-0.000 000 000 742 147 676 646 741 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 741 8(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
-0.000 000 000 742 147 676 646 741 8(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:

|-0.000 000 000 742 147 676 646 741 8| = 0.000 000 000 742 147 676 646 741 8


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: 0.000 000 000 742 147 676 646 741 8.

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) 0.000 000 000 742 147 676 646 741 8 × 2 = 0 + 0.000 000 001 484 295 353 293 483 6;
  • 2) 0.000 000 001 484 295 353 293 483 6 × 2 = 0 + 0.000 000 002 968 590 706 586 967 2;
  • 3) 0.000 000 002 968 590 706 586 967 2 × 2 = 0 + 0.000 000 005 937 181 413 173 934 4;
  • 4) 0.000 000 005 937 181 413 173 934 4 × 2 = 0 + 0.000 000 011 874 362 826 347 868 8;
  • 5) 0.000 000 011 874 362 826 347 868 8 × 2 = 0 + 0.000 000 023 748 725 652 695 737 6;
  • 6) 0.000 000 023 748 725 652 695 737 6 × 2 = 0 + 0.000 000 047 497 451 305 391 475 2;
  • 7) 0.000 000 047 497 451 305 391 475 2 × 2 = 0 + 0.000 000 094 994 902 610 782 950 4;
  • 8) 0.000 000 094 994 902 610 782 950 4 × 2 = 0 + 0.000 000 189 989 805 221 565 900 8;
  • 9) 0.000 000 189 989 805 221 565 900 8 × 2 = 0 + 0.000 000 379 979 610 443 131 801 6;
  • 10) 0.000 000 379 979 610 443 131 801 6 × 2 = 0 + 0.000 000 759 959 220 886 263 603 2;
  • 11) 0.000 000 759 959 220 886 263 603 2 × 2 = 0 + 0.000 001 519 918 441 772 527 206 4;
  • 12) 0.000 001 519 918 441 772 527 206 4 × 2 = 0 + 0.000 003 039 836 883 545 054 412 8;
  • 13) 0.000 003 039 836 883 545 054 412 8 × 2 = 0 + 0.000 006 079 673 767 090 108 825 6;
  • 14) 0.000 006 079 673 767 090 108 825 6 × 2 = 0 + 0.000 012 159 347 534 180 217 651 2;
  • 15) 0.000 012 159 347 534 180 217 651 2 × 2 = 0 + 0.000 024 318 695 068 360 435 302 4;
  • 16) 0.000 024 318 695 068 360 435 302 4 × 2 = 0 + 0.000 048 637 390 136 720 870 604 8;
  • 17) 0.000 048 637 390 136 720 870 604 8 × 2 = 0 + 0.000 097 274 780 273 441 741 209 6;
  • 18) 0.000 097 274 780 273 441 741 209 6 × 2 = 0 + 0.000 194 549 560 546 883 482 419 2;
  • 19) 0.000 194 549 560 546 883 482 419 2 × 2 = 0 + 0.000 389 099 121 093 766 964 838 4;
  • 20) 0.000 389 099 121 093 766 964 838 4 × 2 = 0 + 0.000 778 198 242 187 533 929 676 8;
  • 21) 0.000 778 198 242 187 533 929 676 8 × 2 = 0 + 0.001 556 396 484 375 067 859 353 6;
  • 22) 0.001 556 396 484 375 067 859 353 6 × 2 = 0 + 0.003 112 792 968 750 135 718 707 2;
  • 23) 0.003 112 792 968 750 135 718 707 2 × 2 = 0 + 0.006 225 585 937 500 271 437 414 4;
  • 24) 0.006 225 585 937 500 271 437 414 4 × 2 = 0 + 0.012 451 171 875 000 542 874 828 8;
  • 25) 0.012 451 171 875 000 542 874 828 8 × 2 = 0 + 0.024 902 343 750 001 085 749 657 6;
  • 26) 0.024 902 343 750 001 085 749 657 6 × 2 = 0 + 0.049 804 687 500 002 171 499 315 2;
  • 27) 0.049 804 687 500 002 171 499 315 2 × 2 = 0 + 0.099 609 375 000 004 342 998 630 4;
  • 28) 0.099 609 375 000 004 342 998 630 4 × 2 = 0 + 0.199 218 750 000 008 685 997 260 8;
  • 29) 0.199 218 750 000 008 685 997 260 8 × 2 = 0 + 0.398 437 500 000 017 371 994 521 6;
  • 30) 0.398 437 500 000 017 371 994 521 6 × 2 = 0 + 0.796 875 000 000 034 743 989 043 2;
  • 31) 0.796 875 000 000 034 743 989 043 2 × 2 = 1 + 0.593 750 000 000 069 487 978 086 4;
  • 32) 0.593 750 000 000 069 487 978 086 4 × 2 = 1 + 0.187 500 000 000 138 975 956 172 8;
  • 33) 0.187 500 000 000 138 975 956 172 8 × 2 = 0 + 0.375 000 000 000 277 951 912 345 6;
  • 34) 0.375 000 000 000 277 951 912 345 6 × 2 = 0 + 0.750 000 000 000 555 903 824 691 2;
  • 35) 0.750 000 000 000 555 903 824 691 2 × 2 = 1 + 0.500 000 000 001 111 807 649 382 4;
  • 36) 0.500 000 000 001 111 807 649 382 4 × 2 = 1 + 0.000 000 000 002 223 615 298 764 8;
  • 37) 0.000 000 000 002 223 615 298 764 8 × 2 = 0 + 0.000 000 000 004 447 230 597 529 6;
  • 38) 0.000 000 000 004 447 230 597 529 6 × 2 = 0 + 0.000 000 000 008 894 461 195 059 2;
  • 39) 0.000 000 000 008 894 461 195 059 2 × 2 = 0 + 0.000 000 000 017 788 922 390 118 4;
  • 40) 0.000 000 000 017 788 922 390 118 4 × 2 = 0 + 0.000 000 000 035 577 844 780 236 8;
  • 41) 0.000 000 000 035 577 844 780 236 8 × 2 = 0 + 0.000 000 000 071 155 689 560 473 6;
  • 42) 0.000 000 000 071 155 689 560 473 6 × 2 = 0 + 0.000 000 000 142 311 379 120 947 2;
  • 43) 0.000 000 000 142 311 379 120 947 2 × 2 = 0 + 0.000 000 000 284 622 758 241 894 4;
  • 44) 0.000 000 000 284 622 758 241 894 4 × 2 = 0 + 0.000 000 000 569 245 516 483 788 8;
  • 45) 0.000 000 000 569 245 516 483 788 8 × 2 = 0 + 0.000 000 001 138 491 032 967 577 6;
  • 46) 0.000 000 001 138 491 032 967 577 6 × 2 = 0 + 0.000 000 002 276 982 065 935 155 2;
  • 47) 0.000 000 002 276 982 065 935 155 2 × 2 = 0 + 0.000 000 004 553 964 131 870 310 4;
  • 48) 0.000 000 004 553 964 131 870 310 4 × 2 = 0 + 0.000 000 009 107 928 263 740 620 8;
  • 49) 0.000 000 009 107 928 263 740 620 8 × 2 = 0 + 0.000 000 018 215 856 527 481 241 6;
  • 50) 0.000 000 018 215 856 527 481 241 6 × 2 = 0 + 0.000 000 036 431 713 054 962 483 2;
  • 51) 0.000 000 036 431 713 054 962 483 2 × 2 = 0 + 0.000 000 072 863 426 109 924 966 4;
  • 52) 0.000 000 072 863 426 109 924 966 4 × 2 = 0 + 0.000 000 145 726 852 219 849 932 8;
  • 53) 0.000 000 145 726 852 219 849 932 8 × 2 = 0 + 0.000 000 291 453 704 439 699 865 6;
  • 54) 0.000 000 291 453 704 439 699 865 6 × 2 = 0 + 0.000 000 582 907 408 879 399 731 2;

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:


0.000 000 000 742 147 676 646 741 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 741 8(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:


0.000 000 000 742 147 676 646 741 8(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 -0.000 000 000 742 147 676 646 741 8 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