-0.000 000 000 746 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 746(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
-0.000 000 000 746(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:

|-0.000 000 000 746| = 0.000 000 000 746


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: 0.000 000 000 746.

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) 0.000 000 000 746 × 2 = 0 + 0.000 000 001 492;
  • 2) 0.000 000 001 492 × 2 = 0 + 0.000 000 002 984;
  • 3) 0.000 000 002 984 × 2 = 0 + 0.000 000 005 968;
  • 4) 0.000 000 005 968 × 2 = 0 + 0.000 000 011 936;
  • 5) 0.000 000 011 936 × 2 = 0 + 0.000 000 023 872;
  • 6) 0.000 000 023 872 × 2 = 0 + 0.000 000 047 744;
  • 7) 0.000 000 047 744 × 2 = 0 + 0.000 000 095 488;
  • 8) 0.000 000 095 488 × 2 = 0 + 0.000 000 190 976;
  • 9) 0.000 000 190 976 × 2 = 0 + 0.000 000 381 952;
  • 10) 0.000 000 381 952 × 2 = 0 + 0.000 000 763 904;
  • 11) 0.000 000 763 904 × 2 = 0 + 0.000 001 527 808;
  • 12) 0.000 001 527 808 × 2 = 0 + 0.000 003 055 616;
  • 13) 0.000 003 055 616 × 2 = 0 + 0.000 006 111 232;
  • 14) 0.000 006 111 232 × 2 = 0 + 0.000 012 222 464;
  • 15) 0.000 012 222 464 × 2 = 0 + 0.000 024 444 928;
  • 16) 0.000 024 444 928 × 2 = 0 + 0.000 048 889 856;
  • 17) 0.000 048 889 856 × 2 = 0 + 0.000 097 779 712;
  • 18) 0.000 097 779 712 × 2 = 0 + 0.000 195 559 424;
  • 19) 0.000 195 559 424 × 2 = 0 + 0.000 391 118 848;
  • 20) 0.000 391 118 848 × 2 = 0 + 0.000 782 237 696;
  • 21) 0.000 782 237 696 × 2 = 0 + 0.001 564 475 392;
  • 22) 0.001 564 475 392 × 2 = 0 + 0.003 128 950 784;
  • 23) 0.003 128 950 784 × 2 = 0 + 0.006 257 901 568;
  • 24) 0.006 257 901 568 × 2 = 0 + 0.012 515 803 136;
  • 25) 0.012 515 803 136 × 2 = 0 + 0.025 031 606 272;
  • 26) 0.025 031 606 272 × 2 = 0 + 0.050 063 212 544;
  • 27) 0.050 063 212 544 × 2 = 0 + 0.100 126 425 088;
  • 28) 0.100 126 425 088 × 2 = 0 + 0.200 252 850 176;
  • 29) 0.200 252 850 176 × 2 = 0 + 0.400 505 700 352;
  • 30) 0.400 505 700 352 × 2 = 0 + 0.801 011 400 704;
  • 31) 0.801 011 400 704 × 2 = 1 + 0.602 022 801 408;
  • 32) 0.602 022 801 408 × 2 = 1 + 0.204 045 602 816;
  • 33) 0.204 045 602 816 × 2 = 0 + 0.408 091 205 632;
  • 34) 0.408 091 205 632 × 2 = 0 + 0.816 182 411 264;
  • 35) 0.816 182 411 264 × 2 = 1 + 0.632 364 822 528;
  • 36) 0.632 364 822 528 × 2 = 1 + 0.264 729 645 056;
  • 37) 0.264 729 645 056 × 2 = 0 + 0.529 459 290 112;
  • 38) 0.529 459 290 112 × 2 = 1 + 0.058 918 580 224;
  • 39) 0.058 918 580 224 × 2 = 0 + 0.117 837 160 448;
  • 40) 0.117 837 160 448 × 2 = 0 + 0.235 674 320 896;
  • 41) 0.235 674 320 896 × 2 = 0 + 0.471 348 641 792;
  • 42) 0.471 348 641 792 × 2 = 0 + 0.942 697 283 584;
  • 43) 0.942 697 283 584 × 2 = 1 + 0.885 394 567 168;
  • 44) 0.885 394 567 168 × 2 = 1 + 0.770 789 134 336;
  • 45) 0.770 789 134 336 × 2 = 1 + 0.541 578 268 672;
  • 46) 0.541 578 268 672 × 2 = 1 + 0.083 156 537 344;
  • 47) 0.083 156 537 344 × 2 = 0 + 0.166 313 074 688;
  • 48) 0.166 313 074 688 × 2 = 0 + 0.332 626 149 376;
  • 49) 0.332 626 149 376 × 2 = 0 + 0.665 252 298 752;
  • 50) 0.665 252 298 752 × 2 = 1 + 0.330 504 597 504;
  • 51) 0.330 504 597 504 × 2 = 0 + 0.661 009 195 008;
  • 52) 0.661 009 195 008 × 2 = 1 + 0.322 018 390 016;
  • 53) 0.322 018 390 016 × 2 = 0 + 0.644 036 780 032;
  • 54) 0.644 036 780 032 × 2 = 1 + 0.288 073 560 064;

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:


0.000 000 000 746(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0011 1100 0101 01(2)

6. Positive number before normalization:

0.000 000 000 746(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0011 1100 0101 01(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:


0.000 000 000 746(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0011 1100 0101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0100 0011 1100 0101 01(2) × 20 =

1.1001 1010 0001 1110 0010 101(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 1010 0001 1110 0010 101


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 0000 1111 0001 0101 =

100 1101 0000 1111 0001 0101


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 0000 1111 0001 0101


Decimal number -0.000 000 000 746 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1101 0000 1111 0001 0101

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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