-0.000 000 000 742 147 676 636 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 636 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
-0.000 000 000 742 147 676 636 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:

|-0.000 000 000 742 147 676 636 6| = 0.000 000 000 742 147 676 636 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: 0.000 000 000 742 147 676 636 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) 0.000 000 000 742 147 676 636 6 × 2 = 0 + 0.000 000 001 484 295 353 273 2;
  • 2) 0.000 000 001 484 295 353 273 2 × 2 = 0 + 0.000 000 002 968 590 706 546 4;
  • 3) 0.000 000 002 968 590 706 546 4 × 2 = 0 + 0.000 000 005 937 181 413 092 8;
  • 4) 0.000 000 005 937 181 413 092 8 × 2 = 0 + 0.000 000 011 874 362 826 185 6;
  • 5) 0.000 000 011 874 362 826 185 6 × 2 = 0 + 0.000 000 023 748 725 652 371 2;
  • 6) 0.000 000 023 748 725 652 371 2 × 2 = 0 + 0.000 000 047 497 451 304 742 4;
  • 7) 0.000 000 047 497 451 304 742 4 × 2 = 0 + 0.000 000 094 994 902 609 484 8;
  • 8) 0.000 000 094 994 902 609 484 8 × 2 = 0 + 0.000 000 189 989 805 218 969 6;
  • 9) 0.000 000 189 989 805 218 969 6 × 2 = 0 + 0.000 000 379 979 610 437 939 2;
  • 10) 0.000 000 379 979 610 437 939 2 × 2 = 0 + 0.000 000 759 959 220 875 878 4;
  • 11) 0.000 000 759 959 220 875 878 4 × 2 = 0 + 0.000 001 519 918 441 751 756 8;
  • 12) 0.000 001 519 918 441 751 756 8 × 2 = 0 + 0.000 003 039 836 883 503 513 6;
  • 13) 0.000 003 039 836 883 503 513 6 × 2 = 0 + 0.000 006 079 673 767 007 027 2;
  • 14) 0.000 006 079 673 767 007 027 2 × 2 = 0 + 0.000 012 159 347 534 014 054 4;
  • 15) 0.000 012 159 347 534 014 054 4 × 2 = 0 + 0.000 024 318 695 068 028 108 8;
  • 16) 0.000 024 318 695 068 028 108 8 × 2 = 0 + 0.000 048 637 390 136 056 217 6;
  • 17) 0.000 048 637 390 136 056 217 6 × 2 = 0 + 0.000 097 274 780 272 112 435 2;
  • 18) 0.000 097 274 780 272 112 435 2 × 2 = 0 + 0.000 194 549 560 544 224 870 4;
  • 19) 0.000 194 549 560 544 224 870 4 × 2 = 0 + 0.000 389 099 121 088 449 740 8;
  • 20) 0.000 389 099 121 088 449 740 8 × 2 = 0 + 0.000 778 198 242 176 899 481 6;
  • 21) 0.000 778 198 242 176 899 481 6 × 2 = 0 + 0.001 556 396 484 353 798 963 2;
  • 22) 0.001 556 396 484 353 798 963 2 × 2 = 0 + 0.003 112 792 968 707 597 926 4;
  • 23) 0.003 112 792 968 707 597 926 4 × 2 = 0 + 0.006 225 585 937 415 195 852 8;
  • 24) 0.006 225 585 937 415 195 852 8 × 2 = 0 + 0.012 451 171 874 830 391 705 6;
  • 25) 0.012 451 171 874 830 391 705 6 × 2 = 0 + 0.024 902 343 749 660 783 411 2;
  • 26) 0.024 902 343 749 660 783 411 2 × 2 = 0 + 0.049 804 687 499 321 566 822 4;
  • 27) 0.049 804 687 499 321 566 822 4 × 2 = 0 + 0.099 609 374 998 643 133 644 8;
  • 28) 0.099 609 374 998 643 133 644 8 × 2 = 0 + 0.199 218 749 997 286 267 289 6;
  • 29) 0.199 218 749 997 286 267 289 6 × 2 = 0 + 0.398 437 499 994 572 534 579 2;
  • 30) 0.398 437 499 994 572 534 579 2 × 2 = 0 + 0.796 874 999 989 145 069 158 4;
  • 31) 0.796 874 999 989 145 069 158 4 × 2 = 1 + 0.593 749 999 978 290 138 316 8;
  • 32) 0.593 749 999 978 290 138 316 8 × 2 = 1 + 0.187 499 999 956 580 276 633 6;
  • 33) 0.187 499 999 956 580 276 633 6 × 2 = 0 + 0.374 999 999 913 160 553 267 2;
  • 34) 0.374 999 999 913 160 553 267 2 × 2 = 0 + 0.749 999 999 826 321 106 534 4;
  • 35) 0.749 999 999 826 321 106 534 4 × 2 = 1 + 0.499 999 999 652 642 213 068 8;
  • 36) 0.499 999 999 652 642 213 068 8 × 2 = 0 + 0.999 999 999 305 284 426 137 6;
  • 37) 0.999 999 999 305 284 426 137 6 × 2 = 1 + 0.999 999 998 610 568 852 275 2;
  • 38) 0.999 999 998 610 568 852 275 2 × 2 = 1 + 0.999 999 997 221 137 704 550 4;
  • 39) 0.999 999 997 221 137 704 550 4 × 2 = 1 + 0.999 999 994 442 275 409 100 8;
  • 40) 0.999 999 994 442 275 409 100 8 × 2 = 1 + 0.999 999 988 884 550 818 201 6;
  • 41) 0.999 999 988 884 550 818 201 6 × 2 = 1 + 0.999 999 977 769 101 636 403 2;
  • 42) 0.999 999 977 769 101 636 403 2 × 2 = 1 + 0.999 999 955 538 203 272 806 4;
  • 43) 0.999 999 955 538 203 272 806 4 × 2 = 1 + 0.999 999 911 076 406 545 612 8;
  • 44) 0.999 999 911 076 406 545 612 8 × 2 = 1 + 0.999 999 822 152 813 091 225 6;
  • 45) 0.999 999 822 152 813 091 225 6 × 2 = 1 + 0.999 999 644 305 626 182 451 2;
  • 46) 0.999 999 644 305 626 182 451 2 × 2 = 1 + 0.999 999 288 611 252 364 902 4;
  • 47) 0.999 999 288 611 252 364 902 4 × 2 = 1 + 0.999 998 577 222 504 729 804 8;
  • 48) 0.999 998 577 222 504 729 804 8 × 2 = 1 + 0.999 997 154 445 009 459 609 6;
  • 49) 0.999 997 154 445 009 459 609 6 × 2 = 1 + 0.999 994 308 890 018 919 219 2;
  • 50) 0.999 994 308 890 018 919 219 2 × 2 = 1 + 0.999 988 617 780 037 838 438 4;
  • 51) 0.999 988 617 780 037 838 438 4 × 2 = 1 + 0.999 977 235 560 075 676 876 8;
  • 52) 0.999 977 235 560 075 676 876 8 × 2 = 1 + 0.999 954 471 120 151 353 753 6;
  • 53) 0.999 954 471 120 151 353 753 6 × 2 = 1 + 0.999 908 942 240 302 707 507 2;
  • 54) 0.999 908 942 240 302 707 507 2 × 2 = 1 + 0.999 817 884 480 605 415 014 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:


0.000 000 000 742 147 676 636 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 636 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:


0.000 000 000 742 147 676 636 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 -0.000 000 000 742 147 676 636 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