-0.000 000 000 742 147 833 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 833(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
-0.000 000 000 742 147 833(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:

|-0.000 000 000 742 147 833| = 0.000 000 000 742 147 833


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: 0.000 000 000 742 147 833.

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) 0.000 000 000 742 147 833 × 2 = 0 + 0.000 000 001 484 295 666;
  • 2) 0.000 000 001 484 295 666 × 2 = 0 + 0.000 000 002 968 591 332;
  • 3) 0.000 000 002 968 591 332 × 2 = 0 + 0.000 000 005 937 182 664;
  • 4) 0.000 000 005 937 182 664 × 2 = 0 + 0.000 000 011 874 365 328;
  • 5) 0.000 000 011 874 365 328 × 2 = 0 + 0.000 000 023 748 730 656;
  • 6) 0.000 000 023 748 730 656 × 2 = 0 + 0.000 000 047 497 461 312;
  • 7) 0.000 000 047 497 461 312 × 2 = 0 + 0.000 000 094 994 922 624;
  • 8) 0.000 000 094 994 922 624 × 2 = 0 + 0.000 000 189 989 845 248;
  • 9) 0.000 000 189 989 845 248 × 2 = 0 + 0.000 000 379 979 690 496;
  • 10) 0.000 000 379 979 690 496 × 2 = 0 + 0.000 000 759 959 380 992;
  • 11) 0.000 000 759 959 380 992 × 2 = 0 + 0.000 001 519 918 761 984;
  • 12) 0.000 001 519 918 761 984 × 2 = 0 + 0.000 003 039 837 523 968;
  • 13) 0.000 003 039 837 523 968 × 2 = 0 + 0.000 006 079 675 047 936;
  • 14) 0.000 006 079 675 047 936 × 2 = 0 + 0.000 012 159 350 095 872;
  • 15) 0.000 012 159 350 095 872 × 2 = 0 + 0.000 024 318 700 191 744;
  • 16) 0.000 024 318 700 191 744 × 2 = 0 + 0.000 048 637 400 383 488;
  • 17) 0.000 048 637 400 383 488 × 2 = 0 + 0.000 097 274 800 766 976;
  • 18) 0.000 097 274 800 766 976 × 2 = 0 + 0.000 194 549 601 533 952;
  • 19) 0.000 194 549 601 533 952 × 2 = 0 + 0.000 389 099 203 067 904;
  • 20) 0.000 389 099 203 067 904 × 2 = 0 + 0.000 778 198 406 135 808;
  • 21) 0.000 778 198 406 135 808 × 2 = 0 + 0.001 556 396 812 271 616;
  • 22) 0.001 556 396 812 271 616 × 2 = 0 + 0.003 112 793 624 543 232;
  • 23) 0.003 112 793 624 543 232 × 2 = 0 + 0.006 225 587 249 086 464;
  • 24) 0.006 225 587 249 086 464 × 2 = 0 + 0.012 451 174 498 172 928;
  • 25) 0.012 451 174 498 172 928 × 2 = 0 + 0.024 902 348 996 345 856;
  • 26) 0.024 902 348 996 345 856 × 2 = 0 + 0.049 804 697 992 691 712;
  • 27) 0.049 804 697 992 691 712 × 2 = 0 + 0.099 609 395 985 383 424;
  • 28) 0.099 609 395 985 383 424 × 2 = 0 + 0.199 218 791 970 766 848;
  • 29) 0.199 218 791 970 766 848 × 2 = 0 + 0.398 437 583 941 533 696;
  • 30) 0.398 437 583 941 533 696 × 2 = 0 + 0.796 875 167 883 067 392;
  • 31) 0.796 875 167 883 067 392 × 2 = 1 + 0.593 750 335 766 134 784;
  • 32) 0.593 750 335 766 134 784 × 2 = 1 + 0.187 500 671 532 269 568;
  • 33) 0.187 500 671 532 269 568 × 2 = 0 + 0.375 001 343 064 539 136;
  • 34) 0.375 001 343 064 539 136 × 2 = 0 + 0.750 002 686 129 078 272;
  • 35) 0.750 002 686 129 078 272 × 2 = 1 + 0.500 005 372 258 156 544;
  • 36) 0.500 005 372 258 156 544 × 2 = 1 + 0.000 010 744 516 313 088;
  • 37) 0.000 010 744 516 313 088 × 2 = 0 + 0.000 021 489 032 626 176;
  • 38) 0.000 021 489 032 626 176 × 2 = 0 + 0.000 042 978 065 252 352;
  • 39) 0.000 042 978 065 252 352 × 2 = 0 + 0.000 085 956 130 504 704;
  • 40) 0.000 085 956 130 504 704 × 2 = 0 + 0.000 171 912 261 009 408;
  • 41) 0.000 171 912 261 009 408 × 2 = 0 + 0.000 343 824 522 018 816;
  • 42) 0.000 343 824 522 018 816 × 2 = 0 + 0.000 687 649 044 037 632;
  • 43) 0.000 687 649 044 037 632 × 2 = 0 + 0.001 375 298 088 075 264;
  • 44) 0.001 375 298 088 075 264 × 2 = 0 + 0.002 750 596 176 150 528;
  • 45) 0.002 750 596 176 150 528 × 2 = 0 + 0.005 501 192 352 301 056;
  • 46) 0.005 501 192 352 301 056 × 2 = 0 + 0.011 002 384 704 602 112;
  • 47) 0.011 002 384 704 602 112 × 2 = 0 + 0.022 004 769 409 204 224;
  • 48) 0.022 004 769 409 204 224 × 2 = 0 + 0.044 009 538 818 408 448;
  • 49) 0.044 009 538 818 408 448 × 2 = 0 + 0.088 019 077 636 816 896;
  • 50) 0.088 019 077 636 816 896 × 2 = 0 + 0.176 038 155 273 633 792;
  • 51) 0.176 038 155 273 633 792 × 2 = 0 + 0.352 076 310 547 267 584;
  • 52) 0.352 076 310 547 267 584 × 2 = 0 + 0.704 152 621 094 535 168;
  • 53) 0.704 152 621 094 535 168 × 2 = 1 + 0.408 305 242 189 070 336;
  • 54) 0.408 305 242 189 070 336 × 2 = 0 + 0.816 610 484 378 140 672;

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:


0.000 000 000 742 147 833(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 833(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(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:


0.000 000 000 742 147 833(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) × 20 =

1.1001 1000 0000 0000 0000 010(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 1000 0000 0000 0000 010


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 1100 0000 0000 0000 0010 =

100 1100 0000 0000 0000 0010


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 1100 0000 0000 0000 0010


Decimal number -0.000 000 000 742 147 833 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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