-0.000 000 000 742 147 676 646 670 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 670 6(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
-0.000 000 000 742 147 676 646 670 6(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:

|-0.000 000 000 742 147 676 646 670 6| = 0.000 000 000 742 147 676 646 670 6


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: 0.000 000 000 742 147 676 646 670 6.

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) 0.000 000 000 742 147 676 646 670 6 × 2 = 0 + 0.000 000 001 484 295 353 293 341 2;
  • 2) 0.000 000 001 484 295 353 293 341 2 × 2 = 0 + 0.000 000 002 968 590 706 586 682 4;
  • 3) 0.000 000 002 968 590 706 586 682 4 × 2 = 0 + 0.000 000 005 937 181 413 173 364 8;
  • 4) 0.000 000 005 937 181 413 173 364 8 × 2 = 0 + 0.000 000 011 874 362 826 346 729 6;
  • 5) 0.000 000 011 874 362 826 346 729 6 × 2 = 0 + 0.000 000 023 748 725 652 693 459 2;
  • 6) 0.000 000 023 748 725 652 693 459 2 × 2 = 0 + 0.000 000 047 497 451 305 386 918 4;
  • 7) 0.000 000 047 497 451 305 386 918 4 × 2 = 0 + 0.000 000 094 994 902 610 773 836 8;
  • 8) 0.000 000 094 994 902 610 773 836 8 × 2 = 0 + 0.000 000 189 989 805 221 547 673 6;
  • 9) 0.000 000 189 989 805 221 547 673 6 × 2 = 0 + 0.000 000 379 979 610 443 095 347 2;
  • 10) 0.000 000 379 979 610 443 095 347 2 × 2 = 0 + 0.000 000 759 959 220 886 190 694 4;
  • 11) 0.000 000 759 959 220 886 190 694 4 × 2 = 0 + 0.000 001 519 918 441 772 381 388 8;
  • 12) 0.000 001 519 918 441 772 381 388 8 × 2 = 0 + 0.000 003 039 836 883 544 762 777 6;
  • 13) 0.000 003 039 836 883 544 762 777 6 × 2 = 0 + 0.000 006 079 673 767 089 525 555 2;
  • 14) 0.000 006 079 673 767 089 525 555 2 × 2 = 0 + 0.000 012 159 347 534 179 051 110 4;
  • 15) 0.000 012 159 347 534 179 051 110 4 × 2 = 0 + 0.000 024 318 695 068 358 102 220 8;
  • 16) 0.000 024 318 695 068 358 102 220 8 × 2 = 0 + 0.000 048 637 390 136 716 204 441 6;
  • 17) 0.000 048 637 390 136 716 204 441 6 × 2 = 0 + 0.000 097 274 780 273 432 408 883 2;
  • 18) 0.000 097 274 780 273 432 408 883 2 × 2 = 0 + 0.000 194 549 560 546 864 817 766 4;
  • 19) 0.000 194 549 560 546 864 817 766 4 × 2 = 0 + 0.000 389 099 121 093 729 635 532 8;
  • 20) 0.000 389 099 121 093 729 635 532 8 × 2 = 0 + 0.000 778 198 242 187 459 271 065 6;
  • 21) 0.000 778 198 242 187 459 271 065 6 × 2 = 0 + 0.001 556 396 484 374 918 542 131 2;
  • 22) 0.001 556 396 484 374 918 542 131 2 × 2 = 0 + 0.003 112 792 968 749 837 084 262 4;
  • 23) 0.003 112 792 968 749 837 084 262 4 × 2 = 0 + 0.006 225 585 937 499 674 168 524 8;
  • 24) 0.006 225 585 937 499 674 168 524 8 × 2 = 0 + 0.012 451 171 874 999 348 337 049 6;
  • 25) 0.012 451 171 874 999 348 337 049 6 × 2 = 0 + 0.024 902 343 749 998 696 674 099 2;
  • 26) 0.024 902 343 749 998 696 674 099 2 × 2 = 0 + 0.049 804 687 499 997 393 348 198 4;
  • 27) 0.049 804 687 499 997 393 348 198 4 × 2 = 0 + 0.099 609 374 999 994 786 696 396 8;
  • 28) 0.099 609 374 999 994 786 696 396 8 × 2 = 0 + 0.199 218 749 999 989 573 392 793 6;
  • 29) 0.199 218 749 999 989 573 392 793 6 × 2 = 0 + 0.398 437 499 999 979 146 785 587 2;
  • 30) 0.398 437 499 999 979 146 785 587 2 × 2 = 0 + 0.796 874 999 999 958 293 571 174 4;
  • 31) 0.796 874 999 999 958 293 571 174 4 × 2 = 1 + 0.593 749 999 999 916 587 142 348 8;
  • 32) 0.593 749 999 999 916 587 142 348 8 × 2 = 1 + 0.187 499 999 999 833 174 284 697 6;
  • 33) 0.187 499 999 999 833 174 284 697 6 × 2 = 0 + 0.374 999 999 999 666 348 569 395 2;
  • 34) 0.374 999 999 999 666 348 569 395 2 × 2 = 0 + 0.749 999 999 999 332 697 138 790 4;
  • 35) 0.749 999 999 999 332 697 138 790 4 × 2 = 1 + 0.499 999 999 998 665 394 277 580 8;
  • 36) 0.499 999 999 998 665 394 277 580 8 × 2 = 0 + 0.999 999 999 997 330 788 555 161 6;
  • 37) 0.999 999 999 997 330 788 555 161 6 × 2 = 1 + 0.999 999 999 994 661 577 110 323 2;
  • 38) 0.999 999 999 994 661 577 110 323 2 × 2 = 1 + 0.999 999 999 989 323 154 220 646 4;
  • 39) 0.999 999 999 989 323 154 220 646 4 × 2 = 1 + 0.999 999 999 978 646 308 441 292 8;
  • 40) 0.999 999 999 978 646 308 441 292 8 × 2 = 1 + 0.999 999 999 957 292 616 882 585 6;
  • 41) 0.999 999 999 957 292 616 882 585 6 × 2 = 1 + 0.999 999 999 914 585 233 765 171 2;
  • 42) 0.999 999 999 914 585 233 765 171 2 × 2 = 1 + 0.999 999 999 829 170 467 530 342 4;
  • 43) 0.999 999 999 829 170 467 530 342 4 × 2 = 1 + 0.999 999 999 658 340 935 060 684 8;
  • 44) 0.999 999 999 658 340 935 060 684 8 × 2 = 1 + 0.999 999 999 316 681 870 121 369 6;
  • 45) 0.999 999 999 316 681 870 121 369 6 × 2 = 1 + 0.999 999 998 633 363 740 242 739 2;
  • 46) 0.999 999 998 633 363 740 242 739 2 × 2 = 1 + 0.999 999 997 266 727 480 485 478 4;
  • 47) 0.999 999 997 266 727 480 485 478 4 × 2 = 1 + 0.999 999 994 533 454 960 970 956 8;
  • 48) 0.999 999 994 533 454 960 970 956 8 × 2 = 1 + 0.999 999 989 066 909 921 941 913 6;
  • 49) 0.999 999 989 066 909 921 941 913 6 × 2 = 1 + 0.999 999 978 133 819 843 883 827 2;
  • 50) 0.999 999 978 133 819 843 883 827 2 × 2 = 1 + 0.999 999 956 267 639 687 767 654 4;
  • 51) 0.999 999 956 267 639 687 767 654 4 × 2 = 1 + 0.999 999 912 535 279 375 535 308 8;
  • 52) 0.999 999 912 535 279 375 535 308 8 × 2 = 1 + 0.999 999 825 070 558 751 070 617 6;
  • 53) 0.999 999 825 070 558 751 070 617 6 × 2 = 1 + 0.999 999 650 141 117 502 141 235 2;
  • 54) 0.999 999 650 141 117 502 141 235 2 × 2 = 1 + 0.999 999 300 282 235 004 282 470 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:


0.000 000 000 742 147 676 646 670 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 670 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 676 646 670 6(10) =

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

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

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 676 646 670 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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