-0.000 000 000 748 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 748 3(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
-0.000 000 000 748 3(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:

|-0.000 000 000 748 3| = 0.000 000 000 748 3


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: 0.000 000 000 748 3.

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) 0.000 000 000 748 3 × 2 = 0 + 0.000 000 001 496 6;
  • 2) 0.000 000 001 496 6 × 2 = 0 + 0.000 000 002 993 2;
  • 3) 0.000 000 002 993 2 × 2 = 0 + 0.000 000 005 986 4;
  • 4) 0.000 000 005 986 4 × 2 = 0 + 0.000 000 011 972 8;
  • 5) 0.000 000 011 972 8 × 2 = 0 + 0.000 000 023 945 6;
  • 6) 0.000 000 023 945 6 × 2 = 0 + 0.000 000 047 891 2;
  • 7) 0.000 000 047 891 2 × 2 = 0 + 0.000 000 095 782 4;
  • 8) 0.000 000 095 782 4 × 2 = 0 + 0.000 000 191 564 8;
  • 9) 0.000 000 191 564 8 × 2 = 0 + 0.000 000 383 129 6;
  • 10) 0.000 000 383 129 6 × 2 = 0 + 0.000 000 766 259 2;
  • 11) 0.000 000 766 259 2 × 2 = 0 + 0.000 001 532 518 4;
  • 12) 0.000 001 532 518 4 × 2 = 0 + 0.000 003 065 036 8;
  • 13) 0.000 003 065 036 8 × 2 = 0 + 0.000 006 130 073 6;
  • 14) 0.000 006 130 073 6 × 2 = 0 + 0.000 012 260 147 2;
  • 15) 0.000 012 260 147 2 × 2 = 0 + 0.000 024 520 294 4;
  • 16) 0.000 024 520 294 4 × 2 = 0 + 0.000 049 040 588 8;
  • 17) 0.000 049 040 588 8 × 2 = 0 + 0.000 098 081 177 6;
  • 18) 0.000 098 081 177 6 × 2 = 0 + 0.000 196 162 355 2;
  • 19) 0.000 196 162 355 2 × 2 = 0 + 0.000 392 324 710 4;
  • 20) 0.000 392 324 710 4 × 2 = 0 + 0.000 784 649 420 8;
  • 21) 0.000 784 649 420 8 × 2 = 0 + 0.001 569 298 841 6;
  • 22) 0.001 569 298 841 6 × 2 = 0 + 0.003 138 597 683 2;
  • 23) 0.003 138 597 683 2 × 2 = 0 + 0.006 277 195 366 4;
  • 24) 0.006 277 195 366 4 × 2 = 0 + 0.012 554 390 732 8;
  • 25) 0.012 554 390 732 8 × 2 = 0 + 0.025 108 781 465 6;
  • 26) 0.025 108 781 465 6 × 2 = 0 + 0.050 217 562 931 2;
  • 27) 0.050 217 562 931 2 × 2 = 0 + 0.100 435 125 862 4;
  • 28) 0.100 435 125 862 4 × 2 = 0 + 0.200 870 251 724 8;
  • 29) 0.200 870 251 724 8 × 2 = 0 + 0.401 740 503 449 6;
  • 30) 0.401 740 503 449 6 × 2 = 0 + 0.803 481 006 899 2;
  • 31) 0.803 481 006 899 2 × 2 = 1 + 0.606 962 013 798 4;
  • 32) 0.606 962 013 798 4 × 2 = 1 + 0.213 924 027 596 8;
  • 33) 0.213 924 027 596 8 × 2 = 0 + 0.427 848 055 193 6;
  • 34) 0.427 848 055 193 6 × 2 = 0 + 0.855 696 110 387 2;
  • 35) 0.855 696 110 387 2 × 2 = 1 + 0.711 392 220 774 4;
  • 36) 0.711 392 220 774 4 × 2 = 1 + 0.422 784 441 548 8;
  • 37) 0.422 784 441 548 8 × 2 = 0 + 0.845 568 883 097 6;
  • 38) 0.845 568 883 097 6 × 2 = 1 + 0.691 137 766 195 2;
  • 39) 0.691 137 766 195 2 × 2 = 1 + 0.382 275 532 390 4;
  • 40) 0.382 275 532 390 4 × 2 = 0 + 0.764 551 064 780 8;
  • 41) 0.764 551 064 780 8 × 2 = 1 + 0.529 102 129 561 6;
  • 42) 0.529 102 129 561 6 × 2 = 1 + 0.058 204 259 123 2;
  • 43) 0.058 204 259 123 2 × 2 = 0 + 0.116 408 518 246 4;
  • 44) 0.116 408 518 246 4 × 2 = 0 + 0.232 817 036 492 8;
  • 45) 0.232 817 036 492 8 × 2 = 0 + 0.465 634 072 985 6;
  • 46) 0.465 634 072 985 6 × 2 = 0 + 0.931 268 145 971 2;
  • 47) 0.931 268 145 971 2 × 2 = 1 + 0.862 536 291 942 4;
  • 48) 0.862 536 291 942 4 × 2 = 1 + 0.725 072 583 884 8;
  • 49) 0.725 072 583 884 8 × 2 = 1 + 0.450 145 167 769 6;
  • 50) 0.450 145 167 769 6 × 2 = 0 + 0.900 290 335 539 2;
  • 51) 0.900 290 335 539 2 × 2 = 1 + 0.800 580 671 078 4;
  • 52) 0.800 580 671 078 4 × 2 = 1 + 0.601 161 342 156 8;
  • 53) 0.601 161 342 156 8 × 2 = 1 + 0.202 322 684 313 6;
  • 54) 0.202 322 684 313 6 × 2 = 0 + 0.404 645 368 627 2;

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:


0.000 000 000 748 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 1100 0011 1011 10(2)

6. Positive number before normalization:

0.000 000 000 748 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 1100 0011 1011 10(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:


0.000 000 000 748 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 1100 0011 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0110 1100 0011 1011 10(2) × 20 =

1.1001 1011 0110 0001 1101 110(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 1011 0110 0001 1101 110


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 1101 1011 0000 1110 1110 =

100 1101 1011 0000 1110 1110


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 1101 1011 0000 1110 1110


Decimal number -0.000 000 000 748 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1101 1011 0000 1110 1110

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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