-0.000 000 000 742 147 433 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 433(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
-0.000 000 000 742 147 433(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:

|-0.000 000 000 742 147 433| = 0.000 000 000 742 147 433


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: 0.000 000 000 742 147 433.

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) 0.000 000 000 742 147 433 × 2 = 0 + 0.000 000 001 484 294 866;
  • 2) 0.000 000 001 484 294 866 × 2 = 0 + 0.000 000 002 968 589 732;
  • 3) 0.000 000 002 968 589 732 × 2 = 0 + 0.000 000 005 937 179 464;
  • 4) 0.000 000 005 937 179 464 × 2 = 0 + 0.000 000 011 874 358 928;
  • 5) 0.000 000 011 874 358 928 × 2 = 0 + 0.000 000 023 748 717 856;
  • 6) 0.000 000 023 748 717 856 × 2 = 0 + 0.000 000 047 497 435 712;
  • 7) 0.000 000 047 497 435 712 × 2 = 0 + 0.000 000 094 994 871 424;
  • 8) 0.000 000 094 994 871 424 × 2 = 0 + 0.000 000 189 989 742 848;
  • 9) 0.000 000 189 989 742 848 × 2 = 0 + 0.000 000 379 979 485 696;
  • 10) 0.000 000 379 979 485 696 × 2 = 0 + 0.000 000 759 958 971 392;
  • 11) 0.000 000 759 958 971 392 × 2 = 0 + 0.000 001 519 917 942 784;
  • 12) 0.000 001 519 917 942 784 × 2 = 0 + 0.000 003 039 835 885 568;
  • 13) 0.000 003 039 835 885 568 × 2 = 0 + 0.000 006 079 671 771 136;
  • 14) 0.000 006 079 671 771 136 × 2 = 0 + 0.000 012 159 343 542 272;
  • 15) 0.000 012 159 343 542 272 × 2 = 0 + 0.000 024 318 687 084 544;
  • 16) 0.000 024 318 687 084 544 × 2 = 0 + 0.000 048 637 374 169 088;
  • 17) 0.000 048 637 374 169 088 × 2 = 0 + 0.000 097 274 748 338 176;
  • 18) 0.000 097 274 748 338 176 × 2 = 0 + 0.000 194 549 496 676 352;
  • 19) 0.000 194 549 496 676 352 × 2 = 0 + 0.000 389 098 993 352 704;
  • 20) 0.000 389 098 993 352 704 × 2 = 0 + 0.000 778 197 986 705 408;
  • 21) 0.000 778 197 986 705 408 × 2 = 0 + 0.001 556 395 973 410 816;
  • 22) 0.001 556 395 973 410 816 × 2 = 0 + 0.003 112 791 946 821 632;
  • 23) 0.003 112 791 946 821 632 × 2 = 0 + 0.006 225 583 893 643 264;
  • 24) 0.006 225 583 893 643 264 × 2 = 0 + 0.012 451 167 787 286 528;
  • 25) 0.012 451 167 787 286 528 × 2 = 0 + 0.024 902 335 574 573 056;
  • 26) 0.024 902 335 574 573 056 × 2 = 0 + 0.049 804 671 149 146 112;
  • 27) 0.049 804 671 149 146 112 × 2 = 0 + 0.099 609 342 298 292 224;
  • 28) 0.099 609 342 298 292 224 × 2 = 0 + 0.199 218 684 596 584 448;
  • 29) 0.199 218 684 596 584 448 × 2 = 0 + 0.398 437 369 193 168 896;
  • 30) 0.398 437 369 193 168 896 × 2 = 0 + 0.796 874 738 386 337 792;
  • 31) 0.796 874 738 386 337 792 × 2 = 1 + 0.593 749 476 772 675 584;
  • 32) 0.593 749 476 772 675 584 × 2 = 1 + 0.187 498 953 545 351 168;
  • 33) 0.187 498 953 545 351 168 × 2 = 0 + 0.374 997 907 090 702 336;
  • 34) 0.374 997 907 090 702 336 × 2 = 0 + 0.749 995 814 181 404 672;
  • 35) 0.749 995 814 181 404 672 × 2 = 1 + 0.499 991 628 362 809 344;
  • 36) 0.499 991 628 362 809 344 × 2 = 0 + 0.999 983 256 725 618 688;
  • 37) 0.999 983 256 725 618 688 × 2 = 1 + 0.999 966 513 451 237 376;
  • 38) 0.999 966 513 451 237 376 × 2 = 1 + 0.999 933 026 902 474 752;
  • 39) 0.999 933 026 902 474 752 × 2 = 1 + 0.999 866 053 804 949 504;
  • 40) 0.999 866 053 804 949 504 × 2 = 1 + 0.999 732 107 609 899 008;
  • 41) 0.999 732 107 609 899 008 × 2 = 1 + 0.999 464 215 219 798 016;
  • 42) 0.999 464 215 219 798 016 × 2 = 1 + 0.998 928 430 439 596 032;
  • 43) 0.998 928 430 439 596 032 × 2 = 1 + 0.997 856 860 879 192 064;
  • 44) 0.997 856 860 879 192 064 × 2 = 1 + 0.995 713 721 758 384 128;
  • 45) 0.995 713 721 758 384 128 × 2 = 1 + 0.991 427 443 516 768 256;
  • 46) 0.991 427 443 516 768 256 × 2 = 1 + 0.982 854 887 033 536 512;
  • 47) 0.982 854 887 033 536 512 × 2 = 1 + 0.965 709 774 067 073 024;
  • 48) 0.965 709 774 067 073 024 × 2 = 1 + 0.931 419 548 134 146 048;
  • 49) 0.931 419 548 134 146 048 × 2 = 1 + 0.862 839 096 268 292 096;
  • 50) 0.862 839 096 268 292 096 × 2 = 1 + 0.725 678 192 536 584 192;
  • 51) 0.725 678 192 536 584 192 × 2 = 1 + 0.451 356 385 073 168 384;
  • 52) 0.451 356 385 073 168 384 × 2 = 0 + 0.902 712 770 146 336 768;
  • 53) 0.902 712 770 146 336 768 × 2 = 1 + 0.805 425 540 292 673 536;
  • 54) 0.805 425 540 292 673 536 × 2 = 1 + 0.610 851 080 585 347 072;

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:


0.000 000 000 742 147 433(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 433(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(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:


0.000 000 000 742 147 433(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 1111 1111 011


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 1111 1111 1011 =

100 1011 1111 1111 1111 1011


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 1111 1111 1011


Decimal number -0.000 000 000 742 147 433 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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