-0.000 000 000 741 955 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 955(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
-0.000 000 000 741 955(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:

|-0.000 000 000 741 955| = 0.000 000 000 741 955


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: 0.000 000 000 741 955.

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) 0.000 000 000 741 955 × 2 = 0 + 0.000 000 001 483 91;
  • 2) 0.000 000 001 483 91 × 2 = 0 + 0.000 000 002 967 82;
  • 3) 0.000 000 002 967 82 × 2 = 0 + 0.000 000 005 935 64;
  • 4) 0.000 000 005 935 64 × 2 = 0 + 0.000 000 011 871 28;
  • 5) 0.000 000 011 871 28 × 2 = 0 + 0.000 000 023 742 56;
  • 6) 0.000 000 023 742 56 × 2 = 0 + 0.000 000 047 485 12;
  • 7) 0.000 000 047 485 12 × 2 = 0 + 0.000 000 094 970 24;
  • 8) 0.000 000 094 970 24 × 2 = 0 + 0.000 000 189 940 48;
  • 9) 0.000 000 189 940 48 × 2 = 0 + 0.000 000 379 880 96;
  • 10) 0.000 000 379 880 96 × 2 = 0 + 0.000 000 759 761 92;
  • 11) 0.000 000 759 761 92 × 2 = 0 + 0.000 001 519 523 84;
  • 12) 0.000 001 519 523 84 × 2 = 0 + 0.000 003 039 047 68;
  • 13) 0.000 003 039 047 68 × 2 = 0 + 0.000 006 078 095 36;
  • 14) 0.000 006 078 095 36 × 2 = 0 + 0.000 012 156 190 72;
  • 15) 0.000 012 156 190 72 × 2 = 0 + 0.000 024 312 381 44;
  • 16) 0.000 024 312 381 44 × 2 = 0 + 0.000 048 624 762 88;
  • 17) 0.000 048 624 762 88 × 2 = 0 + 0.000 097 249 525 76;
  • 18) 0.000 097 249 525 76 × 2 = 0 + 0.000 194 499 051 52;
  • 19) 0.000 194 499 051 52 × 2 = 0 + 0.000 388 998 103 04;
  • 20) 0.000 388 998 103 04 × 2 = 0 + 0.000 777 996 206 08;
  • 21) 0.000 777 996 206 08 × 2 = 0 + 0.001 555 992 412 16;
  • 22) 0.001 555 992 412 16 × 2 = 0 + 0.003 111 984 824 32;
  • 23) 0.003 111 984 824 32 × 2 = 0 + 0.006 223 969 648 64;
  • 24) 0.006 223 969 648 64 × 2 = 0 + 0.012 447 939 297 28;
  • 25) 0.012 447 939 297 28 × 2 = 0 + 0.024 895 878 594 56;
  • 26) 0.024 895 878 594 56 × 2 = 0 + 0.049 791 757 189 12;
  • 27) 0.049 791 757 189 12 × 2 = 0 + 0.099 583 514 378 24;
  • 28) 0.099 583 514 378 24 × 2 = 0 + 0.199 167 028 756 48;
  • 29) 0.199 167 028 756 48 × 2 = 0 + 0.398 334 057 512 96;
  • 30) 0.398 334 057 512 96 × 2 = 0 + 0.796 668 115 025 92;
  • 31) 0.796 668 115 025 92 × 2 = 1 + 0.593 336 230 051 84;
  • 32) 0.593 336 230 051 84 × 2 = 1 + 0.186 672 460 103 68;
  • 33) 0.186 672 460 103 68 × 2 = 0 + 0.373 344 920 207 36;
  • 34) 0.373 344 920 207 36 × 2 = 0 + 0.746 689 840 414 72;
  • 35) 0.746 689 840 414 72 × 2 = 1 + 0.493 379 680 829 44;
  • 36) 0.493 379 680 829 44 × 2 = 0 + 0.986 759 361 658 88;
  • 37) 0.986 759 361 658 88 × 2 = 1 + 0.973 518 723 317 76;
  • 38) 0.973 518 723 317 76 × 2 = 1 + 0.947 037 446 635 52;
  • 39) 0.947 037 446 635 52 × 2 = 1 + 0.894 074 893 271 04;
  • 40) 0.894 074 893 271 04 × 2 = 1 + 0.788 149 786 542 08;
  • 41) 0.788 149 786 542 08 × 2 = 1 + 0.576 299 573 084 16;
  • 42) 0.576 299 573 084 16 × 2 = 1 + 0.152 599 146 168 32;
  • 43) 0.152 599 146 168 32 × 2 = 0 + 0.305 198 292 336 64;
  • 44) 0.305 198 292 336 64 × 2 = 0 + 0.610 396 584 673 28;
  • 45) 0.610 396 584 673 28 × 2 = 1 + 0.220 793 169 346 56;
  • 46) 0.220 793 169 346 56 × 2 = 0 + 0.441 586 338 693 12;
  • 47) 0.441 586 338 693 12 × 2 = 0 + 0.883 172 677 386 24;
  • 48) 0.883 172 677 386 24 × 2 = 1 + 0.766 345 354 772 48;
  • 49) 0.766 345 354 772 48 × 2 = 1 + 0.532 690 709 544 96;
  • 50) 0.532 690 709 544 96 × 2 = 1 + 0.065 381 419 089 92;
  • 51) 0.065 381 419 089 92 × 2 = 0 + 0.130 762 838 179 84;
  • 52) 0.130 762 838 179 84 × 2 = 0 + 0.261 525 676 359 68;
  • 53) 0.261 525 676 359 68 × 2 = 0 + 0.523 051 352 719 36;
  • 54) 0.523 051 352 719 36 × 2 = 1 + 0.046 102 705 438 72;

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:


0.000 000 000 741 955(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1001 1100 01(2)

6. Positive number before normalization:

0.000 000 000 741 955(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1001 1100 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:


0.000 000 000 741 955(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1001 1100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1100 1001 1100 01(2) × 20 =

1.1001 0111 1110 0100 1110 001(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 1110 0100 1110 001


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 0010 0111 0001 =

100 1011 1111 0010 0111 0001


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 0010 0111 0001


Decimal number -0.000 000 000 741 955 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0010 0111 0001

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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