-0.000 000 000 742 147 678 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 678(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
-0.000 000 000 742 147 678(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:

|-0.000 000 000 742 147 678| = 0.000 000 000 742 147 678


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: 0.000 000 000 742 147 678.

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) 0.000 000 000 742 147 678 × 2 = 0 + 0.000 000 001 484 295 356;
  • 2) 0.000 000 001 484 295 356 × 2 = 0 + 0.000 000 002 968 590 712;
  • 3) 0.000 000 002 968 590 712 × 2 = 0 + 0.000 000 005 937 181 424;
  • 4) 0.000 000 005 937 181 424 × 2 = 0 + 0.000 000 011 874 362 848;
  • 5) 0.000 000 011 874 362 848 × 2 = 0 + 0.000 000 023 748 725 696;
  • 6) 0.000 000 023 748 725 696 × 2 = 0 + 0.000 000 047 497 451 392;
  • 7) 0.000 000 047 497 451 392 × 2 = 0 + 0.000 000 094 994 902 784;
  • 8) 0.000 000 094 994 902 784 × 2 = 0 + 0.000 000 189 989 805 568;
  • 9) 0.000 000 189 989 805 568 × 2 = 0 + 0.000 000 379 979 611 136;
  • 10) 0.000 000 379 979 611 136 × 2 = 0 + 0.000 000 759 959 222 272;
  • 11) 0.000 000 759 959 222 272 × 2 = 0 + 0.000 001 519 918 444 544;
  • 12) 0.000 001 519 918 444 544 × 2 = 0 + 0.000 003 039 836 889 088;
  • 13) 0.000 003 039 836 889 088 × 2 = 0 + 0.000 006 079 673 778 176;
  • 14) 0.000 006 079 673 778 176 × 2 = 0 + 0.000 012 159 347 556 352;
  • 15) 0.000 012 159 347 556 352 × 2 = 0 + 0.000 024 318 695 112 704;
  • 16) 0.000 024 318 695 112 704 × 2 = 0 + 0.000 048 637 390 225 408;
  • 17) 0.000 048 637 390 225 408 × 2 = 0 + 0.000 097 274 780 450 816;
  • 18) 0.000 097 274 780 450 816 × 2 = 0 + 0.000 194 549 560 901 632;
  • 19) 0.000 194 549 560 901 632 × 2 = 0 + 0.000 389 099 121 803 264;
  • 20) 0.000 389 099 121 803 264 × 2 = 0 + 0.000 778 198 243 606 528;
  • 21) 0.000 778 198 243 606 528 × 2 = 0 + 0.001 556 396 487 213 056;
  • 22) 0.001 556 396 487 213 056 × 2 = 0 + 0.003 112 792 974 426 112;
  • 23) 0.003 112 792 974 426 112 × 2 = 0 + 0.006 225 585 948 852 224;
  • 24) 0.006 225 585 948 852 224 × 2 = 0 + 0.012 451 171 897 704 448;
  • 25) 0.012 451 171 897 704 448 × 2 = 0 + 0.024 902 343 795 408 896;
  • 26) 0.024 902 343 795 408 896 × 2 = 0 + 0.049 804 687 590 817 792;
  • 27) 0.049 804 687 590 817 792 × 2 = 0 + 0.099 609 375 181 635 584;
  • 28) 0.099 609 375 181 635 584 × 2 = 0 + 0.199 218 750 363 271 168;
  • 29) 0.199 218 750 363 271 168 × 2 = 0 + 0.398 437 500 726 542 336;
  • 30) 0.398 437 500 726 542 336 × 2 = 0 + 0.796 875 001 453 084 672;
  • 31) 0.796 875 001 453 084 672 × 2 = 1 + 0.593 750 002 906 169 344;
  • 32) 0.593 750 002 906 169 344 × 2 = 1 + 0.187 500 005 812 338 688;
  • 33) 0.187 500 005 812 338 688 × 2 = 0 + 0.375 000 011 624 677 376;
  • 34) 0.375 000 011 624 677 376 × 2 = 0 + 0.750 000 023 249 354 752;
  • 35) 0.750 000 023 249 354 752 × 2 = 1 + 0.500 000 046 498 709 504;
  • 36) 0.500 000 046 498 709 504 × 2 = 1 + 0.000 000 092 997 419 008;
  • 37) 0.000 000 092 997 419 008 × 2 = 0 + 0.000 000 185 994 838 016;
  • 38) 0.000 000 185 994 838 016 × 2 = 0 + 0.000 000 371 989 676 032;
  • 39) 0.000 000 371 989 676 032 × 2 = 0 + 0.000 000 743 979 352 064;
  • 40) 0.000 000 743 979 352 064 × 2 = 0 + 0.000 001 487 958 704 128;
  • 41) 0.000 001 487 958 704 128 × 2 = 0 + 0.000 002 975 917 408 256;
  • 42) 0.000 002 975 917 408 256 × 2 = 0 + 0.000 005 951 834 816 512;
  • 43) 0.000 005 951 834 816 512 × 2 = 0 + 0.000 011 903 669 633 024;
  • 44) 0.000 011 903 669 633 024 × 2 = 0 + 0.000 023 807 339 266 048;
  • 45) 0.000 023 807 339 266 048 × 2 = 0 + 0.000 047 614 678 532 096;
  • 46) 0.000 047 614 678 532 096 × 2 = 0 + 0.000 095 229 357 064 192;
  • 47) 0.000 095 229 357 064 192 × 2 = 0 + 0.000 190 458 714 128 384;
  • 48) 0.000 190 458 714 128 384 × 2 = 0 + 0.000 380 917 428 256 768;
  • 49) 0.000 380 917 428 256 768 × 2 = 0 + 0.000 761 834 856 513 536;
  • 50) 0.000 761 834 856 513 536 × 2 = 0 + 0.001 523 669 713 027 072;
  • 51) 0.001 523 669 713 027 072 × 2 = 0 + 0.003 047 339 426 054 144;
  • 52) 0.003 047 339 426 054 144 × 2 = 0 + 0.006 094 678 852 108 288;
  • 53) 0.006 094 678 852 108 288 × 2 = 0 + 0.012 189 357 704 216 576;
  • 54) 0.012 189 357 704 216 576 × 2 = 0 + 0.024 378 715 408 433 152;

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:


0.000 000 000 742 147 678(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 678(10) =

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


0.000 000 000 742 147 678(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 000


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 0000 =

100 1100 0000 0000 0000 0000


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 0000


Decimal number -0.000 000 000 742 147 678 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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