-0.000 000 000 742 86 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 86(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
-0.000 000 000 742 86(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:

|-0.000 000 000 742 86| = 0.000 000 000 742 86


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: 0.000 000 000 742 86.

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) 0.000 000 000 742 86 × 2 = 0 + 0.000 000 001 485 72;
  • 2) 0.000 000 001 485 72 × 2 = 0 + 0.000 000 002 971 44;
  • 3) 0.000 000 002 971 44 × 2 = 0 + 0.000 000 005 942 88;
  • 4) 0.000 000 005 942 88 × 2 = 0 + 0.000 000 011 885 76;
  • 5) 0.000 000 011 885 76 × 2 = 0 + 0.000 000 023 771 52;
  • 6) 0.000 000 023 771 52 × 2 = 0 + 0.000 000 047 543 04;
  • 7) 0.000 000 047 543 04 × 2 = 0 + 0.000 000 095 086 08;
  • 8) 0.000 000 095 086 08 × 2 = 0 + 0.000 000 190 172 16;
  • 9) 0.000 000 190 172 16 × 2 = 0 + 0.000 000 380 344 32;
  • 10) 0.000 000 380 344 32 × 2 = 0 + 0.000 000 760 688 64;
  • 11) 0.000 000 760 688 64 × 2 = 0 + 0.000 001 521 377 28;
  • 12) 0.000 001 521 377 28 × 2 = 0 + 0.000 003 042 754 56;
  • 13) 0.000 003 042 754 56 × 2 = 0 + 0.000 006 085 509 12;
  • 14) 0.000 006 085 509 12 × 2 = 0 + 0.000 012 171 018 24;
  • 15) 0.000 012 171 018 24 × 2 = 0 + 0.000 024 342 036 48;
  • 16) 0.000 024 342 036 48 × 2 = 0 + 0.000 048 684 072 96;
  • 17) 0.000 048 684 072 96 × 2 = 0 + 0.000 097 368 145 92;
  • 18) 0.000 097 368 145 92 × 2 = 0 + 0.000 194 736 291 84;
  • 19) 0.000 194 736 291 84 × 2 = 0 + 0.000 389 472 583 68;
  • 20) 0.000 389 472 583 68 × 2 = 0 + 0.000 778 945 167 36;
  • 21) 0.000 778 945 167 36 × 2 = 0 + 0.001 557 890 334 72;
  • 22) 0.001 557 890 334 72 × 2 = 0 + 0.003 115 780 669 44;
  • 23) 0.003 115 780 669 44 × 2 = 0 + 0.006 231 561 338 88;
  • 24) 0.006 231 561 338 88 × 2 = 0 + 0.012 463 122 677 76;
  • 25) 0.012 463 122 677 76 × 2 = 0 + 0.024 926 245 355 52;
  • 26) 0.024 926 245 355 52 × 2 = 0 + 0.049 852 490 711 04;
  • 27) 0.049 852 490 711 04 × 2 = 0 + 0.099 704 981 422 08;
  • 28) 0.099 704 981 422 08 × 2 = 0 + 0.199 409 962 844 16;
  • 29) 0.199 409 962 844 16 × 2 = 0 + 0.398 819 925 688 32;
  • 30) 0.398 819 925 688 32 × 2 = 0 + 0.797 639 851 376 64;
  • 31) 0.797 639 851 376 64 × 2 = 1 + 0.595 279 702 753 28;
  • 32) 0.595 279 702 753 28 × 2 = 1 + 0.190 559 405 506 56;
  • 33) 0.190 559 405 506 56 × 2 = 0 + 0.381 118 811 013 12;
  • 34) 0.381 118 811 013 12 × 2 = 0 + 0.762 237 622 026 24;
  • 35) 0.762 237 622 026 24 × 2 = 1 + 0.524 475 244 052 48;
  • 36) 0.524 475 244 052 48 × 2 = 1 + 0.048 950 488 104 96;
  • 37) 0.048 950 488 104 96 × 2 = 0 + 0.097 900 976 209 92;
  • 38) 0.097 900 976 209 92 × 2 = 0 + 0.195 801 952 419 84;
  • 39) 0.195 801 952 419 84 × 2 = 0 + 0.391 603 904 839 68;
  • 40) 0.391 603 904 839 68 × 2 = 0 + 0.783 207 809 679 36;
  • 41) 0.783 207 809 679 36 × 2 = 1 + 0.566 415 619 358 72;
  • 42) 0.566 415 619 358 72 × 2 = 1 + 0.132 831 238 717 44;
  • 43) 0.132 831 238 717 44 × 2 = 0 + 0.265 662 477 434 88;
  • 44) 0.265 662 477 434 88 × 2 = 0 + 0.531 324 954 869 76;
  • 45) 0.531 324 954 869 76 × 2 = 1 + 0.062 649 909 739 52;
  • 46) 0.062 649 909 739 52 × 2 = 0 + 0.125 299 819 479 04;
  • 47) 0.125 299 819 479 04 × 2 = 0 + 0.250 599 638 958 08;
  • 48) 0.250 599 638 958 08 × 2 = 0 + 0.501 199 277 916 16;
  • 49) 0.501 199 277 916 16 × 2 = 1 + 0.002 398 555 832 32;
  • 50) 0.002 398 555 832 32 × 2 = 0 + 0.004 797 111 664 64;
  • 51) 0.004 797 111 664 64 × 2 = 0 + 0.009 594 223 329 28;
  • 52) 0.009 594 223 329 28 × 2 = 0 + 0.019 188 446 658 56;
  • 53) 0.019 188 446 658 56 × 2 = 0 + 0.038 376 893 317 12;
  • 54) 0.038 376 893 317 12 × 2 = 0 + 0.076 753 786 634 24;

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:


0.000 000 000 742 86(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1100 1000 1000 00(2)

6. Positive number before normalization:

0.000 000 000 742 86(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1100 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:


0.000 000 000 742 86(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1100 1000 1000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 1100 1000 1000 00(2) × 20 =

1.1001 1000 0110 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 1000 0110 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 1100 0011 0010 0010 0000 =

100 1100 0011 0010 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 1100 0011 0010 0010 0000


Decimal number -0.000 000 000 742 86 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0011 0010 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