-0.000 000 000 742 147 676 645 65 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 645 65(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
-0.000 000 000 742 147 676 645 65(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:

|-0.000 000 000 742 147 676 645 65| = 0.000 000 000 742 147 676 645 65


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: 0.000 000 000 742 147 676 645 65.

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) 0.000 000 000 742 147 676 645 65 × 2 = 0 + 0.000 000 001 484 295 353 291 3;
  • 2) 0.000 000 001 484 295 353 291 3 × 2 = 0 + 0.000 000 002 968 590 706 582 6;
  • 3) 0.000 000 002 968 590 706 582 6 × 2 = 0 + 0.000 000 005 937 181 413 165 2;
  • 4) 0.000 000 005 937 181 413 165 2 × 2 = 0 + 0.000 000 011 874 362 826 330 4;
  • 5) 0.000 000 011 874 362 826 330 4 × 2 = 0 + 0.000 000 023 748 725 652 660 8;
  • 6) 0.000 000 023 748 725 652 660 8 × 2 = 0 + 0.000 000 047 497 451 305 321 6;
  • 7) 0.000 000 047 497 451 305 321 6 × 2 = 0 + 0.000 000 094 994 902 610 643 2;
  • 8) 0.000 000 094 994 902 610 643 2 × 2 = 0 + 0.000 000 189 989 805 221 286 4;
  • 9) 0.000 000 189 989 805 221 286 4 × 2 = 0 + 0.000 000 379 979 610 442 572 8;
  • 10) 0.000 000 379 979 610 442 572 8 × 2 = 0 + 0.000 000 759 959 220 885 145 6;
  • 11) 0.000 000 759 959 220 885 145 6 × 2 = 0 + 0.000 001 519 918 441 770 291 2;
  • 12) 0.000 001 519 918 441 770 291 2 × 2 = 0 + 0.000 003 039 836 883 540 582 4;
  • 13) 0.000 003 039 836 883 540 582 4 × 2 = 0 + 0.000 006 079 673 767 081 164 8;
  • 14) 0.000 006 079 673 767 081 164 8 × 2 = 0 + 0.000 012 159 347 534 162 329 6;
  • 15) 0.000 012 159 347 534 162 329 6 × 2 = 0 + 0.000 024 318 695 068 324 659 2;
  • 16) 0.000 024 318 695 068 324 659 2 × 2 = 0 + 0.000 048 637 390 136 649 318 4;
  • 17) 0.000 048 637 390 136 649 318 4 × 2 = 0 + 0.000 097 274 780 273 298 636 8;
  • 18) 0.000 097 274 780 273 298 636 8 × 2 = 0 + 0.000 194 549 560 546 597 273 6;
  • 19) 0.000 194 549 560 546 597 273 6 × 2 = 0 + 0.000 389 099 121 093 194 547 2;
  • 20) 0.000 389 099 121 093 194 547 2 × 2 = 0 + 0.000 778 198 242 186 389 094 4;
  • 21) 0.000 778 198 242 186 389 094 4 × 2 = 0 + 0.001 556 396 484 372 778 188 8;
  • 22) 0.001 556 396 484 372 778 188 8 × 2 = 0 + 0.003 112 792 968 745 556 377 6;
  • 23) 0.003 112 792 968 745 556 377 6 × 2 = 0 + 0.006 225 585 937 491 112 755 2;
  • 24) 0.006 225 585 937 491 112 755 2 × 2 = 0 + 0.012 451 171 874 982 225 510 4;
  • 25) 0.012 451 171 874 982 225 510 4 × 2 = 0 + 0.024 902 343 749 964 451 020 8;
  • 26) 0.024 902 343 749 964 451 020 8 × 2 = 0 + 0.049 804 687 499 928 902 041 6;
  • 27) 0.049 804 687 499 928 902 041 6 × 2 = 0 + 0.099 609 374 999 857 804 083 2;
  • 28) 0.099 609 374 999 857 804 083 2 × 2 = 0 + 0.199 218 749 999 715 608 166 4;
  • 29) 0.199 218 749 999 715 608 166 4 × 2 = 0 + 0.398 437 499 999 431 216 332 8;
  • 30) 0.398 437 499 999 431 216 332 8 × 2 = 0 + 0.796 874 999 998 862 432 665 6;
  • 31) 0.796 874 999 998 862 432 665 6 × 2 = 1 + 0.593 749 999 997 724 865 331 2;
  • 32) 0.593 749 999 997 724 865 331 2 × 2 = 1 + 0.187 499 999 995 449 730 662 4;
  • 33) 0.187 499 999 995 449 730 662 4 × 2 = 0 + 0.374 999 999 990 899 461 324 8;
  • 34) 0.374 999 999 990 899 461 324 8 × 2 = 0 + 0.749 999 999 981 798 922 649 6;
  • 35) 0.749 999 999 981 798 922 649 6 × 2 = 1 + 0.499 999 999 963 597 845 299 2;
  • 36) 0.499 999 999 963 597 845 299 2 × 2 = 0 + 0.999 999 999 927 195 690 598 4;
  • 37) 0.999 999 999 927 195 690 598 4 × 2 = 1 + 0.999 999 999 854 391 381 196 8;
  • 38) 0.999 999 999 854 391 381 196 8 × 2 = 1 + 0.999 999 999 708 782 762 393 6;
  • 39) 0.999 999 999 708 782 762 393 6 × 2 = 1 + 0.999 999 999 417 565 524 787 2;
  • 40) 0.999 999 999 417 565 524 787 2 × 2 = 1 + 0.999 999 998 835 131 049 574 4;
  • 41) 0.999 999 998 835 131 049 574 4 × 2 = 1 + 0.999 999 997 670 262 099 148 8;
  • 42) 0.999 999 997 670 262 099 148 8 × 2 = 1 + 0.999 999 995 340 524 198 297 6;
  • 43) 0.999 999 995 340 524 198 297 6 × 2 = 1 + 0.999 999 990 681 048 396 595 2;
  • 44) 0.999 999 990 681 048 396 595 2 × 2 = 1 + 0.999 999 981 362 096 793 190 4;
  • 45) 0.999 999 981 362 096 793 190 4 × 2 = 1 + 0.999 999 962 724 193 586 380 8;
  • 46) 0.999 999 962 724 193 586 380 8 × 2 = 1 + 0.999 999 925 448 387 172 761 6;
  • 47) 0.999 999 925 448 387 172 761 6 × 2 = 1 + 0.999 999 850 896 774 345 523 2;
  • 48) 0.999 999 850 896 774 345 523 2 × 2 = 1 + 0.999 999 701 793 548 691 046 4;
  • 49) 0.999 999 701 793 548 691 046 4 × 2 = 1 + 0.999 999 403 587 097 382 092 8;
  • 50) 0.999 999 403 587 097 382 092 8 × 2 = 1 + 0.999 998 807 174 194 764 185 6;
  • 51) 0.999 998 807 174 194 764 185 6 × 2 = 1 + 0.999 997 614 348 389 528 371 2;
  • 52) 0.999 997 614 348 389 528 371 2 × 2 = 1 + 0.999 995 228 696 779 056 742 4;
  • 53) 0.999 995 228 696 779 056 742 4 × 2 = 1 + 0.999 990 457 393 558 113 484 8;
  • 54) 0.999 990 457 393 558 113 484 8 × 2 = 1 + 0.999 980 914 787 116 226 969 6;

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:


0.000 000 000 742 147 676 645 65(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 645 65(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:


0.000 000 000 742 147 676 645 65(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 -0.000 000 000 742 147 676 645 65 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