-0.000 000 000 740 29 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 29(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 29(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 29| = 0.000 000 000 740 29


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 29.

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 29 × 2 = 0 + 0.000 000 001 480 58;
  • 2) 0.000 000 001 480 58 × 2 = 0 + 0.000 000 002 961 16;
  • 3) 0.000 000 002 961 16 × 2 = 0 + 0.000 000 005 922 32;
  • 4) 0.000 000 005 922 32 × 2 = 0 + 0.000 000 011 844 64;
  • 5) 0.000 000 011 844 64 × 2 = 0 + 0.000 000 023 689 28;
  • 6) 0.000 000 023 689 28 × 2 = 0 + 0.000 000 047 378 56;
  • 7) 0.000 000 047 378 56 × 2 = 0 + 0.000 000 094 757 12;
  • 8) 0.000 000 094 757 12 × 2 = 0 + 0.000 000 189 514 24;
  • 9) 0.000 000 189 514 24 × 2 = 0 + 0.000 000 379 028 48;
  • 10) 0.000 000 379 028 48 × 2 = 0 + 0.000 000 758 056 96;
  • 11) 0.000 000 758 056 96 × 2 = 0 + 0.000 001 516 113 92;
  • 12) 0.000 001 516 113 92 × 2 = 0 + 0.000 003 032 227 84;
  • 13) 0.000 003 032 227 84 × 2 = 0 + 0.000 006 064 455 68;
  • 14) 0.000 006 064 455 68 × 2 = 0 + 0.000 012 128 911 36;
  • 15) 0.000 012 128 911 36 × 2 = 0 + 0.000 024 257 822 72;
  • 16) 0.000 024 257 822 72 × 2 = 0 + 0.000 048 515 645 44;
  • 17) 0.000 048 515 645 44 × 2 = 0 + 0.000 097 031 290 88;
  • 18) 0.000 097 031 290 88 × 2 = 0 + 0.000 194 062 581 76;
  • 19) 0.000 194 062 581 76 × 2 = 0 + 0.000 388 125 163 52;
  • 20) 0.000 388 125 163 52 × 2 = 0 + 0.000 776 250 327 04;
  • 21) 0.000 776 250 327 04 × 2 = 0 + 0.001 552 500 654 08;
  • 22) 0.001 552 500 654 08 × 2 = 0 + 0.003 105 001 308 16;
  • 23) 0.003 105 001 308 16 × 2 = 0 + 0.006 210 002 616 32;
  • 24) 0.006 210 002 616 32 × 2 = 0 + 0.012 420 005 232 64;
  • 25) 0.012 420 005 232 64 × 2 = 0 + 0.024 840 010 465 28;
  • 26) 0.024 840 010 465 28 × 2 = 0 + 0.049 680 020 930 56;
  • 27) 0.049 680 020 930 56 × 2 = 0 + 0.099 360 041 861 12;
  • 28) 0.099 360 041 861 12 × 2 = 0 + 0.198 720 083 722 24;
  • 29) 0.198 720 083 722 24 × 2 = 0 + 0.397 440 167 444 48;
  • 30) 0.397 440 167 444 48 × 2 = 0 + 0.794 880 334 888 96;
  • 31) 0.794 880 334 888 96 × 2 = 1 + 0.589 760 669 777 92;
  • 32) 0.589 760 669 777 92 × 2 = 1 + 0.179 521 339 555 84;
  • 33) 0.179 521 339 555 84 × 2 = 0 + 0.359 042 679 111 68;
  • 34) 0.359 042 679 111 68 × 2 = 0 + 0.718 085 358 223 36;
  • 35) 0.718 085 358 223 36 × 2 = 1 + 0.436 170 716 446 72;
  • 36) 0.436 170 716 446 72 × 2 = 0 + 0.872 341 432 893 44;
  • 37) 0.872 341 432 893 44 × 2 = 1 + 0.744 682 865 786 88;
  • 38) 0.744 682 865 786 88 × 2 = 1 + 0.489 365 731 573 76;
  • 39) 0.489 365 731 573 76 × 2 = 0 + 0.978 731 463 147 52;
  • 40) 0.978 731 463 147 52 × 2 = 1 + 0.957 462 926 295 04;
  • 41) 0.957 462 926 295 04 × 2 = 1 + 0.914 925 852 590 08;
  • 42) 0.914 925 852 590 08 × 2 = 1 + 0.829 851 705 180 16;
  • 43) 0.829 851 705 180 16 × 2 = 1 + 0.659 703 410 360 32;
  • 44) 0.659 703 410 360 32 × 2 = 1 + 0.319 406 820 720 64;
  • 45) 0.319 406 820 720 64 × 2 = 0 + 0.638 813 641 441 28;
  • 46) 0.638 813 641 441 28 × 2 = 1 + 0.277 627 282 882 56;
  • 47) 0.277 627 282 882 56 × 2 = 0 + 0.555 254 565 765 12;
  • 48) 0.555 254 565 765 12 × 2 = 1 + 0.110 509 131 530 24;
  • 49) 0.110 509 131 530 24 × 2 = 0 + 0.221 018 263 060 48;
  • 50) 0.221 018 263 060 48 × 2 = 0 + 0.442 036 526 120 96;
  • 51) 0.442 036 526 120 96 × 2 = 0 + 0.884 073 052 241 92;
  • 52) 0.884 073 052 241 92 × 2 = 1 + 0.768 146 104 483 84;
  • 53) 0.768 146 104 483 84 × 2 = 1 + 0.536 292 208 967 68;
  • 54) 0.536 292 208 967 68 × 2 = 1 + 0.072 584 417 935 36;

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 29(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1111 0101 0001 11(2)

6. Positive number before normalization:

0.000 000 000 740 29(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1111 0101 0001 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:


0.000 000 000 740 29(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1111 0101 0001 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1101 1111 0101 0001 11(2) × 20 =

1.1001 0110 1111 1010 1000 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 0110 1111 1010 1000 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 0111 1101 0100 0111 =

100 1011 0111 1101 0100 0111


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 0111 1101 0100 0111


Decimal number -0.000 000 000 740 29 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 0111 1101 0100 0111

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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