-0.000 000 000 742 147 747 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 747(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
-0.000 000 000 742 147 747(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:

|-0.000 000 000 742 147 747| = 0.000 000 000 742 147 747


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: 0.000 000 000 742 147 747.

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) 0.000 000 000 742 147 747 × 2 = 0 + 0.000 000 001 484 295 494;
  • 2) 0.000 000 001 484 295 494 × 2 = 0 + 0.000 000 002 968 590 988;
  • 3) 0.000 000 002 968 590 988 × 2 = 0 + 0.000 000 005 937 181 976;
  • 4) 0.000 000 005 937 181 976 × 2 = 0 + 0.000 000 011 874 363 952;
  • 5) 0.000 000 011 874 363 952 × 2 = 0 + 0.000 000 023 748 727 904;
  • 6) 0.000 000 023 748 727 904 × 2 = 0 + 0.000 000 047 497 455 808;
  • 7) 0.000 000 047 497 455 808 × 2 = 0 + 0.000 000 094 994 911 616;
  • 8) 0.000 000 094 994 911 616 × 2 = 0 + 0.000 000 189 989 823 232;
  • 9) 0.000 000 189 989 823 232 × 2 = 0 + 0.000 000 379 979 646 464;
  • 10) 0.000 000 379 979 646 464 × 2 = 0 + 0.000 000 759 959 292 928;
  • 11) 0.000 000 759 959 292 928 × 2 = 0 + 0.000 001 519 918 585 856;
  • 12) 0.000 001 519 918 585 856 × 2 = 0 + 0.000 003 039 837 171 712;
  • 13) 0.000 003 039 837 171 712 × 2 = 0 + 0.000 006 079 674 343 424;
  • 14) 0.000 006 079 674 343 424 × 2 = 0 + 0.000 012 159 348 686 848;
  • 15) 0.000 012 159 348 686 848 × 2 = 0 + 0.000 024 318 697 373 696;
  • 16) 0.000 024 318 697 373 696 × 2 = 0 + 0.000 048 637 394 747 392;
  • 17) 0.000 048 637 394 747 392 × 2 = 0 + 0.000 097 274 789 494 784;
  • 18) 0.000 097 274 789 494 784 × 2 = 0 + 0.000 194 549 578 989 568;
  • 19) 0.000 194 549 578 989 568 × 2 = 0 + 0.000 389 099 157 979 136;
  • 20) 0.000 389 099 157 979 136 × 2 = 0 + 0.000 778 198 315 958 272;
  • 21) 0.000 778 198 315 958 272 × 2 = 0 + 0.001 556 396 631 916 544;
  • 22) 0.001 556 396 631 916 544 × 2 = 0 + 0.003 112 793 263 833 088;
  • 23) 0.003 112 793 263 833 088 × 2 = 0 + 0.006 225 586 527 666 176;
  • 24) 0.006 225 586 527 666 176 × 2 = 0 + 0.012 451 173 055 332 352;
  • 25) 0.012 451 173 055 332 352 × 2 = 0 + 0.024 902 346 110 664 704;
  • 26) 0.024 902 346 110 664 704 × 2 = 0 + 0.049 804 692 221 329 408;
  • 27) 0.049 804 692 221 329 408 × 2 = 0 + 0.099 609 384 442 658 816;
  • 28) 0.099 609 384 442 658 816 × 2 = 0 + 0.199 218 768 885 317 632;
  • 29) 0.199 218 768 885 317 632 × 2 = 0 + 0.398 437 537 770 635 264;
  • 30) 0.398 437 537 770 635 264 × 2 = 0 + 0.796 875 075 541 270 528;
  • 31) 0.796 875 075 541 270 528 × 2 = 1 + 0.593 750 151 082 541 056;
  • 32) 0.593 750 151 082 541 056 × 2 = 1 + 0.187 500 302 165 082 112;
  • 33) 0.187 500 302 165 082 112 × 2 = 0 + 0.375 000 604 330 164 224;
  • 34) 0.375 000 604 330 164 224 × 2 = 0 + 0.750 001 208 660 328 448;
  • 35) 0.750 001 208 660 328 448 × 2 = 1 + 0.500 002 417 320 656 896;
  • 36) 0.500 002 417 320 656 896 × 2 = 1 + 0.000 004 834 641 313 792;
  • 37) 0.000 004 834 641 313 792 × 2 = 0 + 0.000 009 669 282 627 584;
  • 38) 0.000 009 669 282 627 584 × 2 = 0 + 0.000 019 338 565 255 168;
  • 39) 0.000 019 338 565 255 168 × 2 = 0 + 0.000 038 677 130 510 336;
  • 40) 0.000 038 677 130 510 336 × 2 = 0 + 0.000 077 354 261 020 672;
  • 41) 0.000 077 354 261 020 672 × 2 = 0 + 0.000 154 708 522 041 344;
  • 42) 0.000 154 708 522 041 344 × 2 = 0 + 0.000 309 417 044 082 688;
  • 43) 0.000 309 417 044 082 688 × 2 = 0 + 0.000 618 834 088 165 376;
  • 44) 0.000 618 834 088 165 376 × 2 = 0 + 0.001 237 668 176 330 752;
  • 45) 0.001 237 668 176 330 752 × 2 = 0 + 0.002 475 336 352 661 504;
  • 46) 0.002 475 336 352 661 504 × 2 = 0 + 0.004 950 672 705 323 008;
  • 47) 0.004 950 672 705 323 008 × 2 = 0 + 0.009 901 345 410 646 016;
  • 48) 0.009 901 345 410 646 016 × 2 = 0 + 0.019 802 690 821 292 032;
  • 49) 0.019 802 690 821 292 032 × 2 = 0 + 0.039 605 381 642 584 064;
  • 50) 0.039 605 381 642 584 064 × 2 = 0 + 0.079 210 763 285 168 128;
  • 51) 0.079 210 763 285 168 128 × 2 = 0 + 0.158 421 526 570 336 256;
  • 52) 0.158 421 526 570 336 256 × 2 = 0 + 0.316 843 053 140 672 512;
  • 53) 0.316 843 053 140 672 512 × 2 = 0 + 0.633 686 106 281 345 024;
  • 54) 0.633 686 106 281 345 024 × 2 = 1 + 0.267 372 212 562 690 048;

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:


0.000 000 000 742 147 747(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 747(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:


0.000 000 000 742 147 747(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 -0.000 000 000 742 147 747 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