-0.000 000 000 742 147 749 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 749(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
-0.000 000 000 742 147 749(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:

|-0.000 000 000 742 147 749| = 0.000 000 000 742 147 749


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: 0.000 000 000 742 147 749.

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) 0.000 000 000 742 147 749 × 2 = 0 + 0.000 000 001 484 295 498;
  • 2) 0.000 000 001 484 295 498 × 2 = 0 + 0.000 000 002 968 590 996;
  • 3) 0.000 000 002 968 590 996 × 2 = 0 + 0.000 000 005 937 181 992;
  • 4) 0.000 000 005 937 181 992 × 2 = 0 + 0.000 000 011 874 363 984;
  • 5) 0.000 000 011 874 363 984 × 2 = 0 + 0.000 000 023 748 727 968;
  • 6) 0.000 000 023 748 727 968 × 2 = 0 + 0.000 000 047 497 455 936;
  • 7) 0.000 000 047 497 455 936 × 2 = 0 + 0.000 000 094 994 911 872;
  • 8) 0.000 000 094 994 911 872 × 2 = 0 + 0.000 000 189 989 823 744;
  • 9) 0.000 000 189 989 823 744 × 2 = 0 + 0.000 000 379 979 647 488;
  • 10) 0.000 000 379 979 647 488 × 2 = 0 + 0.000 000 759 959 294 976;
  • 11) 0.000 000 759 959 294 976 × 2 = 0 + 0.000 001 519 918 589 952;
  • 12) 0.000 001 519 918 589 952 × 2 = 0 + 0.000 003 039 837 179 904;
  • 13) 0.000 003 039 837 179 904 × 2 = 0 + 0.000 006 079 674 359 808;
  • 14) 0.000 006 079 674 359 808 × 2 = 0 + 0.000 012 159 348 719 616;
  • 15) 0.000 012 159 348 719 616 × 2 = 0 + 0.000 024 318 697 439 232;
  • 16) 0.000 024 318 697 439 232 × 2 = 0 + 0.000 048 637 394 878 464;
  • 17) 0.000 048 637 394 878 464 × 2 = 0 + 0.000 097 274 789 756 928;
  • 18) 0.000 097 274 789 756 928 × 2 = 0 + 0.000 194 549 579 513 856;
  • 19) 0.000 194 549 579 513 856 × 2 = 0 + 0.000 389 099 159 027 712;
  • 20) 0.000 389 099 159 027 712 × 2 = 0 + 0.000 778 198 318 055 424;
  • 21) 0.000 778 198 318 055 424 × 2 = 0 + 0.001 556 396 636 110 848;
  • 22) 0.001 556 396 636 110 848 × 2 = 0 + 0.003 112 793 272 221 696;
  • 23) 0.003 112 793 272 221 696 × 2 = 0 + 0.006 225 586 544 443 392;
  • 24) 0.006 225 586 544 443 392 × 2 = 0 + 0.012 451 173 088 886 784;
  • 25) 0.012 451 173 088 886 784 × 2 = 0 + 0.024 902 346 177 773 568;
  • 26) 0.024 902 346 177 773 568 × 2 = 0 + 0.049 804 692 355 547 136;
  • 27) 0.049 804 692 355 547 136 × 2 = 0 + 0.099 609 384 711 094 272;
  • 28) 0.099 609 384 711 094 272 × 2 = 0 + 0.199 218 769 422 188 544;
  • 29) 0.199 218 769 422 188 544 × 2 = 0 + 0.398 437 538 844 377 088;
  • 30) 0.398 437 538 844 377 088 × 2 = 0 + 0.796 875 077 688 754 176;
  • 31) 0.796 875 077 688 754 176 × 2 = 1 + 0.593 750 155 377 508 352;
  • 32) 0.593 750 155 377 508 352 × 2 = 1 + 0.187 500 310 755 016 704;
  • 33) 0.187 500 310 755 016 704 × 2 = 0 + 0.375 000 621 510 033 408;
  • 34) 0.375 000 621 510 033 408 × 2 = 0 + 0.750 001 243 020 066 816;
  • 35) 0.750 001 243 020 066 816 × 2 = 1 + 0.500 002 486 040 133 632;
  • 36) 0.500 002 486 040 133 632 × 2 = 1 + 0.000 004 972 080 267 264;
  • 37) 0.000 004 972 080 267 264 × 2 = 0 + 0.000 009 944 160 534 528;
  • 38) 0.000 009 944 160 534 528 × 2 = 0 + 0.000 019 888 321 069 056;
  • 39) 0.000 019 888 321 069 056 × 2 = 0 + 0.000 039 776 642 138 112;
  • 40) 0.000 039 776 642 138 112 × 2 = 0 + 0.000 079 553 284 276 224;
  • 41) 0.000 079 553 284 276 224 × 2 = 0 + 0.000 159 106 568 552 448;
  • 42) 0.000 159 106 568 552 448 × 2 = 0 + 0.000 318 213 137 104 896;
  • 43) 0.000 318 213 137 104 896 × 2 = 0 + 0.000 636 426 274 209 792;
  • 44) 0.000 636 426 274 209 792 × 2 = 0 + 0.001 272 852 548 419 584;
  • 45) 0.001 272 852 548 419 584 × 2 = 0 + 0.002 545 705 096 839 168;
  • 46) 0.002 545 705 096 839 168 × 2 = 0 + 0.005 091 410 193 678 336;
  • 47) 0.005 091 410 193 678 336 × 2 = 0 + 0.010 182 820 387 356 672;
  • 48) 0.010 182 820 387 356 672 × 2 = 0 + 0.020 365 640 774 713 344;
  • 49) 0.020 365 640 774 713 344 × 2 = 0 + 0.040 731 281 549 426 688;
  • 50) 0.040 731 281 549 426 688 × 2 = 0 + 0.081 462 563 098 853 376;
  • 51) 0.081 462 563 098 853 376 × 2 = 0 + 0.162 925 126 197 706 752;
  • 52) 0.162 925 126 197 706 752 × 2 = 0 + 0.325 850 252 395 413 504;
  • 53) 0.325 850 252 395 413 504 × 2 = 0 + 0.651 700 504 790 827 008;
  • 54) 0.651 700 504 790 827 008 × 2 = 1 + 0.303 401 009 581 654 016;

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:


0.000 000 000 742 147 749(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 749(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 749(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2) × 20 =

1.1001 1000 0000 0000 0000 001(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 001


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 0001 =

100 1100 0000 0000 0000 0001


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 0001


Decimal number -0.000 000 000 742 147 749 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0001

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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