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

Convert decimal -0.000 000 000 742 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 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 07| = 0.000 000 000 742 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 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 07 × 2 = 0 + 0.000 000 001 484 14;
  • 2) 0.000 000 001 484 14 × 2 = 0 + 0.000 000 002 968 28;
  • 3) 0.000 000 002 968 28 × 2 = 0 + 0.000 000 005 936 56;
  • 4) 0.000 000 005 936 56 × 2 = 0 + 0.000 000 011 873 12;
  • 5) 0.000 000 011 873 12 × 2 = 0 + 0.000 000 023 746 24;
  • 6) 0.000 000 023 746 24 × 2 = 0 + 0.000 000 047 492 48;
  • 7) 0.000 000 047 492 48 × 2 = 0 + 0.000 000 094 984 96;
  • 8) 0.000 000 094 984 96 × 2 = 0 + 0.000 000 189 969 92;
  • 9) 0.000 000 189 969 92 × 2 = 0 + 0.000 000 379 939 84;
  • 10) 0.000 000 379 939 84 × 2 = 0 + 0.000 000 759 879 68;
  • 11) 0.000 000 759 879 68 × 2 = 0 + 0.000 001 519 759 36;
  • 12) 0.000 001 519 759 36 × 2 = 0 + 0.000 003 039 518 72;
  • 13) 0.000 003 039 518 72 × 2 = 0 + 0.000 006 079 037 44;
  • 14) 0.000 006 079 037 44 × 2 = 0 + 0.000 012 158 074 88;
  • 15) 0.000 012 158 074 88 × 2 = 0 + 0.000 024 316 149 76;
  • 16) 0.000 024 316 149 76 × 2 = 0 + 0.000 048 632 299 52;
  • 17) 0.000 048 632 299 52 × 2 = 0 + 0.000 097 264 599 04;
  • 18) 0.000 097 264 599 04 × 2 = 0 + 0.000 194 529 198 08;
  • 19) 0.000 194 529 198 08 × 2 = 0 + 0.000 389 058 396 16;
  • 20) 0.000 389 058 396 16 × 2 = 0 + 0.000 778 116 792 32;
  • 21) 0.000 778 116 792 32 × 2 = 0 + 0.001 556 233 584 64;
  • 22) 0.001 556 233 584 64 × 2 = 0 + 0.003 112 467 169 28;
  • 23) 0.003 112 467 169 28 × 2 = 0 + 0.006 224 934 338 56;
  • 24) 0.006 224 934 338 56 × 2 = 0 + 0.012 449 868 677 12;
  • 25) 0.012 449 868 677 12 × 2 = 0 + 0.024 899 737 354 24;
  • 26) 0.024 899 737 354 24 × 2 = 0 + 0.049 799 474 708 48;
  • 27) 0.049 799 474 708 48 × 2 = 0 + 0.099 598 949 416 96;
  • 28) 0.099 598 949 416 96 × 2 = 0 + 0.199 197 898 833 92;
  • 29) 0.199 197 898 833 92 × 2 = 0 + 0.398 395 797 667 84;
  • 30) 0.398 395 797 667 84 × 2 = 0 + 0.796 791 595 335 68;
  • 31) 0.796 791 595 335 68 × 2 = 1 + 0.593 583 190 671 36;
  • 32) 0.593 583 190 671 36 × 2 = 1 + 0.187 166 381 342 72;
  • 33) 0.187 166 381 342 72 × 2 = 0 + 0.374 332 762 685 44;
  • 34) 0.374 332 762 685 44 × 2 = 0 + 0.748 665 525 370 88;
  • 35) 0.748 665 525 370 88 × 2 = 1 + 0.497 331 050 741 76;
  • 36) 0.497 331 050 741 76 × 2 = 0 + 0.994 662 101 483 52;
  • 37) 0.994 662 101 483 52 × 2 = 1 + 0.989 324 202 967 04;
  • 38) 0.989 324 202 967 04 × 2 = 1 + 0.978 648 405 934 08;
  • 39) 0.978 648 405 934 08 × 2 = 1 + 0.957 296 811 868 16;
  • 40) 0.957 296 811 868 16 × 2 = 1 + 0.914 593 623 736 32;
  • 41) 0.914 593 623 736 32 × 2 = 1 + 0.829 187 247 472 64;
  • 42) 0.829 187 247 472 64 × 2 = 1 + 0.658 374 494 945 28;
  • 43) 0.658 374 494 945 28 × 2 = 1 + 0.316 748 989 890 56;
  • 44) 0.316 748 989 890 56 × 2 = 0 + 0.633 497 979 781 12;
  • 45) 0.633 497 979 781 12 × 2 = 1 + 0.266 995 959 562 24;
  • 46) 0.266 995 959 562 24 × 2 = 0 + 0.533 991 919 124 48;
  • 47) 0.533 991 919 124 48 × 2 = 1 + 0.067 983 838 248 96;
  • 48) 0.067 983 838 248 96 × 2 = 0 + 0.135 967 676 497 92;
  • 49) 0.135 967 676 497 92 × 2 = 0 + 0.271 935 352 995 84;
  • 50) 0.271 935 352 995 84 × 2 = 0 + 0.543 870 705 991 68;
  • 51) 0.543 870 705 991 68 × 2 = 1 + 0.087 741 411 983 36;
  • 52) 0.087 741 411 983 36 × 2 = 0 + 0.175 482 823 966 72;
  • 53) 0.175 482 823 966 72 × 2 = 0 + 0.350 965 647 933 44;
  • 54) 0.350 965 647 933 44 × 2 = 0 + 0.701 931 295 866 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 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0010 00(2)

6. Positive number before normalization:

0.000 000 000 742 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0010 00(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 07(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 1010 0010 00(2) × 20 =

1.1001 0111 1111 0101 0001 000(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 0101 0001 000


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 1010 1000 1000 =

100 1011 1111 1010 1000 1000


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 1010 1000 1000


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

1 - 0110 0000 - 100 1011 1111 1010 1000 1000

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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