-0.000 000 000 741 873 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 741 873(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
-0.000 000 000 741 873(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:

|-0.000 000 000 741 873| = 0.000 000 000 741 873


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: 0.000 000 000 741 873.

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) 0.000 000 000 741 873 × 2 = 0 + 0.000 000 001 483 746;
  • 2) 0.000 000 001 483 746 × 2 = 0 + 0.000 000 002 967 492;
  • 3) 0.000 000 002 967 492 × 2 = 0 + 0.000 000 005 934 984;
  • 4) 0.000 000 005 934 984 × 2 = 0 + 0.000 000 011 869 968;
  • 5) 0.000 000 011 869 968 × 2 = 0 + 0.000 000 023 739 936;
  • 6) 0.000 000 023 739 936 × 2 = 0 + 0.000 000 047 479 872;
  • 7) 0.000 000 047 479 872 × 2 = 0 + 0.000 000 094 959 744;
  • 8) 0.000 000 094 959 744 × 2 = 0 + 0.000 000 189 919 488;
  • 9) 0.000 000 189 919 488 × 2 = 0 + 0.000 000 379 838 976;
  • 10) 0.000 000 379 838 976 × 2 = 0 + 0.000 000 759 677 952;
  • 11) 0.000 000 759 677 952 × 2 = 0 + 0.000 001 519 355 904;
  • 12) 0.000 001 519 355 904 × 2 = 0 + 0.000 003 038 711 808;
  • 13) 0.000 003 038 711 808 × 2 = 0 + 0.000 006 077 423 616;
  • 14) 0.000 006 077 423 616 × 2 = 0 + 0.000 012 154 847 232;
  • 15) 0.000 012 154 847 232 × 2 = 0 + 0.000 024 309 694 464;
  • 16) 0.000 024 309 694 464 × 2 = 0 + 0.000 048 619 388 928;
  • 17) 0.000 048 619 388 928 × 2 = 0 + 0.000 097 238 777 856;
  • 18) 0.000 097 238 777 856 × 2 = 0 + 0.000 194 477 555 712;
  • 19) 0.000 194 477 555 712 × 2 = 0 + 0.000 388 955 111 424;
  • 20) 0.000 388 955 111 424 × 2 = 0 + 0.000 777 910 222 848;
  • 21) 0.000 777 910 222 848 × 2 = 0 + 0.001 555 820 445 696;
  • 22) 0.001 555 820 445 696 × 2 = 0 + 0.003 111 640 891 392;
  • 23) 0.003 111 640 891 392 × 2 = 0 + 0.006 223 281 782 784;
  • 24) 0.006 223 281 782 784 × 2 = 0 + 0.012 446 563 565 568;
  • 25) 0.012 446 563 565 568 × 2 = 0 + 0.024 893 127 131 136;
  • 26) 0.024 893 127 131 136 × 2 = 0 + 0.049 786 254 262 272;
  • 27) 0.049 786 254 262 272 × 2 = 0 + 0.099 572 508 524 544;
  • 28) 0.099 572 508 524 544 × 2 = 0 + 0.199 145 017 049 088;
  • 29) 0.199 145 017 049 088 × 2 = 0 + 0.398 290 034 098 176;
  • 30) 0.398 290 034 098 176 × 2 = 0 + 0.796 580 068 196 352;
  • 31) 0.796 580 068 196 352 × 2 = 1 + 0.593 160 136 392 704;
  • 32) 0.593 160 136 392 704 × 2 = 1 + 0.186 320 272 785 408;
  • 33) 0.186 320 272 785 408 × 2 = 0 + 0.372 640 545 570 816;
  • 34) 0.372 640 545 570 816 × 2 = 0 + 0.745 281 091 141 632;
  • 35) 0.745 281 091 141 632 × 2 = 1 + 0.490 562 182 283 264;
  • 36) 0.490 562 182 283 264 × 2 = 0 + 0.981 124 364 566 528;
  • 37) 0.981 124 364 566 528 × 2 = 1 + 0.962 248 729 133 056;
  • 38) 0.962 248 729 133 056 × 2 = 1 + 0.924 497 458 266 112;
  • 39) 0.924 497 458 266 112 × 2 = 1 + 0.848 994 916 532 224;
  • 40) 0.848 994 916 532 224 × 2 = 1 + 0.697 989 833 064 448;
  • 41) 0.697 989 833 064 448 × 2 = 1 + 0.395 979 666 128 896;
  • 42) 0.395 979 666 128 896 × 2 = 0 + 0.791 959 332 257 792;
  • 43) 0.791 959 332 257 792 × 2 = 1 + 0.583 918 664 515 584;
  • 44) 0.583 918 664 515 584 × 2 = 1 + 0.167 837 329 031 168;
  • 45) 0.167 837 329 031 168 × 2 = 0 + 0.335 674 658 062 336;
  • 46) 0.335 674 658 062 336 × 2 = 0 + 0.671 349 316 124 672;
  • 47) 0.671 349 316 124 672 × 2 = 1 + 0.342 698 632 249 344;
  • 48) 0.342 698 632 249 344 × 2 = 0 + 0.685 397 264 498 688;
  • 49) 0.685 397 264 498 688 × 2 = 1 + 0.370 794 528 997 376;
  • 50) 0.370 794 528 997 376 × 2 = 0 + 0.741 589 057 994 752;
  • 51) 0.741 589 057 994 752 × 2 = 1 + 0.483 178 115 989 504;
  • 52) 0.483 178 115 989 504 × 2 = 0 + 0.966 356 231 979 008;
  • 53) 0.966 356 231 979 008 × 2 = 1 + 0.932 712 463 958 016;
  • 54) 0.932 712 463 958 016 × 2 = 1 + 0.865 424 927 916 032;

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:


0.000 000 000 741 873(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 0010 1010 11(2)

6. Positive number before normalization:

0.000 000 000 741 873(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 0010 1010 11(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:


0.000 000 000 741 873(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 0010 1010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1011 0010 1010 11(2) × 20 =

1.1001 0111 1101 1001 0101 011(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 0111 1101 1001 0101 011


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 1011 1110 1100 1010 1011 =

100 1011 1110 1100 1010 1011


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 1011 1110 1100 1010 1011


Decimal number -0.000 000 000 741 873 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1110 1100 1010 1011

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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