-0.000 000 000 740 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 740 7(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 7(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 7| = 0.000 000 000 740 7


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 7.

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 7 × 2 = 0 + 0.000 000 001 481 4;
  • 2) 0.000 000 001 481 4 × 2 = 0 + 0.000 000 002 962 8;
  • 3) 0.000 000 002 962 8 × 2 = 0 + 0.000 000 005 925 6;
  • 4) 0.000 000 005 925 6 × 2 = 0 + 0.000 000 011 851 2;
  • 5) 0.000 000 011 851 2 × 2 = 0 + 0.000 000 023 702 4;
  • 6) 0.000 000 023 702 4 × 2 = 0 + 0.000 000 047 404 8;
  • 7) 0.000 000 047 404 8 × 2 = 0 + 0.000 000 094 809 6;
  • 8) 0.000 000 094 809 6 × 2 = 0 + 0.000 000 189 619 2;
  • 9) 0.000 000 189 619 2 × 2 = 0 + 0.000 000 379 238 4;
  • 10) 0.000 000 379 238 4 × 2 = 0 + 0.000 000 758 476 8;
  • 11) 0.000 000 758 476 8 × 2 = 0 + 0.000 001 516 953 6;
  • 12) 0.000 001 516 953 6 × 2 = 0 + 0.000 003 033 907 2;
  • 13) 0.000 003 033 907 2 × 2 = 0 + 0.000 006 067 814 4;
  • 14) 0.000 006 067 814 4 × 2 = 0 + 0.000 012 135 628 8;
  • 15) 0.000 012 135 628 8 × 2 = 0 + 0.000 024 271 257 6;
  • 16) 0.000 024 271 257 6 × 2 = 0 + 0.000 048 542 515 2;
  • 17) 0.000 048 542 515 2 × 2 = 0 + 0.000 097 085 030 4;
  • 18) 0.000 097 085 030 4 × 2 = 0 + 0.000 194 170 060 8;
  • 19) 0.000 194 170 060 8 × 2 = 0 + 0.000 388 340 121 6;
  • 20) 0.000 388 340 121 6 × 2 = 0 + 0.000 776 680 243 2;
  • 21) 0.000 776 680 243 2 × 2 = 0 + 0.001 553 360 486 4;
  • 22) 0.001 553 360 486 4 × 2 = 0 + 0.003 106 720 972 8;
  • 23) 0.003 106 720 972 8 × 2 = 0 + 0.006 213 441 945 6;
  • 24) 0.006 213 441 945 6 × 2 = 0 + 0.012 426 883 891 2;
  • 25) 0.012 426 883 891 2 × 2 = 0 + 0.024 853 767 782 4;
  • 26) 0.024 853 767 782 4 × 2 = 0 + 0.049 707 535 564 8;
  • 27) 0.049 707 535 564 8 × 2 = 0 + 0.099 415 071 129 6;
  • 28) 0.099 415 071 129 6 × 2 = 0 + 0.198 830 142 259 2;
  • 29) 0.198 830 142 259 2 × 2 = 0 + 0.397 660 284 518 4;
  • 30) 0.397 660 284 518 4 × 2 = 0 + 0.795 320 569 036 8;
  • 31) 0.795 320 569 036 8 × 2 = 1 + 0.590 641 138 073 6;
  • 32) 0.590 641 138 073 6 × 2 = 1 + 0.181 282 276 147 2;
  • 33) 0.181 282 276 147 2 × 2 = 0 + 0.362 564 552 294 4;
  • 34) 0.362 564 552 294 4 × 2 = 0 + 0.725 129 104 588 8;
  • 35) 0.725 129 104 588 8 × 2 = 1 + 0.450 258 209 177 6;
  • 36) 0.450 258 209 177 6 × 2 = 0 + 0.900 516 418 355 2;
  • 37) 0.900 516 418 355 2 × 2 = 1 + 0.801 032 836 710 4;
  • 38) 0.801 032 836 710 4 × 2 = 1 + 0.602 065 673 420 8;
  • 39) 0.602 065 673 420 8 × 2 = 1 + 0.204 131 346 841 6;
  • 40) 0.204 131 346 841 6 × 2 = 0 + 0.408 262 693 683 2;
  • 41) 0.408 262 693 683 2 × 2 = 0 + 0.816 525 387 366 4;
  • 42) 0.816 525 387 366 4 × 2 = 1 + 0.633 050 774 732 8;
  • 43) 0.633 050 774 732 8 × 2 = 1 + 0.266 101 549 465 6;
  • 44) 0.266 101 549 465 6 × 2 = 0 + 0.532 203 098 931 2;
  • 45) 0.532 203 098 931 2 × 2 = 1 + 0.064 406 197 862 4;
  • 46) 0.064 406 197 862 4 × 2 = 0 + 0.128 812 395 724 8;
  • 47) 0.128 812 395 724 8 × 2 = 0 + 0.257 624 791 449 6;
  • 48) 0.257 624 791 449 6 × 2 = 0 + 0.515 249 582 899 2;
  • 49) 0.515 249 582 899 2 × 2 = 1 + 0.030 499 165 798 4;
  • 50) 0.030 499 165 798 4 × 2 = 0 + 0.060 998 331 596 8;
  • 51) 0.060 998 331 596 8 × 2 = 0 + 0.121 996 663 193 6;
  • 52) 0.121 996 663 193 6 × 2 = 0 + 0.243 993 326 387 2;
  • 53) 0.243 993 326 387 2 × 2 = 0 + 0.487 986 652 774 4;
  • 54) 0.487 986 652 774 4 × 2 = 0 + 0.975 973 305 548 8;

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 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0110 1000 1000 00(2)

6. Positive number before normalization:

0.000 000 000 740 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0110 1000 1000 00(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 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0110 1000 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1110 0110 1000 1000 00(2) × 20 =

1.1001 0111 0011 0100 0100 000(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 0011 0100 0100 000


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 1001 1010 0010 0000 =

100 1011 1001 1010 0010 0000


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 1001 1010 0010 0000


Decimal number -0.000 000 000 740 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1001 1010 0010 0000

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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