-0.000 000 000 742 147 676 646 693 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 693 2(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
-0.000 000 000 742 147 676 646 693 2(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:

|-0.000 000 000 742 147 676 646 693 2| = 0.000 000 000 742 147 676 646 693 2


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: 0.000 000 000 742 147 676 646 693 2.

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) 0.000 000 000 742 147 676 646 693 2 × 2 = 0 + 0.000 000 001 484 295 353 293 386 4;
  • 2) 0.000 000 001 484 295 353 293 386 4 × 2 = 0 + 0.000 000 002 968 590 706 586 772 8;
  • 3) 0.000 000 002 968 590 706 586 772 8 × 2 = 0 + 0.000 000 005 937 181 413 173 545 6;
  • 4) 0.000 000 005 937 181 413 173 545 6 × 2 = 0 + 0.000 000 011 874 362 826 347 091 2;
  • 5) 0.000 000 011 874 362 826 347 091 2 × 2 = 0 + 0.000 000 023 748 725 652 694 182 4;
  • 6) 0.000 000 023 748 725 652 694 182 4 × 2 = 0 + 0.000 000 047 497 451 305 388 364 8;
  • 7) 0.000 000 047 497 451 305 388 364 8 × 2 = 0 + 0.000 000 094 994 902 610 776 729 6;
  • 8) 0.000 000 094 994 902 610 776 729 6 × 2 = 0 + 0.000 000 189 989 805 221 553 459 2;
  • 9) 0.000 000 189 989 805 221 553 459 2 × 2 = 0 + 0.000 000 379 979 610 443 106 918 4;
  • 10) 0.000 000 379 979 610 443 106 918 4 × 2 = 0 + 0.000 000 759 959 220 886 213 836 8;
  • 11) 0.000 000 759 959 220 886 213 836 8 × 2 = 0 + 0.000 001 519 918 441 772 427 673 6;
  • 12) 0.000 001 519 918 441 772 427 673 6 × 2 = 0 + 0.000 003 039 836 883 544 855 347 2;
  • 13) 0.000 003 039 836 883 544 855 347 2 × 2 = 0 + 0.000 006 079 673 767 089 710 694 4;
  • 14) 0.000 006 079 673 767 089 710 694 4 × 2 = 0 + 0.000 012 159 347 534 179 421 388 8;
  • 15) 0.000 012 159 347 534 179 421 388 8 × 2 = 0 + 0.000 024 318 695 068 358 842 777 6;
  • 16) 0.000 024 318 695 068 358 842 777 6 × 2 = 0 + 0.000 048 637 390 136 717 685 555 2;
  • 17) 0.000 048 637 390 136 717 685 555 2 × 2 = 0 + 0.000 097 274 780 273 435 371 110 4;
  • 18) 0.000 097 274 780 273 435 371 110 4 × 2 = 0 + 0.000 194 549 560 546 870 742 220 8;
  • 19) 0.000 194 549 560 546 870 742 220 8 × 2 = 0 + 0.000 389 099 121 093 741 484 441 6;
  • 20) 0.000 389 099 121 093 741 484 441 6 × 2 = 0 + 0.000 778 198 242 187 482 968 883 2;
  • 21) 0.000 778 198 242 187 482 968 883 2 × 2 = 0 + 0.001 556 396 484 374 965 937 766 4;
  • 22) 0.001 556 396 484 374 965 937 766 4 × 2 = 0 + 0.003 112 792 968 749 931 875 532 8;
  • 23) 0.003 112 792 968 749 931 875 532 8 × 2 = 0 + 0.006 225 585 937 499 863 751 065 6;
  • 24) 0.006 225 585 937 499 863 751 065 6 × 2 = 0 + 0.012 451 171 874 999 727 502 131 2;
  • 25) 0.012 451 171 874 999 727 502 131 2 × 2 = 0 + 0.024 902 343 749 999 455 004 262 4;
  • 26) 0.024 902 343 749 999 455 004 262 4 × 2 = 0 + 0.049 804 687 499 998 910 008 524 8;
  • 27) 0.049 804 687 499 998 910 008 524 8 × 2 = 0 + 0.099 609 374 999 997 820 017 049 6;
  • 28) 0.099 609 374 999 997 820 017 049 6 × 2 = 0 + 0.199 218 749 999 995 640 034 099 2;
  • 29) 0.199 218 749 999 995 640 034 099 2 × 2 = 0 + 0.398 437 499 999 991 280 068 198 4;
  • 30) 0.398 437 499 999 991 280 068 198 4 × 2 = 0 + 0.796 874 999 999 982 560 136 396 8;
  • 31) 0.796 874 999 999 982 560 136 396 8 × 2 = 1 + 0.593 749 999 999 965 120 272 793 6;
  • 32) 0.593 749 999 999 965 120 272 793 6 × 2 = 1 + 0.187 499 999 999 930 240 545 587 2;
  • 33) 0.187 499 999 999 930 240 545 587 2 × 2 = 0 + 0.374 999 999 999 860 481 091 174 4;
  • 34) 0.374 999 999 999 860 481 091 174 4 × 2 = 0 + 0.749 999 999 999 720 962 182 348 8;
  • 35) 0.749 999 999 999 720 962 182 348 8 × 2 = 1 + 0.499 999 999 999 441 924 364 697 6;
  • 36) 0.499 999 999 999 441 924 364 697 6 × 2 = 0 + 0.999 999 999 998 883 848 729 395 2;
  • 37) 0.999 999 999 998 883 848 729 395 2 × 2 = 1 + 0.999 999 999 997 767 697 458 790 4;
  • 38) 0.999 999 999 997 767 697 458 790 4 × 2 = 1 + 0.999 999 999 995 535 394 917 580 8;
  • 39) 0.999 999 999 995 535 394 917 580 8 × 2 = 1 + 0.999 999 999 991 070 789 835 161 6;
  • 40) 0.999 999 999 991 070 789 835 161 6 × 2 = 1 + 0.999 999 999 982 141 579 670 323 2;
  • 41) 0.999 999 999 982 141 579 670 323 2 × 2 = 1 + 0.999 999 999 964 283 159 340 646 4;
  • 42) 0.999 999 999 964 283 159 340 646 4 × 2 = 1 + 0.999 999 999 928 566 318 681 292 8;
  • 43) 0.999 999 999 928 566 318 681 292 8 × 2 = 1 + 0.999 999 999 857 132 637 362 585 6;
  • 44) 0.999 999 999 857 132 637 362 585 6 × 2 = 1 + 0.999 999 999 714 265 274 725 171 2;
  • 45) 0.999 999 999 714 265 274 725 171 2 × 2 = 1 + 0.999 999 999 428 530 549 450 342 4;
  • 46) 0.999 999 999 428 530 549 450 342 4 × 2 = 1 + 0.999 999 998 857 061 098 900 684 8;
  • 47) 0.999 999 998 857 061 098 900 684 8 × 2 = 1 + 0.999 999 997 714 122 197 801 369 6;
  • 48) 0.999 999 997 714 122 197 801 369 6 × 2 = 1 + 0.999 999 995 428 244 395 602 739 2;
  • 49) 0.999 999 995 428 244 395 602 739 2 × 2 = 1 + 0.999 999 990 856 488 791 205 478 4;
  • 50) 0.999 999 990 856 488 791 205 478 4 × 2 = 1 + 0.999 999 981 712 977 582 410 956 8;
  • 51) 0.999 999 981 712 977 582 410 956 8 × 2 = 1 + 0.999 999 963 425 955 164 821 913 6;
  • 52) 0.999 999 963 425 955 164 821 913 6 × 2 = 1 + 0.999 999 926 851 910 329 643 827 2;
  • 53) 0.999 999 926 851 910 329 643 827 2 × 2 = 1 + 0.999 999 853 703 820 659 287 654 4;
  • 54) 0.999 999 853 703 820 659 287 654 4 × 2 = 1 + 0.999 999 707 407 641 318 575 308 8;

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:


0.000 000 000 742 147 676 646 693 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 693 2(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:


0.000 000 000 742 147 676 646 693 2(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 -0.000 000 000 742 147 676 646 693 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 698 7 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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