-0.000 000 000 742 147 676 677 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 677(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
-0.000 000 000 742 147 676 677(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:

|-0.000 000 000 742 147 676 677| = 0.000 000 000 742 147 676 677


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: 0.000 000 000 742 147 676 677.

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) 0.000 000 000 742 147 676 677 × 2 = 0 + 0.000 000 001 484 295 353 354;
  • 2) 0.000 000 001 484 295 353 354 × 2 = 0 + 0.000 000 002 968 590 706 708;
  • 3) 0.000 000 002 968 590 706 708 × 2 = 0 + 0.000 000 005 937 181 413 416;
  • 4) 0.000 000 005 937 181 413 416 × 2 = 0 + 0.000 000 011 874 362 826 832;
  • 5) 0.000 000 011 874 362 826 832 × 2 = 0 + 0.000 000 023 748 725 653 664;
  • 6) 0.000 000 023 748 725 653 664 × 2 = 0 + 0.000 000 047 497 451 307 328;
  • 7) 0.000 000 047 497 451 307 328 × 2 = 0 + 0.000 000 094 994 902 614 656;
  • 8) 0.000 000 094 994 902 614 656 × 2 = 0 + 0.000 000 189 989 805 229 312;
  • 9) 0.000 000 189 989 805 229 312 × 2 = 0 + 0.000 000 379 979 610 458 624;
  • 10) 0.000 000 379 979 610 458 624 × 2 = 0 + 0.000 000 759 959 220 917 248;
  • 11) 0.000 000 759 959 220 917 248 × 2 = 0 + 0.000 001 519 918 441 834 496;
  • 12) 0.000 001 519 918 441 834 496 × 2 = 0 + 0.000 003 039 836 883 668 992;
  • 13) 0.000 003 039 836 883 668 992 × 2 = 0 + 0.000 006 079 673 767 337 984;
  • 14) 0.000 006 079 673 767 337 984 × 2 = 0 + 0.000 012 159 347 534 675 968;
  • 15) 0.000 012 159 347 534 675 968 × 2 = 0 + 0.000 024 318 695 069 351 936;
  • 16) 0.000 024 318 695 069 351 936 × 2 = 0 + 0.000 048 637 390 138 703 872;
  • 17) 0.000 048 637 390 138 703 872 × 2 = 0 + 0.000 097 274 780 277 407 744;
  • 18) 0.000 097 274 780 277 407 744 × 2 = 0 + 0.000 194 549 560 554 815 488;
  • 19) 0.000 194 549 560 554 815 488 × 2 = 0 + 0.000 389 099 121 109 630 976;
  • 20) 0.000 389 099 121 109 630 976 × 2 = 0 + 0.000 778 198 242 219 261 952;
  • 21) 0.000 778 198 242 219 261 952 × 2 = 0 + 0.001 556 396 484 438 523 904;
  • 22) 0.001 556 396 484 438 523 904 × 2 = 0 + 0.003 112 792 968 877 047 808;
  • 23) 0.003 112 792 968 877 047 808 × 2 = 0 + 0.006 225 585 937 754 095 616;
  • 24) 0.006 225 585 937 754 095 616 × 2 = 0 + 0.012 451 171 875 508 191 232;
  • 25) 0.012 451 171 875 508 191 232 × 2 = 0 + 0.024 902 343 751 016 382 464;
  • 26) 0.024 902 343 751 016 382 464 × 2 = 0 + 0.049 804 687 502 032 764 928;
  • 27) 0.049 804 687 502 032 764 928 × 2 = 0 + 0.099 609 375 004 065 529 856;
  • 28) 0.099 609 375 004 065 529 856 × 2 = 0 + 0.199 218 750 008 131 059 712;
  • 29) 0.199 218 750 008 131 059 712 × 2 = 0 + 0.398 437 500 016 262 119 424;
  • 30) 0.398 437 500 016 262 119 424 × 2 = 0 + 0.796 875 000 032 524 238 848;
  • 31) 0.796 875 000 032 524 238 848 × 2 = 1 + 0.593 750 000 065 048 477 696;
  • 32) 0.593 750 000 065 048 477 696 × 2 = 1 + 0.187 500 000 130 096 955 392;
  • 33) 0.187 500 000 130 096 955 392 × 2 = 0 + 0.375 000 000 260 193 910 784;
  • 34) 0.375 000 000 260 193 910 784 × 2 = 0 + 0.750 000 000 520 387 821 568;
  • 35) 0.750 000 000 520 387 821 568 × 2 = 1 + 0.500 000 001 040 775 643 136;
  • 36) 0.500 000 001 040 775 643 136 × 2 = 1 + 0.000 000 002 081 551 286 272;
  • 37) 0.000 000 002 081 551 286 272 × 2 = 0 + 0.000 000 004 163 102 572 544;
  • 38) 0.000 000 004 163 102 572 544 × 2 = 0 + 0.000 000 008 326 205 145 088;
  • 39) 0.000 000 008 326 205 145 088 × 2 = 0 + 0.000 000 016 652 410 290 176;
  • 40) 0.000 000 016 652 410 290 176 × 2 = 0 + 0.000 000 033 304 820 580 352;
  • 41) 0.000 000 033 304 820 580 352 × 2 = 0 + 0.000 000 066 609 641 160 704;
  • 42) 0.000 000 066 609 641 160 704 × 2 = 0 + 0.000 000 133 219 282 321 408;
  • 43) 0.000 000 133 219 282 321 408 × 2 = 0 + 0.000 000 266 438 564 642 816;
  • 44) 0.000 000 266 438 564 642 816 × 2 = 0 + 0.000 000 532 877 129 285 632;
  • 45) 0.000 000 532 877 129 285 632 × 2 = 0 + 0.000 001 065 754 258 571 264;
  • 46) 0.000 001 065 754 258 571 264 × 2 = 0 + 0.000 002 131 508 517 142 528;
  • 47) 0.000 002 131 508 517 142 528 × 2 = 0 + 0.000 004 263 017 034 285 056;
  • 48) 0.000 004 263 017 034 285 056 × 2 = 0 + 0.000 008 526 034 068 570 112;
  • 49) 0.000 008 526 034 068 570 112 × 2 = 0 + 0.000 017 052 068 137 140 224;
  • 50) 0.000 017 052 068 137 140 224 × 2 = 0 + 0.000 034 104 136 274 280 448;
  • 51) 0.000 034 104 136 274 280 448 × 2 = 0 + 0.000 068 208 272 548 560 896;
  • 52) 0.000 068 208 272 548 560 896 × 2 = 0 + 0.000 136 416 545 097 121 792;
  • 53) 0.000 136 416 545 097 121 792 × 2 = 0 + 0.000 272 833 090 194 243 584;
  • 54) 0.000 272 833 090 194 243 584 × 2 = 0 + 0.000 545 666 180 388 487 168;

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:


0.000 000 000 742 147 676 677(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 677(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:


0.000 000 000 742 147 676 677(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 -0.000 000 000 742 147 676 677 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