-0.000 000 000 742 147 676 651 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 651 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
-0.000 000 000 742 147 676 651 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:

|-0.000 000 000 742 147 676 651 2| = 0.000 000 000 742 147 676 651 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: 0.000 000 000 742 147 676 651 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) 0.000 000 000 742 147 676 651 2 × 2 = 0 + 0.000 000 001 484 295 353 302 4;
  • 2) 0.000 000 001 484 295 353 302 4 × 2 = 0 + 0.000 000 002 968 590 706 604 8;
  • 3) 0.000 000 002 968 590 706 604 8 × 2 = 0 + 0.000 000 005 937 181 413 209 6;
  • 4) 0.000 000 005 937 181 413 209 6 × 2 = 0 + 0.000 000 011 874 362 826 419 2;
  • 5) 0.000 000 011 874 362 826 419 2 × 2 = 0 + 0.000 000 023 748 725 652 838 4;
  • 6) 0.000 000 023 748 725 652 838 4 × 2 = 0 + 0.000 000 047 497 451 305 676 8;
  • 7) 0.000 000 047 497 451 305 676 8 × 2 = 0 + 0.000 000 094 994 902 611 353 6;
  • 8) 0.000 000 094 994 902 611 353 6 × 2 = 0 + 0.000 000 189 989 805 222 707 2;
  • 9) 0.000 000 189 989 805 222 707 2 × 2 = 0 + 0.000 000 379 979 610 445 414 4;
  • 10) 0.000 000 379 979 610 445 414 4 × 2 = 0 + 0.000 000 759 959 220 890 828 8;
  • 11) 0.000 000 759 959 220 890 828 8 × 2 = 0 + 0.000 001 519 918 441 781 657 6;
  • 12) 0.000 001 519 918 441 781 657 6 × 2 = 0 + 0.000 003 039 836 883 563 315 2;
  • 13) 0.000 003 039 836 883 563 315 2 × 2 = 0 + 0.000 006 079 673 767 126 630 4;
  • 14) 0.000 006 079 673 767 126 630 4 × 2 = 0 + 0.000 012 159 347 534 253 260 8;
  • 15) 0.000 012 159 347 534 253 260 8 × 2 = 0 + 0.000 024 318 695 068 506 521 6;
  • 16) 0.000 024 318 695 068 506 521 6 × 2 = 0 + 0.000 048 637 390 137 013 043 2;
  • 17) 0.000 048 637 390 137 013 043 2 × 2 = 0 + 0.000 097 274 780 274 026 086 4;
  • 18) 0.000 097 274 780 274 026 086 4 × 2 = 0 + 0.000 194 549 560 548 052 172 8;
  • 19) 0.000 194 549 560 548 052 172 8 × 2 = 0 + 0.000 389 099 121 096 104 345 6;
  • 20) 0.000 389 099 121 096 104 345 6 × 2 = 0 + 0.000 778 198 242 192 208 691 2;
  • 21) 0.000 778 198 242 192 208 691 2 × 2 = 0 + 0.001 556 396 484 384 417 382 4;
  • 22) 0.001 556 396 484 384 417 382 4 × 2 = 0 + 0.003 112 792 968 768 834 764 8;
  • 23) 0.003 112 792 968 768 834 764 8 × 2 = 0 + 0.006 225 585 937 537 669 529 6;
  • 24) 0.006 225 585 937 537 669 529 6 × 2 = 0 + 0.012 451 171 875 075 339 059 2;
  • 25) 0.012 451 171 875 075 339 059 2 × 2 = 0 + 0.024 902 343 750 150 678 118 4;
  • 26) 0.024 902 343 750 150 678 118 4 × 2 = 0 + 0.049 804 687 500 301 356 236 8;
  • 27) 0.049 804 687 500 301 356 236 8 × 2 = 0 + 0.099 609 375 000 602 712 473 6;
  • 28) 0.099 609 375 000 602 712 473 6 × 2 = 0 + 0.199 218 750 001 205 424 947 2;
  • 29) 0.199 218 750 001 205 424 947 2 × 2 = 0 + 0.398 437 500 002 410 849 894 4;
  • 30) 0.398 437 500 002 410 849 894 4 × 2 = 0 + 0.796 875 000 004 821 699 788 8;
  • 31) 0.796 875 000 004 821 699 788 8 × 2 = 1 + 0.593 750 000 009 643 399 577 6;
  • 32) 0.593 750 000 009 643 399 577 6 × 2 = 1 + 0.187 500 000 019 286 799 155 2;
  • 33) 0.187 500 000 019 286 799 155 2 × 2 = 0 + 0.375 000 000 038 573 598 310 4;
  • 34) 0.375 000 000 038 573 598 310 4 × 2 = 0 + 0.750 000 000 077 147 196 620 8;
  • 35) 0.750 000 000 077 147 196 620 8 × 2 = 1 + 0.500 000 000 154 294 393 241 6;
  • 36) 0.500 000 000 154 294 393 241 6 × 2 = 1 + 0.000 000 000 308 588 786 483 2;
  • 37) 0.000 000 000 308 588 786 483 2 × 2 = 0 + 0.000 000 000 617 177 572 966 4;
  • 38) 0.000 000 000 617 177 572 966 4 × 2 = 0 + 0.000 000 001 234 355 145 932 8;
  • 39) 0.000 000 001 234 355 145 932 8 × 2 = 0 + 0.000 000 002 468 710 291 865 6;
  • 40) 0.000 000 002 468 710 291 865 6 × 2 = 0 + 0.000 000 004 937 420 583 731 2;
  • 41) 0.000 000 004 937 420 583 731 2 × 2 = 0 + 0.000 000 009 874 841 167 462 4;
  • 42) 0.000 000 009 874 841 167 462 4 × 2 = 0 + 0.000 000 019 749 682 334 924 8;
  • 43) 0.000 000 019 749 682 334 924 8 × 2 = 0 + 0.000 000 039 499 364 669 849 6;
  • 44) 0.000 000 039 499 364 669 849 6 × 2 = 0 + 0.000 000 078 998 729 339 699 2;
  • 45) 0.000 000 078 998 729 339 699 2 × 2 = 0 + 0.000 000 157 997 458 679 398 4;
  • 46) 0.000 000 157 997 458 679 398 4 × 2 = 0 + 0.000 000 315 994 917 358 796 8;
  • 47) 0.000 000 315 994 917 358 796 8 × 2 = 0 + 0.000 000 631 989 834 717 593 6;
  • 48) 0.000 000 631 989 834 717 593 6 × 2 = 0 + 0.000 001 263 979 669 435 187 2;
  • 49) 0.000 001 263 979 669 435 187 2 × 2 = 0 + 0.000 002 527 959 338 870 374 4;
  • 50) 0.000 002 527 959 338 870 374 4 × 2 = 0 + 0.000 005 055 918 677 740 748 8;
  • 51) 0.000 005 055 918 677 740 748 8 × 2 = 0 + 0.000 010 111 837 355 481 497 6;
  • 52) 0.000 010 111 837 355 481 497 6 × 2 = 0 + 0.000 020 223 674 710 962 995 2;
  • 53) 0.000 020 223 674 710 962 995 2 × 2 = 0 + 0.000 040 447 349 421 925 990 4;
  • 54) 0.000 040 447 349 421 925 990 4 × 2 = 0 + 0.000 080 894 698 843 851 980 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:


0.000 000 000 742 147 676 651 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 651 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 651 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 651 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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