-0.000 000 000 749 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 749(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
-0.000 000 000 749(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:

|-0.000 000 000 749| = 0.000 000 000 749


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: 0.000 000 000 749.

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) 0.000 000 000 749 × 2 = 0 + 0.000 000 001 498;
  • 2) 0.000 000 001 498 × 2 = 0 + 0.000 000 002 996;
  • 3) 0.000 000 002 996 × 2 = 0 + 0.000 000 005 992;
  • 4) 0.000 000 005 992 × 2 = 0 + 0.000 000 011 984;
  • 5) 0.000 000 011 984 × 2 = 0 + 0.000 000 023 968;
  • 6) 0.000 000 023 968 × 2 = 0 + 0.000 000 047 936;
  • 7) 0.000 000 047 936 × 2 = 0 + 0.000 000 095 872;
  • 8) 0.000 000 095 872 × 2 = 0 + 0.000 000 191 744;
  • 9) 0.000 000 191 744 × 2 = 0 + 0.000 000 383 488;
  • 10) 0.000 000 383 488 × 2 = 0 + 0.000 000 766 976;
  • 11) 0.000 000 766 976 × 2 = 0 + 0.000 001 533 952;
  • 12) 0.000 001 533 952 × 2 = 0 + 0.000 003 067 904;
  • 13) 0.000 003 067 904 × 2 = 0 + 0.000 006 135 808;
  • 14) 0.000 006 135 808 × 2 = 0 + 0.000 012 271 616;
  • 15) 0.000 012 271 616 × 2 = 0 + 0.000 024 543 232;
  • 16) 0.000 024 543 232 × 2 = 0 + 0.000 049 086 464;
  • 17) 0.000 049 086 464 × 2 = 0 + 0.000 098 172 928;
  • 18) 0.000 098 172 928 × 2 = 0 + 0.000 196 345 856;
  • 19) 0.000 196 345 856 × 2 = 0 + 0.000 392 691 712;
  • 20) 0.000 392 691 712 × 2 = 0 + 0.000 785 383 424;
  • 21) 0.000 785 383 424 × 2 = 0 + 0.001 570 766 848;
  • 22) 0.001 570 766 848 × 2 = 0 + 0.003 141 533 696;
  • 23) 0.003 141 533 696 × 2 = 0 + 0.006 283 067 392;
  • 24) 0.006 283 067 392 × 2 = 0 + 0.012 566 134 784;
  • 25) 0.012 566 134 784 × 2 = 0 + 0.025 132 269 568;
  • 26) 0.025 132 269 568 × 2 = 0 + 0.050 264 539 136;
  • 27) 0.050 264 539 136 × 2 = 0 + 0.100 529 078 272;
  • 28) 0.100 529 078 272 × 2 = 0 + 0.201 058 156 544;
  • 29) 0.201 058 156 544 × 2 = 0 + 0.402 116 313 088;
  • 30) 0.402 116 313 088 × 2 = 0 + 0.804 232 626 176;
  • 31) 0.804 232 626 176 × 2 = 1 + 0.608 465 252 352;
  • 32) 0.608 465 252 352 × 2 = 1 + 0.216 930 504 704;
  • 33) 0.216 930 504 704 × 2 = 0 + 0.433 861 009 408;
  • 34) 0.433 861 009 408 × 2 = 0 + 0.867 722 018 816;
  • 35) 0.867 722 018 816 × 2 = 1 + 0.735 444 037 632;
  • 36) 0.735 444 037 632 × 2 = 1 + 0.470 888 075 264;
  • 37) 0.470 888 075 264 × 2 = 0 + 0.941 776 150 528;
  • 38) 0.941 776 150 528 × 2 = 1 + 0.883 552 301 056;
  • 39) 0.883 552 301 056 × 2 = 1 + 0.767 104 602 112;
  • 40) 0.767 104 602 112 × 2 = 1 + 0.534 209 204 224;
  • 41) 0.534 209 204 224 × 2 = 1 + 0.068 418 408 448;
  • 42) 0.068 418 408 448 × 2 = 0 + 0.136 836 816 896;
  • 43) 0.136 836 816 896 × 2 = 0 + 0.273 673 633 792;
  • 44) 0.273 673 633 792 × 2 = 0 + 0.547 347 267 584;
  • 45) 0.547 347 267 584 × 2 = 1 + 0.094 694 535 168;
  • 46) 0.094 694 535 168 × 2 = 0 + 0.189 389 070 336;
  • 47) 0.189 389 070 336 × 2 = 0 + 0.378 778 140 672;
  • 48) 0.378 778 140 672 × 2 = 0 + 0.757 556 281 344;
  • 49) 0.757 556 281 344 × 2 = 1 + 0.515 112 562 688;
  • 50) 0.515 112 562 688 × 2 = 1 + 0.030 225 125 376;
  • 51) 0.030 225 125 376 × 2 = 0 + 0.060 450 250 752;
  • 52) 0.060 450 250 752 × 2 = 0 + 0.120 900 501 504;
  • 53) 0.120 900 501 504 × 2 = 0 + 0.241 801 003 008;
  • 54) 0.241 801 003 008 × 2 = 0 + 0.483 602 006 016;

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:


0.000 000 000 749(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0111 1000 1000 1100 00(2)

6. Positive number before normalization:

0.000 000 000 749(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0111 1000 1000 1100 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:


0.000 000 000 749(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0111 1000 1000 1100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0111 1000 1000 1100 00(2) × 20 =

1.1001 1011 1100 0100 0110 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 1011 1100 0100 0110 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 1101 1110 0010 0011 0000 =

100 1101 1110 0010 0011 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 1101 1110 0010 0011 0000


Decimal number -0.000 000 000 749 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1101 1110 0010 0011 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