-0.000 000 000 742 052 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 052(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
-0.000 000 000 742 052(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:

|-0.000 000 000 742 052| = 0.000 000 000 742 052


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: 0.000 000 000 742 052.

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) 0.000 000 000 742 052 × 2 = 0 + 0.000 000 001 484 104;
  • 2) 0.000 000 001 484 104 × 2 = 0 + 0.000 000 002 968 208;
  • 3) 0.000 000 002 968 208 × 2 = 0 + 0.000 000 005 936 416;
  • 4) 0.000 000 005 936 416 × 2 = 0 + 0.000 000 011 872 832;
  • 5) 0.000 000 011 872 832 × 2 = 0 + 0.000 000 023 745 664;
  • 6) 0.000 000 023 745 664 × 2 = 0 + 0.000 000 047 491 328;
  • 7) 0.000 000 047 491 328 × 2 = 0 + 0.000 000 094 982 656;
  • 8) 0.000 000 094 982 656 × 2 = 0 + 0.000 000 189 965 312;
  • 9) 0.000 000 189 965 312 × 2 = 0 + 0.000 000 379 930 624;
  • 10) 0.000 000 379 930 624 × 2 = 0 + 0.000 000 759 861 248;
  • 11) 0.000 000 759 861 248 × 2 = 0 + 0.000 001 519 722 496;
  • 12) 0.000 001 519 722 496 × 2 = 0 + 0.000 003 039 444 992;
  • 13) 0.000 003 039 444 992 × 2 = 0 + 0.000 006 078 889 984;
  • 14) 0.000 006 078 889 984 × 2 = 0 + 0.000 012 157 779 968;
  • 15) 0.000 012 157 779 968 × 2 = 0 + 0.000 024 315 559 936;
  • 16) 0.000 024 315 559 936 × 2 = 0 + 0.000 048 631 119 872;
  • 17) 0.000 048 631 119 872 × 2 = 0 + 0.000 097 262 239 744;
  • 18) 0.000 097 262 239 744 × 2 = 0 + 0.000 194 524 479 488;
  • 19) 0.000 194 524 479 488 × 2 = 0 + 0.000 389 048 958 976;
  • 20) 0.000 389 048 958 976 × 2 = 0 + 0.000 778 097 917 952;
  • 21) 0.000 778 097 917 952 × 2 = 0 + 0.001 556 195 835 904;
  • 22) 0.001 556 195 835 904 × 2 = 0 + 0.003 112 391 671 808;
  • 23) 0.003 112 391 671 808 × 2 = 0 + 0.006 224 783 343 616;
  • 24) 0.006 224 783 343 616 × 2 = 0 + 0.012 449 566 687 232;
  • 25) 0.012 449 566 687 232 × 2 = 0 + 0.024 899 133 374 464;
  • 26) 0.024 899 133 374 464 × 2 = 0 + 0.049 798 266 748 928;
  • 27) 0.049 798 266 748 928 × 2 = 0 + 0.099 596 533 497 856;
  • 28) 0.099 596 533 497 856 × 2 = 0 + 0.199 193 066 995 712;
  • 29) 0.199 193 066 995 712 × 2 = 0 + 0.398 386 133 991 424;
  • 30) 0.398 386 133 991 424 × 2 = 0 + 0.796 772 267 982 848;
  • 31) 0.796 772 267 982 848 × 2 = 1 + 0.593 544 535 965 696;
  • 32) 0.593 544 535 965 696 × 2 = 1 + 0.187 089 071 931 392;
  • 33) 0.187 089 071 931 392 × 2 = 0 + 0.374 178 143 862 784;
  • 34) 0.374 178 143 862 784 × 2 = 0 + 0.748 356 287 725 568;
  • 35) 0.748 356 287 725 568 × 2 = 1 + 0.496 712 575 451 136;
  • 36) 0.496 712 575 451 136 × 2 = 0 + 0.993 425 150 902 272;
  • 37) 0.993 425 150 902 272 × 2 = 1 + 0.986 850 301 804 544;
  • 38) 0.986 850 301 804 544 × 2 = 1 + 0.973 700 603 609 088;
  • 39) 0.973 700 603 609 088 × 2 = 1 + 0.947 401 207 218 176;
  • 40) 0.947 401 207 218 176 × 2 = 1 + 0.894 802 414 436 352;
  • 41) 0.894 802 414 436 352 × 2 = 1 + 0.789 604 828 872 704;
  • 42) 0.789 604 828 872 704 × 2 = 1 + 0.579 209 657 745 408;
  • 43) 0.579 209 657 745 408 × 2 = 1 + 0.158 419 315 490 816;
  • 44) 0.158 419 315 490 816 × 2 = 0 + 0.316 838 630 981 632;
  • 45) 0.316 838 630 981 632 × 2 = 0 + 0.633 677 261 963 264;
  • 46) 0.633 677 261 963 264 × 2 = 1 + 0.267 354 523 926 528;
  • 47) 0.267 354 523 926 528 × 2 = 0 + 0.534 709 047 853 056;
  • 48) 0.534 709 047 853 056 × 2 = 1 + 0.069 418 095 706 112;
  • 49) 0.069 418 095 706 112 × 2 = 0 + 0.138 836 191 412 224;
  • 50) 0.138 836 191 412 224 × 2 = 0 + 0.277 672 382 824 448;
  • 51) 0.277 672 382 824 448 × 2 = 0 + 0.555 344 765 648 896;
  • 52) 0.555 344 765 648 896 × 2 = 1 + 0.110 689 531 297 792;
  • 53) 0.110 689 531 297 792 × 2 = 0 + 0.221 379 062 595 584;
  • 54) 0.221 379 062 595 584 × 2 = 0 + 0.442 758 125 191 168;

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:


0.000 000 000 742 052(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0101 0001 00(2)

6. Positive number before normalization:

0.000 000 000 742 052(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0101 0001 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:


0.000 000 000 742 052(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0101 0001 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1110 0101 0001 00(2) × 20 =

1.1001 0111 1111 0010 1000 100(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 1111 0010 1000 100


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 1111 1001 0100 0100 =

100 1011 1111 1001 0100 0100


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 1111 1001 0100 0100


Decimal number -0.000 000 000 742 052 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1001 0100 0100

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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