-0.000 000 000 742 147 676 646 699 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 699 3(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
-0.000 000 000 742 147 676 646 699 3(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:

|-0.000 000 000 742 147 676 646 699 3| = 0.000 000 000 742 147 676 646 699 3


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: 0.000 000 000 742 147 676 646 699 3.

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) 0.000 000 000 742 147 676 646 699 3 × 2 = 0 + 0.000 000 001 484 295 353 293 398 6;
  • 2) 0.000 000 001 484 295 353 293 398 6 × 2 = 0 + 0.000 000 002 968 590 706 586 797 2;
  • 3) 0.000 000 002 968 590 706 586 797 2 × 2 = 0 + 0.000 000 005 937 181 413 173 594 4;
  • 4) 0.000 000 005 937 181 413 173 594 4 × 2 = 0 + 0.000 000 011 874 362 826 347 188 8;
  • 5) 0.000 000 011 874 362 826 347 188 8 × 2 = 0 + 0.000 000 023 748 725 652 694 377 6;
  • 6) 0.000 000 023 748 725 652 694 377 6 × 2 = 0 + 0.000 000 047 497 451 305 388 755 2;
  • 7) 0.000 000 047 497 451 305 388 755 2 × 2 = 0 + 0.000 000 094 994 902 610 777 510 4;
  • 8) 0.000 000 094 994 902 610 777 510 4 × 2 = 0 + 0.000 000 189 989 805 221 555 020 8;
  • 9) 0.000 000 189 989 805 221 555 020 8 × 2 = 0 + 0.000 000 379 979 610 443 110 041 6;
  • 10) 0.000 000 379 979 610 443 110 041 6 × 2 = 0 + 0.000 000 759 959 220 886 220 083 2;
  • 11) 0.000 000 759 959 220 886 220 083 2 × 2 = 0 + 0.000 001 519 918 441 772 440 166 4;
  • 12) 0.000 001 519 918 441 772 440 166 4 × 2 = 0 + 0.000 003 039 836 883 544 880 332 8;
  • 13) 0.000 003 039 836 883 544 880 332 8 × 2 = 0 + 0.000 006 079 673 767 089 760 665 6;
  • 14) 0.000 006 079 673 767 089 760 665 6 × 2 = 0 + 0.000 012 159 347 534 179 521 331 2;
  • 15) 0.000 012 159 347 534 179 521 331 2 × 2 = 0 + 0.000 024 318 695 068 359 042 662 4;
  • 16) 0.000 024 318 695 068 359 042 662 4 × 2 = 0 + 0.000 048 637 390 136 718 085 324 8;
  • 17) 0.000 048 637 390 136 718 085 324 8 × 2 = 0 + 0.000 097 274 780 273 436 170 649 6;
  • 18) 0.000 097 274 780 273 436 170 649 6 × 2 = 0 + 0.000 194 549 560 546 872 341 299 2;
  • 19) 0.000 194 549 560 546 872 341 299 2 × 2 = 0 + 0.000 389 099 121 093 744 682 598 4;
  • 20) 0.000 389 099 121 093 744 682 598 4 × 2 = 0 + 0.000 778 198 242 187 489 365 196 8;
  • 21) 0.000 778 198 242 187 489 365 196 8 × 2 = 0 + 0.001 556 396 484 374 978 730 393 6;
  • 22) 0.001 556 396 484 374 978 730 393 6 × 2 = 0 + 0.003 112 792 968 749 957 460 787 2;
  • 23) 0.003 112 792 968 749 957 460 787 2 × 2 = 0 + 0.006 225 585 937 499 914 921 574 4;
  • 24) 0.006 225 585 937 499 914 921 574 4 × 2 = 0 + 0.012 451 171 874 999 829 843 148 8;
  • 25) 0.012 451 171 874 999 829 843 148 8 × 2 = 0 + 0.024 902 343 749 999 659 686 297 6;
  • 26) 0.024 902 343 749 999 659 686 297 6 × 2 = 0 + 0.049 804 687 499 999 319 372 595 2;
  • 27) 0.049 804 687 499 999 319 372 595 2 × 2 = 0 + 0.099 609 374 999 998 638 745 190 4;
  • 28) 0.099 609 374 999 998 638 745 190 4 × 2 = 0 + 0.199 218 749 999 997 277 490 380 8;
  • 29) 0.199 218 749 999 997 277 490 380 8 × 2 = 0 + 0.398 437 499 999 994 554 980 761 6;
  • 30) 0.398 437 499 999 994 554 980 761 6 × 2 = 0 + 0.796 874 999 999 989 109 961 523 2;
  • 31) 0.796 874 999 999 989 109 961 523 2 × 2 = 1 + 0.593 749 999 999 978 219 923 046 4;
  • 32) 0.593 749 999 999 978 219 923 046 4 × 2 = 1 + 0.187 499 999 999 956 439 846 092 8;
  • 33) 0.187 499 999 999 956 439 846 092 8 × 2 = 0 + 0.374 999 999 999 912 879 692 185 6;
  • 34) 0.374 999 999 999 912 879 692 185 6 × 2 = 0 + 0.749 999 999 999 825 759 384 371 2;
  • 35) 0.749 999 999 999 825 759 384 371 2 × 2 = 1 + 0.499 999 999 999 651 518 768 742 4;
  • 36) 0.499 999 999 999 651 518 768 742 4 × 2 = 0 + 0.999 999 999 999 303 037 537 484 8;
  • 37) 0.999 999 999 999 303 037 537 484 8 × 2 = 1 + 0.999 999 999 998 606 075 074 969 6;
  • 38) 0.999 999 999 998 606 075 074 969 6 × 2 = 1 + 0.999 999 999 997 212 150 149 939 2;
  • 39) 0.999 999 999 997 212 150 149 939 2 × 2 = 1 + 0.999 999 999 994 424 300 299 878 4;
  • 40) 0.999 999 999 994 424 300 299 878 4 × 2 = 1 + 0.999 999 999 988 848 600 599 756 8;
  • 41) 0.999 999 999 988 848 600 599 756 8 × 2 = 1 + 0.999 999 999 977 697 201 199 513 6;
  • 42) 0.999 999 999 977 697 201 199 513 6 × 2 = 1 + 0.999 999 999 955 394 402 399 027 2;
  • 43) 0.999 999 999 955 394 402 399 027 2 × 2 = 1 + 0.999 999 999 910 788 804 798 054 4;
  • 44) 0.999 999 999 910 788 804 798 054 4 × 2 = 1 + 0.999 999 999 821 577 609 596 108 8;
  • 45) 0.999 999 999 821 577 609 596 108 8 × 2 = 1 + 0.999 999 999 643 155 219 192 217 6;
  • 46) 0.999 999 999 643 155 219 192 217 6 × 2 = 1 + 0.999 999 999 286 310 438 384 435 2;
  • 47) 0.999 999 999 286 310 438 384 435 2 × 2 = 1 + 0.999 999 998 572 620 876 768 870 4;
  • 48) 0.999 999 998 572 620 876 768 870 4 × 2 = 1 + 0.999 999 997 145 241 753 537 740 8;
  • 49) 0.999 999 997 145 241 753 537 740 8 × 2 = 1 + 0.999 999 994 290 483 507 075 481 6;
  • 50) 0.999 999 994 290 483 507 075 481 6 × 2 = 1 + 0.999 999 988 580 967 014 150 963 2;
  • 51) 0.999 999 988 580 967 014 150 963 2 × 2 = 1 + 0.999 999 977 161 934 028 301 926 4;
  • 52) 0.999 999 977 161 934 028 301 926 4 × 2 = 1 + 0.999 999 954 323 868 056 603 852 8;
  • 53) 0.999 999 954 323 868 056 603 852 8 × 2 = 1 + 0.999 999 908 647 736 113 207 705 6;
  • 54) 0.999 999 908 647 736 113 207 705 6 × 2 = 1 + 0.999 999 817 295 472 226 415 411 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:


0.000 000 000 742 147 676 646 699 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 699 3(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:


0.000 000 000 742 147 676 646 699 3(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 -0.000 000 000 742 147 676 646 699 3 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