-0.000 000 000 742 147 676 646 742 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 742 1(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
-0.000 000 000 742 147 676 646 742 1(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:

|-0.000 000 000 742 147 676 646 742 1| = 0.000 000 000 742 147 676 646 742 1


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: 0.000 000 000 742 147 676 646 742 1.

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) 0.000 000 000 742 147 676 646 742 1 × 2 = 0 + 0.000 000 001 484 295 353 293 484 2;
  • 2) 0.000 000 001 484 295 353 293 484 2 × 2 = 0 + 0.000 000 002 968 590 706 586 968 4;
  • 3) 0.000 000 002 968 590 706 586 968 4 × 2 = 0 + 0.000 000 005 937 181 413 173 936 8;
  • 4) 0.000 000 005 937 181 413 173 936 8 × 2 = 0 + 0.000 000 011 874 362 826 347 873 6;
  • 5) 0.000 000 011 874 362 826 347 873 6 × 2 = 0 + 0.000 000 023 748 725 652 695 747 2;
  • 6) 0.000 000 023 748 725 652 695 747 2 × 2 = 0 + 0.000 000 047 497 451 305 391 494 4;
  • 7) 0.000 000 047 497 451 305 391 494 4 × 2 = 0 + 0.000 000 094 994 902 610 782 988 8;
  • 8) 0.000 000 094 994 902 610 782 988 8 × 2 = 0 + 0.000 000 189 989 805 221 565 977 6;
  • 9) 0.000 000 189 989 805 221 565 977 6 × 2 = 0 + 0.000 000 379 979 610 443 131 955 2;
  • 10) 0.000 000 379 979 610 443 131 955 2 × 2 = 0 + 0.000 000 759 959 220 886 263 910 4;
  • 11) 0.000 000 759 959 220 886 263 910 4 × 2 = 0 + 0.000 001 519 918 441 772 527 820 8;
  • 12) 0.000 001 519 918 441 772 527 820 8 × 2 = 0 + 0.000 003 039 836 883 545 055 641 6;
  • 13) 0.000 003 039 836 883 545 055 641 6 × 2 = 0 + 0.000 006 079 673 767 090 111 283 2;
  • 14) 0.000 006 079 673 767 090 111 283 2 × 2 = 0 + 0.000 012 159 347 534 180 222 566 4;
  • 15) 0.000 012 159 347 534 180 222 566 4 × 2 = 0 + 0.000 024 318 695 068 360 445 132 8;
  • 16) 0.000 024 318 695 068 360 445 132 8 × 2 = 0 + 0.000 048 637 390 136 720 890 265 6;
  • 17) 0.000 048 637 390 136 720 890 265 6 × 2 = 0 + 0.000 097 274 780 273 441 780 531 2;
  • 18) 0.000 097 274 780 273 441 780 531 2 × 2 = 0 + 0.000 194 549 560 546 883 561 062 4;
  • 19) 0.000 194 549 560 546 883 561 062 4 × 2 = 0 + 0.000 389 099 121 093 767 122 124 8;
  • 20) 0.000 389 099 121 093 767 122 124 8 × 2 = 0 + 0.000 778 198 242 187 534 244 249 6;
  • 21) 0.000 778 198 242 187 534 244 249 6 × 2 = 0 + 0.001 556 396 484 375 068 488 499 2;
  • 22) 0.001 556 396 484 375 068 488 499 2 × 2 = 0 + 0.003 112 792 968 750 136 976 998 4;
  • 23) 0.003 112 792 968 750 136 976 998 4 × 2 = 0 + 0.006 225 585 937 500 273 953 996 8;
  • 24) 0.006 225 585 937 500 273 953 996 8 × 2 = 0 + 0.012 451 171 875 000 547 907 993 6;
  • 25) 0.012 451 171 875 000 547 907 993 6 × 2 = 0 + 0.024 902 343 750 001 095 815 987 2;
  • 26) 0.024 902 343 750 001 095 815 987 2 × 2 = 0 + 0.049 804 687 500 002 191 631 974 4;
  • 27) 0.049 804 687 500 002 191 631 974 4 × 2 = 0 + 0.099 609 375 000 004 383 263 948 8;
  • 28) 0.099 609 375 000 004 383 263 948 8 × 2 = 0 + 0.199 218 750 000 008 766 527 897 6;
  • 29) 0.199 218 750 000 008 766 527 897 6 × 2 = 0 + 0.398 437 500 000 017 533 055 795 2;
  • 30) 0.398 437 500 000 017 533 055 795 2 × 2 = 0 + 0.796 875 000 000 035 066 111 590 4;
  • 31) 0.796 875 000 000 035 066 111 590 4 × 2 = 1 + 0.593 750 000 000 070 132 223 180 8;
  • 32) 0.593 750 000 000 070 132 223 180 8 × 2 = 1 + 0.187 500 000 000 140 264 446 361 6;
  • 33) 0.187 500 000 000 140 264 446 361 6 × 2 = 0 + 0.375 000 000 000 280 528 892 723 2;
  • 34) 0.375 000 000 000 280 528 892 723 2 × 2 = 0 + 0.750 000 000 000 561 057 785 446 4;
  • 35) 0.750 000 000 000 561 057 785 446 4 × 2 = 1 + 0.500 000 000 001 122 115 570 892 8;
  • 36) 0.500 000 000 001 122 115 570 892 8 × 2 = 1 + 0.000 000 000 002 244 231 141 785 6;
  • 37) 0.000 000 000 002 244 231 141 785 6 × 2 = 0 + 0.000 000 000 004 488 462 283 571 2;
  • 38) 0.000 000 000 004 488 462 283 571 2 × 2 = 0 + 0.000 000 000 008 976 924 567 142 4;
  • 39) 0.000 000 000 008 976 924 567 142 4 × 2 = 0 + 0.000 000 000 017 953 849 134 284 8;
  • 40) 0.000 000 000 017 953 849 134 284 8 × 2 = 0 + 0.000 000 000 035 907 698 268 569 6;
  • 41) 0.000 000 000 035 907 698 268 569 6 × 2 = 0 + 0.000 000 000 071 815 396 537 139 2;
  • 42) 0.000 000 000 071 815 396 537 139 2 × 2 = 0 + 0.000 000 000 143 630 793 074 278 4;
  • 43) 0.000 000 000 143 630 793 074 278 4 × 2 = 0 + 0.000 000 000 287 261 586 148 556 8;
  • 44) 0.000 000 000 287 261 586 148 556 8 × 2 = 0 + 0.000 000 000 574 523 172 297 113 6;
  • 45) 0.000 000 000 574 523 172 297 113 6 × 2 = 0 + 0.000 000 001 149 046 344 594 227 2;
  • 46) 0.000 000 001 149 046 344 594 227 2 × 2 = 0 + 0.000 000 002 298 092 689 188 454 4;
  • 47) 0.000 000 002 298 092 689 188 454 4 × 2 = 0 + 0.000 000 004 596 185 378 376 908 8;
  • 48) 0.000 000 004 596 185 378 376 908 8 × 2 = 0 + 0.000 000 009 192 370 756 753 817 6;
  • 49) 0.000 000 009 192 370 756 753 817 6 × 2 = 0 + 0.000 000 018 384 741 513 507 635 2;
  • 50) 0.000 000 018 384 741 513 507 635 2 × 2 = 0 + 0.000 000 036 769 483 027 015 270 4;
  • 51) 0.000 000 036 769 483 027 015 270 4 × 2 = 0 + 0.000 000 073 538 966 054 030 540 8;
  • 52) 0.000 000 073 538 966 054 030 540 8 × 2 = 0 + 0.000 000 147 077 932 108 061 081 6;
  • 53) 0.000 000 147 077 932 108 061 081 6 × 2 = 0 + 0.000 000 294 155 864 216 122 163 2;
  • 54) 0.000 000 294 155 864 216 122 163 2 × 2 = 0 + 0.000 000 588 311 728 432 244 326 4;

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:


0.000 000 000 742 147 676 646 742 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 742 1(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:


0.000 000 000 742 147 676 646 742 1(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 -0.000 000 000 742 147 676 646 742 1 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