68 × 2 = 0 + 0.633 953 052 705 423 36;
  • 52) 0.633 953 052 705 423 36 × 2 = 1 + 0.267 906 105 410 846 72;
  • 53) 0.267 906 105 410 846 72 × 2 = 0 + 0.535 812 210 821 693 44;
  • 54) 0.535 812 210 821 693 44 × 2 = 1 + 0.071 624 421 643 386 88;

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 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 01(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 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1101 01(2) × 20 =

1.1001 0111 1111 1111 1110 101(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 101


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 0101 =

100 1011 1111 1111 1111 0101


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 0101


Decimal number -0.000 000 000 742 147 07 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 0101

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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