-0.000 000 047 684 745 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 047 684 745(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
-0.000 000 047 684 745(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:

|-0.000 000 047 684 745| = 0.000 000 047 684 745


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: 0.000 000 047 684 745.

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) 0.000 000 047 684 745 × 2 = 0 + 0.000 000 095 369 49;
  • 2) 0.000 000 095 369 49 × 2 = 0 + 0.000 000 190 738 98;
  • 3) 0.000 000 190 738 98 × 2 = 0 + 0.000 000 381 477 96;
  • 4) 0.000 000 381 477 96 × 2 = 0 + 0.000 000 762 955 92;
  • 5) 0.000 000 762 955 92 × 2 = 0 + 0.000 001 525 911 84;
  • 6) 0.000 001 525 911 84 × 2 = 0 + 0.000 003 051 823 68;
  • 7) 0.000 003 051 823 68 × 2 = 0 + 0.000 006 103 647 36;
  • 8) 0.000 006 103 647 36 × 2 = 0 + 0.000 012 207 294 72;
  • 9) 0.000 012 207 294 72 × 2 = 0 + 0.000 024 414 589 44;
  • 10) 0.000 024 414 589 44 × 2 = 0 + 0.000 048 829 178 88;
  • 11) 0.000 048 829 178 88 × 2 = 0 + 0.000 097 658 357 76;
  • 12) 0.000 097 658 357 76 × 2 = 0 + 0.000 195 316 715 52;
  • 13) 0.000 195 316 715 52 × 2 = 0 + 0.000 390 633 431 04;
  • 14) 0.000 390 633 431 04 × 2 = 0 + 0.000 781 266 862 08;
  • 15) 0.000 781 266 862 08 × 2 = 0 + 0.001 562 533 724 16;
  • 16) 0.001 562 533 724 16 × 2 = 0 + 0.003 125 067 448 32;
  • 17) 0.003 125 067 448 32 × 2 = 0 + 0.006 250 134 896 64;
  • 18) 0.006 250 134 896 64 × 2 = 0 + 0.012 500 269 793 28;
  • 19) 0.012 500 269 793 28 × 2 = 0 + 0.025 000 539 586 56;
  • 20) 0.025 000 539 586 56 × 2 = 0 + 0.050 001 079 173 12;
  • 21) 0.050 001 079 173 12 × 2 = 0 + 0.100 002 158 346 24;
  • 22) 0.100 002 158 346 24 × 2 = 0 + 0.200 004 316 692 48;
  • 23) 0.200 004 316 692 48 × 2 = 0 + 0.400 008 633 384 96;
  • 24) 0.400 008 633 384 96 × 2 = 0 + 0.800 017 266 769 92;
  • 25) 0.800 017 266 769 92 × 2 = 1 + 0.600 034 533 539 84;
  • 26) 0.600 034 533 539 84 × 2 = 1 + 0.200 069 067 079 68;
  • 27) 0.200 069 067 079 68 × 2 = 0 + 0.400 138 134 159 36;
  • 28) 0.400 138 134 159 36 × 2 = 0 + 0.800 276 268 318 72;
  • 29) 0.800 276 268 318 72 × 2 = 1 + 0.600 552 536 637 44;
  • 30) 0.600 552 536 637 44 × 2 = 1 + 0.201 105 073 274 88;
  • 31) 0.201 105 073 274 88 × 2 = 0 + 0.402 210 146 549 76;
  • 32) 0.402 210 146 549 76 × 2 = 0 + 0.804 420 293 099 52;
  • 33) 0.804 420 293 099 52 × 2 = 1 + 0.608 840 586 199 04;
  • 34) 0.608 840 586 199 04 × 2 = 1 + 0.217 681 172 398 08;
  • 35) 0.217 681 172 398 08 × 2 = 0 + 0.435 362 344 796 16;
  • 36) 0.435 362 344 796 16 × 2 = 0 + 0.870 724 689 592 32;
  • 37) 0.870 724 689 592 32 × 2 = 1 + 0.741 449 379 184 64;
  • 38) 0.741 449 379 184 64 × 2 = 1 + 0.482 898 758 369 28;
  • 39) 0.482 898 758 369 28 × 2 = 0 + 0.965 797 516 738 56;
  • 40) 0.965 797 516 738 56 × 2 = 1 + 0.931 595 033 477 12;
  • 41) 0.931 595 033 477 12 × 2 = 1 + 0.863 190 066 954 24;
  • 42) 0.863 190 066 954 24 × 2 = 1 + 0.726 380 133 908 48;
  • 43) 0.726 380 133 908 48 × 2 = 1 + 0.452 760 267 816 96;
  • 44) 0.452 760 267 816 96 × 2 = 0 + 0.905 520 535 633 92;
  • 45) 0.905 520 535 633 92 × 2 = 1 + 0.811 041 071 267 84;
  • 46) 0.811 041 071 267 84 × 2 = 1 + 0.622 082 142 535 68;
  • 47) 0.622 082 142 535 68 × 2 = 1 + 0.244 164 285 071 36;
  • 48) 0.244 164 285 071 36 × 2 = 0 + 0.488 328 570 142 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:


0.000 000 047 684 745(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1110 1110(2)

6. Positive number before normalization:

0.000 000 047 684 745(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1110 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 047 684 745(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1110 1110(2) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1110 1110(2) × 20 =

1.1001 1001 1001 1011 1101 110(2) × 2-25


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): -25

Mantissa (not normalized):
1.1001 1001 1001 1011 1101 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(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;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

102(10) =

0110 0110(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 1100 1100 1101 1110 1110 =

100 1100 1100 1101 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 0110

Mantissa (23 bits) =
100 1100 1100 1101 1110 1110


Decimal number -0.000 000 047 684 745 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 100 1100 1100 1101 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