-0.000 000 000 726 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 726(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
-0.000 000 000 726(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:

|-0.000 000 000 726| = 0.000 000 000 726


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: 0.000 000 000 726.

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) 0.000 000 000 726 × 2 = 0 + 0.000 000 001 452;
  • 2) 0.000 000 001 452 × 2 = 0 + 0.000 000 002 904;
  • 3) 0.000 000 002 904 × 2 = 0 + 0.000 000 005 808;
  • 4) 0.000 000 005 808 × 2 = 0 + 0.000 000 011 616;
  • 5) 0.000 000 011 616 × 2 = 0 + 0.000 000 023 232;
  • 6) 0.000 000 023 232 × 2 = 0 + 0.000 000 046 464;
  • 7) 0.000 000 046 464 × 2 = 0 + 0.000 000 092 928;
  • 8) 0.000 000 092 928 × 2 = 0 + 0.000 000 185 856;
  • 9) 0.000 000 185 856 × 2 = 0 + 0.000 000 371 712;
  • 10) 0.000 000 371 712 × 2 = 0 + 0.000 000 743 424;
  • 11) 0.000 000 743 424 × 2 = 0 + 0.000 001 486 848;
  • 12) 0.000 001 486 848 × 2 = 0 + 0.000 002 973 696;
  • 13) 0.000 002 973 696 × 2 = 0 + 0.000 005 947 392;
  • 14) 0.000 005 947 392 × 2 = 0 + 0.000 011 894 784;
  • 15) 0.000 011 894 784 × 2 = 0 + 0.000 023 789 568;
  • 16) 0.000 023 789 568 × 2 = 0 + 0.000 047 579 136;
  • 17) 0.000 047 579 136 × 2 = 0 + 0.000 095 158 272;
  • 18) 0.000 095 158 272 × 2 = 0 + 0.000 190 316 544;
  • 19) 0.000 190 316 544 × 2 = 0 + 0.000 380 633 088;
  • 20) 0.000 380 633 088 × 2 = 0 + 0.000 761 266 176;
  • 21) 0.000 761 266 176 × 2 = 0 + 0.001 522 532 352;
  • 22) 0.001 522 532 352 × 2 = 0 + 0.003 045 064 704;
  • 23) 0.003 045 064 704 × 2 = 0 + 0.006 090 129 408;
  • 24) 0.006 090 129 408 × 2 = 0 + 0.012 180 258 816;
  • 25) 0.012 180 258 816 × 2 = 0 + 0.024 360 517 632;
  • 26) 0.024 360 517 632 × 2 = 0 + 0.048 721 035 264;
  • 27) 0.048 721 035 264 × 2 = 0 + 0.097 442 070 528;
  • 28) 0.097 442 070 528 × 2 = 0 + 0.194 884 141 056;
  • 29) 0.194 884 141 056 × 2 = 0 + 0.389 768 282 112;
  • 30) 0.389 768 282 112 × 2 = 0 + 0.779 536 564 224;
  • 31) 0.779 536 564 224 × 2 = 1 + 0.559 073 128 448;
  • 32) 0.559 073 128 448 × 2 = 1 + 0.118 146 256 896;
  • 33) 0.118 146 256 896 × 2 = 0 + 0.236 292 513 792;
  • 34) 0.236 292 513 792 × 2 = 0 + 0.472 585 027 584;
  • 35) 0.472 585 027 584 × 2 = 0 + 0.945 170 055 168;
  • 36) 0.945 170 055 168 × 2 = 1 + 0.890 340 110 336;
  • 37) 0.890 340 110 336 × 2 = 1 + 0.780 680 220 672;
  • 38) 0.780 680 220 672 × 2 = 1 + 0.561 360 441 344;
  • 39) 0.561 360 441 344 × 2 = 1 + 0.122 720 882 688;
  • 40) 0.122 720 882 688 × 2 = 0 + 0.245 441 765 376;
  • 41) 0.245 441 765 376 × 2 = 0 + 0.490 883 530 752;
  • 42) 0.490 883 530 752 × 2 = 0 + 0.981 767 061 504;
  • 43) 0.981 767 061 504 × 2 = 1 + 0.963 534 123 008;
  • 44) 0.963 534 123 008 × 2 = 1 + 0.927 068 246 016;
  • 45) 0.927 068 246 016 × 2 = 1 + 0.854 136 492 032;
  • 46) 0.854 136 492 032 × 2 = 1 + 0.708 272 984 064;
  • 47) 0.708 272 984 064 × 2 = 1 + 0.416 545 968 128;
  • 48) 0.416 545 968 128 × 2 = 0 + 0.833 091 936 256;
  • 49) 0.833 091 936 256 × 2 = 1 + 0.666 183 872 512;
  • 50) 0.666 183 872 512 × 2 = 1 + 0.332 367 745 024;
  • 51) 0.332 367 745 024 × 2 = 0 + 0.664 735 490 048;
  • 52) 0.664 735 490 048 × 2 = 1 + 0.329 470 980 096;
  • 53) 0.329 470 980 096 × 2 = 0 + 0.658 941 960 192;
  • 54) 0.658 941 960 192 × 2 = 1 + 0.317 883 920 384;

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:


0.000 000 000 726(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1110 0011 1110 1101 01(2)

6. Positive number before normalization:

0.000 000 000 726(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1110 0011 1110 1101 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:


0.000 000 000 726(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1110 0011 1110 1101 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1110 0011 1110 1101 01(2) × 20 =

1.1000 1111 0001 1111 0110 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):
1.1000 1111 0001 1111 0110 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 0111 1000 1111 1011 0101 =

100 0111 1000 1111 1011 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 0111 1000 1111 1011 0101


Decimal number -0.000 000 000 726 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0111 1000 1111 1011 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