-0.000 000 000 893 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 893(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
-0.000 000 000 893(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:

|-0.000 000 000 893| = 0.000 000 000 893


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: 0.000 000 000 893.

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) 0.000 000 000 893 × 2 = 0 + 0.000 000 001 786;
  • 2) 0.000 000 001 786 × 2 = 0 + 0.000 000 003 572;
  • 3) 0.000 000 003 572 × 2 = 0 + 0.000 000 007 144;
  • 4) 0.000 000 007 144 × 2 = 0 + 0.000 000 014 288;
  • 5) 0.000 000 014 288 × 2 = 0 + 0.000 000 028 576;
  • 6) 0.000 000 028 576 × 2 = 0 + 0.000 000 057 152;
  • 7) 0.000 000 057 152 × 2 = 0 + 0.000 000 114 304;
  • 8) 0.000 000 114 304 × 2 = 0 + 0.000 000 228 608;
  • 9) 0.000 000 228 608 × 2 = 0 + 0.000 000 457 216;
  • 10) 0.000 000 457 216 × 2 = 0 + 0.000 000 914 432;
  • 11) 0.000 000 914 432 × 2 = 0 + 0.000 001 828 864;
  • 12) 0.000 001 828 864 × 2 = 0 + 0.000 003 657 728;
  • 13) 0.000 003 657 728 × 2 = 0 + 0.000 007 315 456;
  • 14) 0.000 007 315 456 × 2 = 0 + 0.000 014 630 912;
  • 15) 0.000 014 630 912 × 2 = 0 + 0.000 029 261 824;
  • 16) 0.000 029 261 824 × 2 = 0 + 0.000 058 523 648;
  • 17) 0.000 058 523 648 × 2 = 0 + 0.000 117 047 296;
  • 18) 0.000 117 047 296 × 2 = 0 + 0.000 234 094 592;
  • 19) 0.000 234 094 592 × 2 = 0 + 0.000 468 189 184;
  • 20) 0.000 468 189 184 × 2 = 0 + 0.000 936 378 368;
  • 21) 0.000 936 378 368 × 2 = 0 + 0.001 872 756 736;
  • 22) 0.001 872 756 736 × 2 = 0 + 0.003 745 513 472;
  • 23) 0.003 745 513 472 × 2 = 0 + 0.007 491 026 944;
  • 24) 0.007 491 026 944 × 2 = 0 + 0.014 982 053 888;
  • 25) 0.014 982 053 888 × 2 = 0 + 0.029 964 107 776;
  • 26) 0.029 964 107 776 × 2 = 0 + 0.059 928 215 552;
  • 27) 0.059 928 215 552 × 2 = 0 + 0.119 856 431 104;
  • 28) 0.119 856 431 104 × 2 = 0 + 0.239 712 862 208;
  • 29) 0.239 712 862 208 × 2 = 0 + 0.479 425 724 416;
  • 30) 0.479 425 724 416 × 2 = 0 + 0.958 851 448 832;
  • 31) 0.958 851 448 832 × 2 = 1 + 0.917 702 897 664;
  • 32) 0.917 702 897 664 × 2 = 1 + 0.835 405 795 328;
  • 33) 0.835 405 795 328 × 2 = 1 + 0.670 811 590 656;
  • 34) 0.670 811 590 656 × 2 = 1 + 0.341 623 181 312;
  • 35) 0.341 623 181 312 × 2 = 0 + 0.683 246 362 624;
  • 36) 0.683 246 362 624 × 2 = 1 + 0.366 492 725 248;
  • 37) 0.366 492 725 248 × 2 = 0 + 0.732 985 450 496;
  • 38) 0.732 985 450 496 × 2 = 1 + 0.465 970 900 992;
  • 39) 0.465 970 900 992 × 2 = 0 + 0.931 941 801 984;
  • 40) 0.931 941 801 984 × 2 = 1 + 0.863 883 603 968;
  • 41) 0.863 883 603 968 × 2 = 1 + 0.727 767 207 936;
  • 42) 0.727 767 207 936 × 2 = 1 + 0.455 534 415 872;
  • 43) 0.455 534 415 872 × 2 = 0 + 0.911 068 831 744;
  • 44) 0.911 068 831 744 × 2 = 1 + 0.822 137 663 488;
  • 45) 0.822 137 663 488 × 2 = 1 + 0.644 275 326 976;
  • 46) 0.644 275 326 976 × 2 = 1 + 0.288 550 653 952;
  • 47) 0.288 550 653 952 × 2 = 0 + 0.577 101 307 904;
  • 48) 0.577 101 307 904 × 2 = 1 + 0.154 202 615 808;
  • 49) 0.154 202 615 808 × 2 = 0 + 0.308 405 231 616;
  • 50) 0.308 405 231 616 × 2 = 0 + 0.616 810 463 232;
  • 51) 0.616 810 463 232 × 2 = 1 + 0.233 620 926 464;
  • 52) 0.233 620 926 464 × 2 = 0 + 0.467 241 852 928;
  • 53) 0.467 241 852 928 × 2 = 0 + 0.934 483 705 856;
  • 54) 0.934 483 705 856 × 2 = 1 + 0.868 967 411 712;

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:


0.000 000 000 893(10) =

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

6. Positive number before normalization:

0.000 000 000 893(10) =

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


0.000 000 000 893(10) =

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

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

1.1110 1010 1110 1110 1001 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):
1.1110 1010 1110 1110 1001 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. 111 0101 0111 0111 0100 1001 =

111 0101 0111 0111 0100 1001


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) =
111 0101 0111 0111 0100 1001


Decimal number -0.000 000 000 893 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 0101 0111 0111 0100 1001

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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