-0.000 000 000 740 46 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 46(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
-0.000 000 000 740 46(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:

|-0.000 000 000 740 46| = 0.000 000 000 740 46


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: 0.000 000 000 740 46.

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) 0.000 000 000 740 46 × 2 = 0 + 0.000 000 001 480 92;
  • 2) 0.000 000 001 480 92 × 2 = 0 + 0.000 000 002 961 84;
  • 3) 0.000 000 002 961 84 × 2 = 0 + 0.000 000 005 923 68;
  • 4) 0.000 000 005 923 68 × 2 = 0 + 0.000 000 011 847 36;
  • 5) 0.000 000 011 847 36 × 2 = 0 + 0.000 000 023 694 72;
  • 6) 0.000 000 023 694 72 × 2 = 0 + 0.000 000 047 389 44;
  • 7) 0.000 000 047 389 44 × 2 = 0 + 0.000 000 094 778 88;
  • 8) 0.000 000 094 778 88 × 2 = 0 + 0.000 000 189 557 76;
  • 9) 0.000 000 189 557 76 × 2 = 0 + 0.000 000 379 115 52;
  • 10) 0.000 000 379 115 52 × 2 = 0 + 0.000 000 758 231 04;
  • 11) 0.000 000 758 231 04 × 2 = 0 + 0.000 001 516 462 08;
  • 12) 0.000 001 516 462 08 × 2 = 0 + 0.000 003 032 924 16;
  • 13) 0.000 003 032 924 16 × 2 = 0 + 0.000 006 065 848 32;
  • 14) 0.000 006 065 848 32 × 2 = 0 + 0.000 012 131 696 64;
  • 15) 0.000 012 131 696 64 × 2 = 0 + 0.000 024 263 393 28;
  • 16) 0.000 024 263 393 28 × 2 = 0 + 0.000 048 526 786 56;
  • 17) 0.000 048 526 786 56 × 2 = 0 + 0.000 097 053 573 12;
  • 18) 0.000 097 053 573 12 × 2 = 0 + 0.000 194 107 146 24;
  • 19) 0.000 194 107 146 24 × 2 = 0 + 0.000 388 214 292 48;
  • 20) 0.000 388 214 292 48 × 2 = 0 + 0.000 776 428 584 96;
  • 21) 0.000 776 428 584 96 × 2 = 0 + 0.001 552 857 169 92;
  • 22) 0.001 552 857 169 92 × 2 = 0 + 0.003 105 714 339 84;
  • 23) 0.003 105 714 339 84 × 2 = 0 + 0.006 211 428 679 68;
  • 24) 0.006 211 428 679 68 × 2 = 0 + 0.012 422 857 359 36;
  • 25) 0.012 422 857 359 36 × 2 = 0 + 0.024 845 714 718 72;
  • 26) 0.024 845 714 718 72 × 2 = 0 + 0.049 691 429 437 44;
  • 27) 0.049 691 429 437 44 × 2 = 0 + 0.099 382 858 874 88;
  • 28) 0.099 382 858 874 88 × 2 = 0 + 0.198 765 717 749 76;
  • 29) 0.198 765 717 749 76 × 2 = 0 + 0.397 531 435 499 52;
  • 30) 0.397 531 435 499 52 × 2 = 0 + 0.795 062 870 999 04;
  • 31) 0.795 062 870 999 04 × 2 = 1 + 0.590 125 741 998 08;
  • 32) 0.590 125 741 998 08 × 2 = 1 + 0.180 251 483 996 16;
  • 33) 0.180 251 483 996 16 × 2 = 0 + 0.360 502 967 992 32;
  • 34) 0.360 502 967 992 32 × 2 = 0 + 0.721 005 935 984 64;
  • 35) 0.721 005 935 984 64 × 2 = 1 + 0.442 011 871 969 28;
  • 36) 0.442 011 871 969 28 × 2 = 0 + 0.884 023 743 938 56;
  • 37) 0.884 023 743 938 56 × 2 = 1 + 0.768 047 487 877 12;
  • 38) 0.768 047 487 877 12 × 2 = 1 + 0.536 094 975 754 24;
  • 39) 0.536 094 975 754 24 × 2 = 1 + 0.072 189 951 508 48;
  • 40) 0.072 189 951 508 48 × 2 = 0 + 0.144 379 903 016 96;
  • 41) 0.144 379 903 016 96 × 2 = 0 + 0.288 759 806 033 92;
  • 42) 0.288 759 806 033 92 × 2 = 0 + 0.577 519 612 067 84;
  • 43) 0.577 519 612 067 84 × 2 = 1 + 0.155 039 224 135 68;
  • 44) 0.155 039 224 135 68 × 2 = 0 + 0.310 078 448 271 36;
  • 45) 0.310 078 448 271 36 × 2 = 0 + 0.620 156 896 542 72;
  • 46) 0.620 156 896 542 72 × 2 = 1 + 0.240 313 793 085 44;
  • 47) 0.240 313 793 085 44 × 2 = 0 + 0.480 627 586 170 88;
  • 48) 0.480 627 586 170 88 × 2 = 0 + 0.961 255 172 341 76;
  • 49) 0.961 255 172 341 76 × 2 = 1 + 0.922 510 344 683 52;
  • 50) 0.922 510 344 683 52 × 2 = 1 + 0.845 020 689 367 04;
  • 51) 0.845 020 689 367 04 × 2 = 1 + 0.690 041 378 734 08;
  • 52) 0.690 041 378 734 08 × 2 = 1 + 0.380 082 757 468 16;
  • 53) 0.380 082 757 468 16 × 2 = 0 + 0.760 165 514 936 32;
  • 54) 0.760 165 514 936 32 × 2 = 1 + 0.520 331 029 872 64;

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:


0.000 000 000 740 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0100 1111 01(2)

6. Positive number before normalization:

0.000 000 000 740 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0100 1111 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:


0.000 000 000 740 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0100 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0010 0100 1111 01(2) × 20 =

1.1001 0111 0001 0010 0111 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 0001 0010 0111 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 1000 1001 0011 1101 =

100 1011 1000 1001 0011 1101


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 1000 1001 0011 1101


Decimal number -0.000 000 000 740 46 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1000 1001 0011 1101

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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