-0.000 000 000 741 995 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 995(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
-0.000 000 000 741 995(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:

|-0.000 000 000 741 995| = 0.000 000 000 741 995


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: 0.000 000 000 741 995.

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) 0.000 000 000 741 995 × 2 = 0 + 0.000 000 001 483 99;
  • 2) 0.000 000 001 483 99 × 2 = 0 + 0.000 000 002 967 98;
  • 3) 0.000 000 002 967 98 × 2 = 0 + 0.000 000 005 935 96;
  • 4) 0.000 000 005 935 96 × 2 = 0 + 0.000 000 011 871 92;
  • 5) 0.000 000 011 871 92 × 2 = 0 + 0.000 000 023 743 84;
  • 6) 0.000 000 023 743 84 × 2 = 0 + 0.000 000 047 487 68;
  • 7) 0.000 000 047 487 68 × 2 = 0 + 0.000 000 094 975 36;
  • 8) 0.000 000 094 975 36 × 2 = 0 + 0.000 000 189 950 72;
  • 9) 0.000 000 189 950 72 × 2 = 0 + 0.000 000 379 901 44;
  • 10) 0.000 000 379 901 44 × 2 = 0 + 0.000 000 759 802 88;
  • 11) 0.000 000 759 802 88 × 2 = 0 + 0.000 001 519 605 76;
  • 12) 0.000 001 519 605 76 × 2 = 0 + 0.000 003 039 211 52;
  • 13) 0.000 003 039 211 52 × 2 = 0 + 0.000 006 078 423 04;
  • 14) 0.000 006 078 423 04 × 2 = 0 + 0.000 012 156 846 08;
  • 15) 0.000 012 156 846 08 × 2 = 0 + 0.000 024 313 692 16;
  • 16) 0.000 024 313 692 16 × 2 = 0 + 0.000 048 627 384 32;
  • 17) 0.000 048 627 384 32 × 2 = 0 + 0.000 097 254 768 64;
  • 18) 0.000 097 254 768 64 × 2 = 0 + 0.000 194 509 537 28;
  • 19) 0.000 194 509 537 28 × 2 = 0 + 0.000 389 019 074 56;
  • 20) 0.000 389 019 074 56 × 2 = 0 + 0.000 778 038 149 12;
  • 21) 0.000 778 038 149 12 × 2 = 0 + 0.001 556 076 298 24;
  • 22) 0.001 556 076 298 24 × 2 = 0 + 0.003 112 152 596 48;
  • 23) 0.003 112 152 596 48 × 2 = 0 + 0.006 224 305 192 96;
  • 24) 0.006 224 305 192 96 × 2 = 0 + 0.012 448 610 385 92;
  • 25) 0.012 448 610 385 92 × 2 = 0 + 0.024 897 220 771 84;
  • 26) 0.024 897 220 771 84 × 2 = 0 + 0.049 794 441 543 68;
  • 27) 0.049 794 441 543 68 × 2 = 0 + 0.099 588 883 087 36;
  • 28) 0.099 588 883 087 36 × 2 = 0 + 0.199 177 766 174 72;
  • 29) 0.199 177 766 174 72 × 2 = 0 + 0.398 355 532 349 44;
  • 30) 0.398 355 532 349 44 × 2 = 0 + 0.796 711 064 698 88;
  • 31) 0.796 711 064 698 88 × 2 = 1 + 0.593 422 129 397 76;
  • 32) 0.593 422 129 397 76 × 2 = 1 + 0.186 844 258 795 52;
  • 33) 0.186 844 258 795 52 × 2 = 0 + 0.373 688 517 591 04;
  • 34) 0.373 688 517 591 04 × 2 = 0 + 0.747 377 035 182 08;
  • 35) 0.747 377 035 182 08 × 2 = 1 + 0.494 754 070 364 16;
  • 36) 0.494 754 070 364 16 × 2 = 0 + 0.989 508 140 728 32;
  • 37) 0.989 508 140 728 32 × 2 = 1 + 0.979 016 281 456 64;
  • 38) 0.979 016 281 456 64 × 2 = 1 + 0.958 032 562 913 28;
  • 39) 0.958 032 562 913 28 × 2 = 1 + 0.916 065 125 826 56;
  • 40) 0.916 065 125 826 56 × 2 = 1 + 0.832 130 251 653 12;
  • 41) 0.832 130 251 653 12 × 2 = 1 + 0.664 260 503 306 24;
  • 42) 0.664 260 503 306 24 × 2 = 1 + 0.328 521 006 612 48;
  • 43) 0.328 521 006 612 48 × 2 = 0 + 0.657 042 013 224 96;
  • 44) 0.657 042 013 224 96 × 2 = 1 + 0.314 084 026 449 92;
  • 45) 0.314 084 026 449 92 × 2 = 0 + 0.628 168 052 899 84;
  • 46) 0.628 168 052 899 84 × 2 = 1 + 0.256 336 105 799 68;
  • 47) 0.256 336 105 799 68 × 2 = 0 + 0.512 672 211 599 36;
  • 48) 0.512 672 211 599 36 × 2 = 1 + 0.025 344 423 198 72;
  • 49) 0.025 344 423 198 72 × 2 = 0 + 0.050 688 846 397 44;
  • 50) 0.050 688 846 397 44 × 2 = 0 + 0.101 377 692 794 88;
  • 51) 0.101 377 692 794 88 × 2 = 0 + 0.202 755 385 589 76;
  • 52) 0.202 755 385 589 76 × 2 = 0 + 0.405 510 771 179 52;
  • 53) 0.405 510 771 179 52 × 2 = 0 + 0.811 021 542 359 04;
  • 54) 0.811 021 542 359 04 × 2 = 1 + 0.622 043 084 718 08;

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:


0.000 000 000 741 995(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0101 0000 01(2)

6. Positive number before normalization:

0.000 000 000 741 995(10) =

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


0.000 000 000 741 995(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0101 0000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1101 0101 0000 01(2) × 20 =

1.1001 0111 1110 1010 1000 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 1010 1000 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 0101 0100 0001 =

100 1011 1111 0101 0100 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 0101 0100 0001


Decimal number -0.000 000 000 741 995 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 0101 0100 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